Data size definition in C programming while using mathematical expressions - c

When using mathematical operators in c programming, it is very important to use casts or define size of variable properly. I need help in it.
#include <stdio.h>
#include <stdint.h>
int main(void)
{
uint32_t a;
uint8_t b;
uint8_t d;
uint64_t c;
float cd;
a = 4294967295;
b = 2;
d = 2;
c = a * b * d;
cd = c;
printf("%f\n", cd);
return 0;
}
The result variable is large enough to store the 2 * 2 * uint32_max. However I noticed that the b or d variable need to be 64 bit wide (or use cast) to get the proper result. For this time I thought the mathematical operations takes place in the result variable, but it looks like is not true. Can somebody explain me which variable need to be widened (b or d) and what is the theoretical background behind it?
What is the situation in division? Should I consider whether I want to divide a 32 bit number with a 8bit long. will the result be in that case 8 bit only? Does any rule about the type of the denominator?


When you perform the multiplication a * b * d, what will happen is that b and d will get promoted to uint32_t (or int if int is wider than uint32_t) to match the type of a. However, this operation might overflow. So what you need to do is to cast at least one of them to uint_64_t to prevent this from happening.
Do note that (uint64_t)(a * b * d) will NOT work. Type casts has lower priority than parenthesis.


In C, the evaluation of an expression is determined by the operator and its operands, not by where the result will eventually be stored.
The expression a * b * d is structured as (a * b) * d. So a * b is evaluated, and then the result is multiplied by d.
One of the rules for * is in C 2018 6.5.5 3:
The usual arithmetic conversions are performed on the operands.
The usual arithmetic conversions are defined in 6.3.1.8 1. They are a bit complicated, and I give most of the details below. Applying them to your example:
In a * b, a is a uint32_t , and b is a uint8_t.
The integer promotions convert b to an int—essentially all arithmetic in C is done in a width of at least int.
If int is 32 bits or narrower, a remains uint32_t. Otherwise, a is converted to int.
If converted types of a and b are both int, the conversions are done, and the multiplication is performed.
If the converted type of a is uint32_t, b is converted to uint32_t, and the multiplication is performed.
Then the multiplication with d is performed the same way.
So, if int is 32 bits or narrower, the multiplications are performed with uint32_t, and the result is uint32_t. If int is wider, the multiplications are performed with int, and the result is int.
Casting either operand to uint64_t would cause the arithmetic to be done with uint64_t. (Except it is theoretically possible that int is wider than uint64_t, in which case the arithmetic would be done with int, but that is still satisfactory—performing a cast guarantees the arithmetic will be done with at least that width.)
For real numbers, the usual arithmetic conversions are largely:
If either operand is long double, the other is converted to long double.
Otherwise, if either is double, the other is converted to double.
Otherwise, if either is float, the other is converted to float.
Otherwise, the integer promotions are performed on both operands.
Then, if both have the same type, no further conversion is performed.
Otherwise, if both are signed or both are unsigned, the narrower (actually “lesser rank”) operand is converted to the type of the other.
Otherwise, if the unsigned operand is the same width or wider (greater or equal rank), the signed operand is converted to the type of the unsigned operand.
Otherwise, if the type of the signed operand can represent all the values of the type of the unsigned operand, the unsigned operand is converted to the type of the signed operand.
Otherwise, both operands are converted to the unsigned type that has the same width as the signed operand.
The integer promotions are defined in 6.3.1.1 2. They apply to all integer types as wide as or narrower than int or unsigned int (technically of rank less than or equal to the rank of int and unsigned int), including bit-fields of type _Bool, int, signed int, or unsigned int.
If an int can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions. All other types are unchanged by the integer promotions.


a * b * d is an expression of type uint32_t, or int if int is wider than uint32_t (due to the conversion rule for uint8_t).
The fact that this expression is assigned to a wider type is not a factor. That's the crux of the issue.
Writing c = 1ULL * a * b * d is a tractable fix.

Related

Comparisons with bit shifting [duplicate]

This post is meant to be used as a FAQ regarding implicit integer promotion in C, particularly implicit promotion caused by the usual arithmetic conversions and/or the integer promotions.
Example 1)
Why does this give a strange, large integer number and not 255?
unsigned char x = 0;
unsigned char y = 1;
printf("%u\n", x - y);
Example 2)
Why does this give "-1 is larger than 0"?
unsigned int a = 1;
signed int b = -2;
if(a + b > 0)
puts("-1 is larger than 0");
Example 3)
Why does changing the type in the above example to short fix the problem?
unsigned short a = 1;
signed short b = -2;
if(a + b > 0)
puts("-1 is larger than 0"); // will not print
(These examples were intended for a 32 or 64 bit computer with 16 bit short.)

C was designed to implicitly and silently change the integer types of the operands used in expressions. There exist several cases where the language forces the compiler to either change the operands to a larger type, or to change their signedness.
The rationale behind this is to prevent accidental overflows during arithmetic, but also to allow operands with different signedness to co-exist in the same expression.
Unfortunately, the rules for implicit type promotion cause much more harm than good, to the point where they might be one of the biggest flaws in the C language. These rules are often not even known by the average C programmer and therefore cause all manner of very subtle bugs.
Typically you see scenarios where the programmer says "just cast to type x and it works" - but they don't know why. Or such bugs manifest themselves as rare, intermittent phenomena striking from within seemingly simple and straight-forward code. Implicit promotion is particularly troublesome in code doing bit manipulations, since most bit-wise operators in C come with poorly-defined behavior when given a signed operand.
Integer types and conversion rank
The integer types in C are char, short, int, long, long long and enum.
_Bool/bool is also treated as an integer type when it comes to type promotions.
All integers have a specified conversion rank. C11 6.3.1.1, emphasis mine on the most important parts:
Every integer type has an integer conversion rank defined as follows:
— No two signed integer types shall have the same rank, even if they have the same representation.
— The rank of a signed integer type shall be greater than the rank of any signed integer type with less precision.
— The rank of long long int shall be greater than the rank of long int, which shall be greater than the rank of int, which shall be greater than the rank of short int, which shall be greater than the rank of signed char.
— The rank of any unsigned integer type shall equal the rank of the corresponding signed integer type, if any.
— The rank of any standard integer type shall be greater than the rank of any extended integer type with the same width.
— The rank of char shall equal the rank of signed char and unsigned char.
— The rank of _Bool shall be less than the rank of all other standard integer types.
— The rank of any enumerated type shall equal the rank of the compatible integer type (see 6.7.2.2).
The types from stdint.h sort in here too, with the same rank as whatever type they happen to correspond to on the given system. For example, int32_t has the same rank as int on a 32 bit system.
Further, C11 6.3.1.1 specifies which types are regarded as the small integer types (not a formal term):
The following may be used in an expression wherever an int or unsigned int may
be used:
— An object or expression with an integer type (other than int or unsigned int) whose integer conversion rank is less than or equal to the rank of int and unsigned int.
What this somewhat cryptic text means in practice, is that _Bool, char and short (and also int8_t, uint8_t etc) are the "small integer types". These are treated in special ways and subject to implicit promotion, as explained below.
The integer promotions
Whenever a small integer type is used in an expression, it is implicitly converted to int which is always signed. This is known as the integer promotions or the integer promotion rule.
Formally, the rule says (C11 6.3.1.1):
If an int can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions.
This means that all small integer types, no matter signedness, get implicitly converted to (signed) int when used in most expressions.
This text is often misunderstood as: "all small signed integer types are converted to signed int and all small, unsigned integer types are converted to unsigned int". This is incorrect. The unsigned part here only means that if we have for example an unsigned short operand, and int happens to have the same size as short on the given system, then the unsigned short operand is converted to unsigned int. As in, nothing of note really happens. But in case short is a smaller type than int, it is always converted to (signed) int, regardless of it the short was signed or unsigned!
The harsh reality caused by the integer promotions means that almost no operation in C can be carried out on small types like char or short. Operations are always carried out on int or larger types.
This might sound like nonsense, but luckily the compiler is allowed to optimize the code. For example, an expression containing two unsigned char operands would get the operands promoted to int and the operation carried out as int. But the compiler is allowed to optimize the expression to actually get carried out as an 8-bit operation, as would be expected. However, here comes the problem: the compiler is not allowed to optimize out the implicit change of signedness caused by the integer promotion because there is no way for the compiler to tell if the programmer is purposely relying on implicit promotion to happen, or if it is unintentional.
This is why example 1 in the question fails. Both unsigned char operands are promoted to type int, the operation is carried out on type int, and the result of x - y is of type int. Meaning that we get -1 instead of 255 which might have been expected. The compiler may generate machine code that executes the code with 8 bit instructions instead of int, but it may not optimize out the change of signedness. Meaning that we end up with a negative result, that in turn results in a weird number when printf("%u is invoked. Example 1 could be fixed by casting the result of the operation back to type unsigned char.
With the exception of a few special cases like ++ and sizeof operators, the integer promotions apply to almost all operations in C, no matter if unary, binary (or ternary) operators are used.
The usual arithmetic conversions
Whenever a binary operation (an operation with 2 operands) is done in C, both operands of the operator have to be of the same type. Therefore, in case the operands are of different types, C enforces an implicit conversion of one operand to the type of the other operand. The rules for how this is done are named the usual artihmetic conversions (sometimes informally referred to as "balancing"). These are specified in C11 6.3.18:
(Think of this rule as a long, nested if-else if statement and it might be easier to read :) )
6.3.1.8 Usual arithmetic conversions
Many operators that expect operands of arithmetic type cause conversions and yield result
types in a similar way. The purpose is to determine a common real type for the operands
and result. For the specified operands, each operand is converted, without change of type
domain, to a type whose corresponding real type is the common real type. Unless
explicitly stated otherwise, the common real type is also the corresponding real type of
the result, whose type domain is the type domain of the operands if they are the same,
and complex otherwise. This pattern is called the usual arithmetic conversions:
First, if the corresponding real type of either operand is long double, the other operand is converted, without change of type domain, to a type whose corresponding real type is long double.
Otherwise, if the corresponding real type of either operand is double, the other operand is converted, without change of type domain, to a type whose corresponding real type is double.
Otherwise, if the corresponding real type of either operand is float, the other operand is converted, without change of type domain, to a type whose corresponding real type is float.
Otherwise, the integer promotions are performed on both operands. Then the
following rules are applied to the promoted operands:
If both operands have the same type, then no further conversion is needed.
Otherwise, if both operands have signed integer types or both have unsigned
integer types, the operand with the type of lesser integer conversion rank is
converted to the type of the operand with greater rank.
Otherwise, if the operand that has unsigned integer type has rank greater or
equal to the rank of the type of the other operand, then the operand with
signed integer type is converted to the type of the operand with unsigned
integer type.
Otherwise, if the type of the operand with signed integer type can represent
all of the values of the type of the operand with unsigned integer type, then
the operand with unsigned integer type is converted to the type of the
operand with signed integer type.
Otherwise, both operands are converted to the unsigned integer type
corresponding to the type of the operand with signed integer type.
Notable here is that the usual arithmetic conversions apply to both floating point and integer variables. In the case of integers, we can also note that the integer promotions are invoked from within the usual arithmetic conversions. And after that, when both operands have at least the rank of int, the operators are balanced to the same type, with the same signedness.
This is the reason why a + b in example 2 gives a strange result. Both operands are integers and they are at least of rank int, so the integer promotions do not apply. The operands are not of the same type - a is unsigned int and b is signed int. Therefore the operator b is temporarily converted to type unsigned int. During this conversion, it loses the sign information and ends up as a large value.
The reason why changing type to short in example 3 fixes the problem, is because short is a small integer type. Meaning that both operands are integer promoted to type int which is signed. After integer promotion, both operands have the same type (int), no further conversion is needed. And then the operation can be carried out on a signed type as expected.

According to the previous post, I want to give more information about each example.
Example 1)
int main(){
unsigned char x = 0;
unsigned char y = 1;
printf("%u\n", x - y);
printf("%d\n", x - y);
}
Since unsigned char is smaller than int, we apply the integer promotion on them, then we have (int)x-(int)y = (int)(-1) and unsigned int (-1) = 4294967295.
The output from the above code:(same as what we expected)
4294967295
-1
How to fix it?
I tried what the previous post recommended, but it doesn't really work.
Here is the code based on the previous post:
change one of them to unsigned int
int main(){
unsigned int x = 0;
unsigned char y = 1;
printf("%u\n", x - y);
printf("%d\n", x - y);
}
Since x is already an unsigned integer, we only apply the integer promotion to y. Then we get (unsigned int)x-(int)y. Since they still don't have the same type, we apply the usual arithmetic converions, we get (unsigned int)x-(unsigned int)y = 4294967295.
The output from the above code:(same as what we expected):
4294967295
-1
Similarly, the following code gets the same result:
int main(){
unsigned char x = 0;
unsigned int y = 1;
printf("%u\n", x - y);
printf("%d\n", x - y);
}
change both of them to unsigned int
int main(){
unsigned int x = 0;
unsigned int y = 1;
printf("%u\n", x - y);
printf("%d\n", x - y);
}
Since both of them are unsigned int, no integer promotion is needed. By the usual arithmetic converison(have the same type), (unsigned int)x-(unsigned int)y = 4294967295.
The output from the above code:(same as what we expected):
4294967295
-1
One of possible ways to fix the code:(add a type cast in the end)
int main(){
unsigned char x = 0;
unsigned char y = 1;
printf("%u\n", x - y);
printf("%d\n", x - y);
unsigned char z = x-y;
printf("%u\n", z);
}
The output from the above code:
4294967295
-1
255
Example 2)
int main(){
unsigned int a = 1;
signed int b = -2;
if(a + b > 0)
puts("-1 is larger than 0");
printf("%u\n", a+b);
}
Since both of them are integers, no integer promotion is needed. By the usual arithmetic conversion, we get (unsigned int)a+(unsigned int)b = 1+4294967294 = 4294967295.
The output from the above code:(same as what we expected)
-1 is larger than 0
4294967295
How to fix it?
int main(){
unsigned int a = 1;
signed int b = -2;
signed int c = a+b;
if(c < 0)
puts("-1 is smaller than 0");
printf("%d\n", c);
}
The output from the above code:
-1 is smaller than 0
-1
Example 3)
int main(){
unsigned short a = 1;
signed short b = -2;
if(a + b < 0)
puts("-1 is smaller than 0");
printf("%d\n", a+b);
}
The last example fixed the problem since a and b both converted to int due to the integer promotion.
The output from the above code:
-1 is smaller than 0
-1
If I got some concepts mixed up, please let me know. Thanks~

Integer and floating point rank and promotion rules in C and C++
I'd like to take a stab at this to summarize the rules so I can quickly reference them. I've fully studied the question and both of the other two answers here, including the main one by #Lundin. If you want more examples beyond the ones below, go study that answer in detail as well, while referencing my "rules" and "promotion flow" summaries below.
I've also written my own example and demo code here: integer_promotion_overflow_underflow_undefined_behavior.c.
Despite normally being incredibly verbose myself, I'm going to try to keep this a short summary, since the other two answers plus my test code already have sufficient detail via their necessary verbosity.
Integer and variable promotion quick reference guide and summary
3 simple rules
For any operation where multiple operands (input variables) are involved (ex: mathematical operations, comparisons, or ternary), the variables are promoted as required to the required variable type before the operation is performed.
Therefore, you must manually, explicitly cast the output to any desired type you desire if you do not want it to be implicitly chosen for you. See the example below.
All types smaller than int (int32_t on my 64-bit Linux system) are "small types". They cannot be used in ANY operation. So, if all input variables are "small types", they are ALL first promoted to int (int32_t on my 64-bit Linux system) before performing the operation.
Otherwise, if at least one of the input types is int or larger, the other, smaller input type or types are promoted to this largest-input-type's type.
Example
Example: with this code:
uint8_t x = 0;
uint8_t y = 1;
...if you do x - y, they first get implicitly promoted to int (which is int32_t on my 64-bit
system), and you end up with this: (int)x - (int)y, which results in an int type with value
-1, rather than a uint8_t type of value 255. To get the desired 255 result, manually
cast the result back to uint8_t, by doing this: (uint8_t)(x - y).
Promotion flow
The promotion rules are as follows. Promotion from smallest to largest types is as follows.
Read "-->" as "gets promoted to".
The types in square brackets (ex: [int8_t]) are the typical "fixed-width integer types" for the given standard type on a typical 64-bit Unix (Linux or Mac) architecture. See, for example:
https://www.cs.yale.edu/homes/aspnes/pinewiki/C(2f)IntegerTypes.html
https://www.ibm.com/docs/en/ibm-mq/7.5?topic=platforms-standard-data-types
And even better, test it for yourself on your machine by running my code here!: stdint_sizes.c from my eRCaGuy_hello_world repo.
1. For integer types
Note: "small types" = bool (_Bool), char [int8_t], unsigned char [uint8_t], short [int16_t], unsigned short [uint16_t].
SMALL TYPES: bool (_Bool), char [int8_t], unsigned char [uint8_t], short [int16_t], unsigned short [uint16_t]
--> int [int32_t]
--> unsigned int [uint32_t]
--> long int [int64_t]
--> unsigned long int [uint64_t]
--> long long int [int64_t]
--> unsigned long long int [uint64_t]
Pointers (ex: void*) and size_t are both 64-bits, so I imagine they fit into the uint64_t category above.
2. For floating point types
float [32-bits] --> double [64-bits] --> long double [128-bits]

I would like to add two clarifications to #Lundin's otherwise excellent answer, regarding example 1, where there are two operands of identical integer type, but are "small types" that require integer promotion.
I'm using the N1256 draft since I don't have access to a paid copy of the C standard.
First: (normative)
6.3.1.1's definition of integer promotion isn't the triggering clause of actually doing integer promotion. In reality it is 6.3.1.8 Usual arithmetic conversions.
Most of the time, the "usual arithmetic conversions" apply when the operands are of different types, in which case at least one operand must be promoted. But the catch is that for integer types, integer promotion is required in all cases.
[clauses of floating-point types come first]
Otherwise, the integer promotions are performed on both operands. Then the
following rules are applied to the promoted operands:
If both operands have the same type, then no further conversion is needed.
Otherwise, if both operands have signed integer types or both have unsigned
integer types, the operand with the type of lesser integer conversion rank is
converted to the type of the operand with greater rank.
Otherwise, if the operand that has unsigned integer type has rank greater or
equal to the rank of the type of the other operand, then the operand with
signed integer type is converted to the type of the operand with unsigned
integer type.
Otherwise, if the type of the operand with signed integer type can represent
all of the values of the type of the operand with unsigned integer type, then
the operand with unsigned integer type is converted to the type of the
operand with signed integer type.
Otherwise, both operands are converted to the unsigned integer type
corresponding to the type of the operand with signed integer type.
Second: (non-normative)
There is an explicit example cited by the standard to demonstrate this:
EXAMPLE 2 In executing the fragment
char c1, c2;
/* ... */
c1 = c1 + c2;
the "integer promotions" require that the abstract machine promote the value of each variable to int size
and then add the two ints and truncate the sum. Provided the addition of two chars can be done without
overflow, or with overflow wrapping silently to produce the correct result, the actual execution need only
produce the same result, possibly omitting the promotions.

Mechanics of addition type conversion in C [duplicate]

This post is meant to be used as a FAQ regarding implicit integer promotion in C, particularly implicit promotion caused by the usual arithmetic conversions and/or the integer promotions.
Example 1)
Why does this give a strange, large integer number and not 255?
unsigned char x = 0;
unsigned char y = 1;
printf("%u\n", x - y);
Example 2)
Why does this give "-1 is larger than 0"?
unsigned int a = 1;
signed int b = -2;
if(a + b > 0)
puts("-1 is larger than 0");
Example 3)
Why does changing the type in the above example to short fix the problem?
unsigned short a = 1;
signed short b = -2;
if(a + b > 0)
puts("-1 is larger than 0"); // will not print
(These examples were intended for a 32 or 64 bit computer with 16 bit short.)

C was designed to implicitly and silently change the integer types of the operands used in expressions. There exist several cases where the language forces the compiler to either change the operands to a larger type, or to change their signedness.
The rationale behind this is to prevent accidental overflows during arithmetic, but also to allow operands with different signedness to co-exist in the same expression.
Unfortunately, the rules for implicit type promotion cause much more harm than good, to the point where they might be one of the biggest flaws in the C language. These rules are often not even known by the average C programmer and therefore cause all manner of very subtle bugs.
Typically you see scenarios where the programmer says "just cast to type x and it works" - but they don't know why. Or such bugs manifest themselves as rare, intermittent phenomena striking from within seemingly simple and straight-forward code. Implicit promotion is particularly troublesome in code doing bit manipulations, since most bit-wise operators in C come with poorly-defined behavior when given a signed operand.
Integer types and conversion rank
The integer types in C are char, short, int, long, long long and enum.
_Bool/bool is also treated as an integer type when it comes to type promotions.
All integers have a specified conversion rank. C11 6.3.1.1, emphasis mine on the most important parts:
Every integer type has an integer conversion rank defined as follows:
— No two signed integer types shall have the same rank, even if they have the same representation.
— The rank of a signed integer type shall be greater than the rank of any signed integer type with less precision.
— The rank of long long int shall be greater than the rank of long int, which shall be greater than the rank of int, which shall be greater than the rank of short int, which shall be greater than the rank of signed char.
— The rank of any unsigned integer type shall equal the rank of the corresponding signed integer type, if any.
— The rank of any standard integer type shall be greater than the rank of any extended integer type with the same width.
— The rank of char shall equal the rank of signed char and unsigned char.
— The rank of _Bool shall be less than the rank of all other standard integer types.
— The rank of any enumerated type shall equal the rank of the compatible integer type (see 6.7.2.2).
The types from stdint.h sort in here too, with the same rank as whatever type they happen to correspond to on the given system. For example, int32_t has the same rank as int on a 32 bit system.
Further, C11 6.3.1.1 specifies which types are regarded as the small integer types (not a formal term):
The following may be used in an expression wherever an int or unsigned int may
be used:
— An object or expression with an integer type (other than int or unsigned int) whose integer conversion rank is less than or equal to the rank of int and unsigned int.
What this somewhat cryptic text means in practice, is that _Bool, char and short (and also int8_t, uint8_t etc) are the "small integer types". These are treated in special ways and subject to implicit promotion, as explained below.
The integer promotions
Whenever a small integer type is used in an expression, it is implicitly converted to int which is always signed. This is known as the integer promotions or the integer promotion rule.
Formally, the rule says (C11 6.3.1.1):
If an int can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions.
This means that all small integer types, no matter signedness, get implicitly converted to (signed) int when used in most expressions.
This text is often misunderstood as: "all small signed integer types are converted to signed int and all small, unsigned integer types are converted to unsigned int". This is incorrect. The unsigned part here only means that if we have for example an unsigned short operand, and int happens to have the same size as short on the given system, then the unsigned short operand is converted to unsigned int. As in, nothing of note really happens. But in case short is a smaller type than int, it is always converted to (signed) int, regardless of it the short was signed or unsigned!
The harsh reality caused by the integer promotions means that almost no operation in C can be carried out on small types like char or short. Operations are always carried out on int or larger types.
This might sound like nonsense, but luckily the compiler is allowed to optimize the code. For example, an expression containing two unsigned char operands would get the operands promoted to int and the operation carried out as int. But the compiler is allowed to optimize the expression to actually get carried out as an 8-bit operation, as would be expected. However, here comes the problem: the compiler is not allowed to optimize out the implicit change of signedness caused by the integer promotion because there is no way for the compiler to tell if the programmer is purposely relying on implicit promotion to happen, or if it is unintentional.
This is why example 1 in the question fails. Both unsigned char operands are promoted to type int, the operation is carried out on type int, and the result of x - y is of type int. Meaning that we get -1 instead of 255 which might have been expected. The compiler may generate machine code that executes the code with 8 bit instructions instead of int, but it may not optimize out the change of signedness. Meaning that we end up with a negative result, that in turn results in a weird number when printf("%u is invoked. Example 1 could be fixed by casting the result of the operation back to type unsigned char.
With the exception of a few special cases like ++ and sizeof operators, the integer promotions apply to almost all operations in C, no matter if unary, binary (or ternary) operators are used.
The usual arithmetic conversions
Whenever a binary operation (an operation with 2 operands) is done in C, both operands of the operator have to be of the same type. Therefore, in case the operands are of different types, C enforces an implicit conversion of one operand to the type of the other operand. The rules for how this is done are named the usual artihmetic conversions (sometimes informally referred to as "balancing"). These are specified in C11 6.3.18:
(Think of this rule as a long, nested if-else if statement and it might be easier to read :) )
6.3.1.8 Usual arithmetic conversions
Many operators that expect operands of arithmetic type cause conversions and yield result
types in a similar way. The purpose is to determine a common real type for the operands
and result. For the specified operands, each operand is converted, without change of type
domain, to a type whose corresponding real type is the common real type. Unless
explicitly stated otherwise, the common real type is also the corresponding real type of
the result, whose type domain is the type domain of the operands if they are the same,
and complex otherwise. This pattern is called the usual arithmetic conversions:
First, if the corresponding real type of either operand is long double, the other operand is converted, without change of type domain, to a type whose corresponding real type is long double.
Otherwise, if the corresponding real type of either operand is double, the other operand is converted, without change of type domain, to a type whose corresponding real type is double.
Otherwise, if the corresponding real type of either operand is float, the other operand is converted, without change of type domain, to a type whose corresponding real type is float.
Otherwise, the integer promotions are performed on both operands. Then the
following rules are applied to the promoted operands:
If both operands have the same type, then no further conversion is needed.
Otherwise, if both operands have signed integer types or both have unsigned
integer types, the operand with the type of lesser integer conversion rank is
converted to the type of the operand with greater rank.
Otherwise, if the operand that has unsigned integer type has rank greater or
equal to the rank of the type of the other operand, then the operand with
signed integer type is converted to the type of the operand with unsigned
integer type.
Otherwise, if the type of the operand with signed integer type can represent
all of the values of the type of the operand with unsigned integer type, then
the operand with unsigned integer type is converted to the type of the
operand with signed integer type.
Otherwise, both operands are converted to the unsigned integer type
corresponding to the type of the operand with signed integer type.
Notable here is that the usual arithmetic conversions apply to both floating point and integer variables. In the case of integers, we can also note that the integer promotions are invoked from within the usual arithmetic conversions. And after that, when both operands have at least the rank of int, the operators are balanced to the same type, with the same signedness.
This is the reason why a + b in example 2 gives a strange result. Both operands are integers and they are at least of rank int, so the integer promotions do not apply. The operands are not of the same type - a is unsigned int and b is signed int. Therefore the operator b is temporarily converted to type unsigned int. During this conversion, it loses the sign information and ends up as a large value.
The reason why changing type to short in example 3 fixes the problem, is because short is a small integer type. Meaning that both operands are integer promoted to type int which is signed. After integer promotion, both operands have the same type (int), no further conversion is needed. And then the operation can be carried out on a signed type as expected.

According to the previous post, I want to give more information about each example.
Example 1)
int main(){
unsigned char x = 0;
unsigned char y = 1;
printf("%u\n", x - y);
printf("%d\n", x - y);
}
Since unsigned char is smaller than int, we apply the integer promotion on them, then we have (int)x-(int)y = (int)(-1) and unsigned int (-1) = 4294967295.
The output from the above code:(same as what we expected)
4294967295
-1
How to fix it?
I tried what the previous post recommended, but it doesn't really work.
Here is the code based on the previous post:
change one of them to unsigned int
int main(){
unsigned int x = 0;
unsigned char y = 1;
printf("%u\n", x - y);
printf("%d\n", x - y);
}
Since x is already an unsigned integer, we only apply the integer promotion to y. Then we get (unsigned int)x-(int)y. Since they still don't have the same type, we apply the usual arithmetic converions, we get (unsigned int)x-(unsigned int)y = 4294967295.
The output from the above code:(same as what we expected):
4294967295
-1
Similarly, the following code gets the same result:
int main(){
unsigned char x = 0;
unsigned int y = 1;
printf("%u\n", x - y);
printf("%d\n", x - y);
}
change both of them to unsigned int
int main(){
unsigned int x = 0;
unsigned int y = 1;
printf("%u\n", x - y);
printf("%d\n", x - y);
}
Since both of them are unsigned int, no integer promotion is needed. By the usual arithmetic converison(have the same type), (unsigned int)x-(unsigned int)y = 4294967295.
The output from the above code:(same as what we expected):
4294967295
-1
One of possible ways to fix the code:(add a type cast in the end)
int main(){
unsigned char x = 0;
unsigned char y = 1;
printf("%u\n", x - y);
printf("%d\n", x - y);
unsigned char z = x-y;
printf("%u\n", z);
}
The output from the above code:
4294967295
-1
255
Example 2)
int main(){
unsigned int a = 1;
signed int b = -2;
if(a + b > 0)
puts("-1 is larger than 0");
printf("%u\n", a+b);
}
Since both of them are integers, no integer promotion is needed. By the usual arithmetic conversion, we get (unsigned int)a+(unsigned int)b = 1+4294967294 = 4294967295.
The output from the above code:(same as what we expected)
-1 is larger than 0
4294967295
How to fix it?
int main(){
unsigned int a = 1;
signed int b = -2;
signed int c = a+b;
if(c < 0)
puts("-1 is smaller than 0");
printf("%d\n", c);
}
The output from the above code:
-1 is smaller than 0
-1
Example 3)
int main(){
unsigned short a = 1;
signed short b = -2;
if(a + b < 0)
puts("-1 is smaller than 0");
printf("%d\n", a+b);
}
The last example fixed the problem since a and b both converted to int due to the integer promotion.
The output from the above code:
-1 is smaller than 0
-1
If I got some concepts mixed up, please let me know. Thanks~

Integer and floating point rank and promotion rules in C and C++
I'd like to take a stab at this to summarize the rules so I can quickly reference them. I've fully studied the question and both of the other two answers here, including the main one by #Lundin. If you want more examples beyond the ones below, go study that answer in detail as well, while referencing my "rules" and "promotion flow" summaries below.
I've also written my own example and demo code here: integer_promotion_overflow_underflow_undefined_behavior.c.
Despite normally being incredibly verbose myself, I'm going to try to keep this a short summary, since the other two answers plus my test code already have sufficient detail via their necessary verbosity.
Integer and variable promotion quick reference guide and summary
3 simple rules
For any operation where multiple operands (input variables) are involved (ex: mathematical operations, comparisons, or ternary), the variables are promoted as required to the required variable type before the operation is performed.
Therefore, you must manually, explicitly cast the output to any desired type you desire if you do not want it to be implicitly chosen for you. See the example below.
All types smaller than int (int32_t on my 64-bit Linux system) are "small types". They cannot be used in ANY operation. So, if all input variables are "small types", they are ALL first promoted to int (int32_t on my 64-bit Linux system) before performing the operation.
Otherwise, if at least one of the input types is int or larger, the other, smaller input type or types are promoted to this largest-input-type's type.
Example
Example: with this code:
uint8_t x = 0;
uint8_t y = 1;
...if you do x - y, they first get implicitly promoted to int (which is int32_t on my 64-bit
system), and you end up with this: (int)x - (int)y, which results in an int type with value
-1, rather than a uint8_t type of value 255. To get the desired 255 result, manually
cast the result back to uint8_t, by doing this: (uint8_t)(x - y).
Promotion flow
The promotion rules are as follows. Promotion from smallest to largest types is as follows.
Read "-->" as "gets promoted to".
The types in square brackets (ex: [int8_t]) are the typical "fixed-width integer types" for the given standard type on a typical 64-bit Unix (Linux or Mac) architecture. See, for example:
https://www.cs.yale.edu/homes/aspnes/pinewiki/C(2f)IntegerTypes.html
https://www.ibm.com/docs/en/ibm-mq/7.5?topic=platforms-standard-data-types
And even better, test it for yourself on your machine by running my code here!: stdint_sizes.c from my eRCaGuy_hello_world repo.
1. For integer types
Note: "small types" = bool (_Bool), char [int8_t], unsigned char [uint8_t], short [int16_t], unsigned short [uint16_t].
SMALL TYPES: bool (_Bool), char [int8_t], unsigned char [uint8_t], short [int16_t], unsigned short [uint16_t]
--> int [int32_t]
--> unsigned int [uint32_t]
--> long int [int64_t]
--> unsigned long int [uint64_t]
--> long long int [int64_t]
--> unsigned long long int [uint64_t]
Pointers (ex: void*) and size_t are both 64-bits, so I imagine they fit into the uint64_t category above.
2. For floating point types
float [32-bits] --> double [64-bits] --> long double [128-bits]

I would like to add two clarifications to #Lundin's otherwise excellent answer, regarding example 1, where there are two operands of identical integer type, but are "small types" that require integer promotion.
I'm using the N1256 draft since I don't have access to a paid copy of the C standard.
First: (normative)
6.3.1.1's definition of integer promotion isn't the triggering clause of actually doing integer promotion. In reality it is 6.3.1.8 Usual arithmetic conversions.
Most of the time, the "usual arithmetic conversions" apply when the operands are of different types, in which case at least one operand must be promoted. But the catch is that for integer types, integer promotion is required in all cases.
[clauses of floating-point types come first]
Otherwise, the integer promotions are performed on both operands. Then the
following rules are applied to the promoted operands:
If both operands have the same type, then no further conversion is needed.
Otherwise, if both operands have signed integer types or both have unsigned
integer types, the operand with the type of lesser integer conversion rank is
converted to the type of the operand with greater rank.
Otherwise, if the operand that has unsigned integer type has rank greater or
equal to the rank of the type of the other operand, then the operand with
signed integer type is converted to the type of the operand with unsigned
integer type.
Otherwise, if the type of the operand with signed integer type can represent
all of the values of the type of the operand with unsigned integer type, then
the operand with unsigned integer type is converted to the type of the
operand with signed integer type.
Otherwise, both operands are converted to the unsigned integer type
corresponding to the type of the operand with signed integer type.
Second: (non-normative)
There is an explicit example cited by the standard to demonstrate this:
EXAMPLE 2 In executing the fragment
char c1, c2;
/* ... */
c1 = c1 + c2;
the "integer promotions" require that the abstract machine promote the value of each variable to int size
and then add the two ints and truncate the sum. Provided the addition of two chars can be done without
overflow, or with overflow wrapping silently to produce the correct result, the actual execution need only
produce the same result, possibly omitting the promotions.

Type of expression's result in C... Integer Promotion and Usual Conversions [duplicate]

This post is meant to be used as a FAQ regarding implicit integer promotion in C, particularly implicit promotion caused by the usual arithmetic conversions and/or the integer promotions.
Example 1)
Why does this give a strange, large integer number and not 255?
unsigned char x = 0;
unsigned char y = 1;
printf("%u\n", x - y);
Example 2)
Why does this give "-1 is larger than 0"?
unsigned int a = 1;
signed int b = -2;
if(a + b > 0)
puts("-1 is larger than 0");
Example 3)
Why does changing the type in the above example to short fix the problem?
unsigned short a = 1;
signed short b = -2;
if(a + b > 0)
puts("-1 is larger than 0"); // will not print
(These examples were intended for a 32 or 64 bit computer with 16 bit short.)

C was designed to implicitly and silently change the integer types of the operands used in expressions. There exist several cases where the language forces the compiler to either change the operands to a larger type, or to change their signedness.
The rationale behind this is to prevent accidental overflows during arithmetic, but also to allow operands with different signedness to co-exist in the same expression.
Unfortunately, the rules for implicit type promotion cause much more harm than good, to the point where they might be one of the biggest flaws in the C language. These rules are often not even known by the average C programmer and therefore cause all manner of very subtle bugs.
Typically you see scenarios where the programmer says "just cast to type x and it works" - but they don't know why. Or such bugs manifest themselves as rare, intermittent phenomena striking from within seemingly simple and straight-forward code. Implicit promotion is particularly troublesome in code doing bit manipulations, since most bit-wise operators in C come with poorly-defined behavior when given a signed operand.
Integer types and conversion rank
The integer types in C are char, short, int, long, long long and enum.
_Bool/bool is also treated as an integer type when it comes to type promotions.
All integers have a specified conversion rank. C11 6.3.1.1, emphasis mine on the most important parts:
Every integer type has an integer conversion rank defined as follows:
— No two signed integer types shall have the same rank, even if they have the same representation.
— The rank of a signed integer type shall be greater than the rank of any signed integer type with less precision.
— The rank of long long int shall be greater than the rank of long int, which shall be greater than the rank of int, which shall be greater than the rank of short int, which shall be greater than the rank of signed char.
— The rank of any unsigned integer type shall equal the rank of the corresponding signed integer type, if any.
— The rank of any standard integer type shall be greater than the rank of any extended integer type with the same width.
— The rank of char shall equal the rank of signed char and unsigned char.
— The rank of _Bool shall be less than the rank of all other standard integer types.
— The rank of any enumerated type shall equal the rank of the compatible integer type (see 6.7.2.2).
The types from stdint.h sort in here too, with the same rank as whatever type they happen to correspond to on the given system. For example, int32_t has the same rank as int on a 32 bit system.
Further, C11 6.3.1.1 specifies which types are regarded as the small integer types (not a formal term):
The following may be used in an expression wherever an int or unsigned int may
be used:
— An object or expression with an integer type (other than int or unsigned int) whose integer conversion rank is less than or equal to the rank of int and unsigned int.
What this somewhat cryptic text means in practice, is that _Bool, char and short (and also int8_t, uint8_t etc) are the "small integer types". These are treated in special ways and subject to implicit promotion, as explained below.
The integer promotions
Whenever a small integer type is used in an expression, it is implicitly converted to int which is always signed. This is known as the integer promotions or the integer promotion rule.
Formally, the rule says (C11 6.3.1.1):
If an int can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions.
This means that all small integer types, no matter signedness, get implicitly converted to (signed) int when used in most expressions.
This text is often misunderstood as: "all small signed integer types are converted to signed int and all small, unsigned integer types are converted to unsigned int". This is incorrect. The unsigned part here only means that if we have for example an unsigned short operand, and int happens to have the same size as short on the given system, then the unsigned short operand is converted to unsigned int. As in, nothing of note really happens. But in case short is a smaller type than int, it is always converted to (signed) int, regardless of it the short was signed or unsigned!
The harsh reality caused by the integer promotions means that almost no operation in C can be carried out on small types like char or short. Operations are always carried out on int or larger types.
This might sound like nonsense, but luckily the compiler is allowed to optimize the code. For example, an expression containing two unsigned char operands would get the operands promoted to int and the operation carried out as int. But the compiler is allowed to optimize the expression to actually get carried out as an 8-bit operation, as would be expected. However, here comes the problem: the compiler is not allowed to optimize out the implicit change of signedness caused by the integer promotion because there is no way for the compiler to tell if the programmer is purposely relying on implicit promotion to happen, or if it is unintentional.
This is why example 1 in the question fails. Both unsigned char operands are promoted to type int, the operation is carried out on type int, and the result of x - y is of type int. Meaning that we get -1 instead of 255 which might have been expected. The compiler may generate machine code that executes the code with 8 bit instructions instead of int, but it may not optimize out the change of signedness. Meaning that we end up with a negative result, that in turn results in a weird number when printf("%u is invoked. Example 1 could be fixed by casting the result of the operation back to type unsigned char.
With the exception of a few special cases like ++ and sizeof operators, the integer promotions apply to almost all operations in C, no matter if unary, binary (or ternary) operators are used.
The usual arithmetic conversions
Whenever a binary operation (an operation with 2 operands) is done in C, both operands of the operator have to be of the same type. Therefore, in case the operands are of different types, C enforces an implicit conversion of one operand to the type of the other operand. The rules for how this is done are named the usual artihmetic conversions (sometimes informally referred to as "balancing"). These are specified in C11 6.3.18:
(Think of this rule as a long, nested if-else if statement and it might be easier to read :) )
6.3.1.8 Usual arithmetic conversions
Many operators that expect operands of arithmetic type cause conversions and yield result
types in a similar way. The purpose is to determine a common real type for the operands
and result. For the specified operands, each operand is converted, without change of type
domain, to a type whose corresponding real type is the common real type. Unless
explicitly stated otherwise, the common real type is also the corresponding real type of
the result, whose type domain is the type domain of the operands if they are the same,
and complex otherwise. This pattern is called the usual arithmetic conversions:
First, if the corresponding real type of either operand is long double, the other operand is converted, without change of type domain, to a type whose corresponding real type is long double.
Otherwise, if the corresponding real type of either operand is double, the other operand is converted, without change of type domain, to a type whose corresponding real type is double.
Otherwise, if the corresponding real type of either operand is float, the other operand is converted, without change of type domain, to a type whose corresponding real type is float.
Otherwise, the integer promotions are performed on both operands. Then the
following rules are applied to the promoted operands:
If both operands have the same type, then no further conversion is needed.
Otherwise, if both operands have signed integer types or both have unsigned
integer types, the operand with the type of lesser integer conversion rank is
converted to the type of the operand with greater rank.
Otherwise, if the operand that has unsigned integer type has rank greater or
equal to the rank of the type of the other operand, then the operand with
signed integer type is converted to the type of the operand with unsigned
integer type.
Otherwise, if the type of the operand with signed integer type can represent
all of the values of the type of the operand with unsigned integer type, then
the operand with unsigned integer type is converted to the type of the
operand with signed integer type.
Otherwise, both operands are converted to the unsigned integer type
corresponding to the type of the operand with signed integer type.
Notable here is that the usual arithmetic conversions apply to both floating point and integer variables. In the case of integers, we can also note that the integer promotions are invoked from within the usual arithmetic conversions. And after that, when both operands have at least the rank of int, the operators are balanced to the same type, with the same signedness.
This is the reason why a + b in example 2 gives a strange result. Both operands are integers and they are at least of rank int, so the integer promotions do not apply. The operands are not of the same type - a is unsigned int and b is signed int. Therefore the operator b is temporarily converted to type unsigned int. During this conversion, it loses the sign information and ends up as a large value.
The reason why changing type to short in example 3 fixes the problem, is because short is a small integer type. Meaning that both operands are integer promoted to type int which is signed. After integer promotion, both operands have the same type (int), no further conversion is needed. And then the operation can be carried out on a signed type as expected.

According to the previous post, I want to give more information about each example.
Example 1)
int main(){
unsigned char x = 0;
unsigned char y = 1;
printf("%u\n", x - y);
printf("%d\n", x - y);
}
Since unsigned char is smaller than int, we apply the integer promotion on them, then we have (int)x-(int)y = (int)(-1) and unsigned int (-1) = 4294967295.
The output from the above code:(same as what we expected)
4294967295
-1
How to fix it?
I tried what the previous post recommended, but it doesn't really work.
Here is the code based on the previous post:
change one of them to unsigned int
int main(){
unsigned int x = 0;
unsigned char y = 1;
printf("%u\n", x - y);
printf("%d\n", x - y);
}
Since x is already an unsigned integer, we only apply the integer promotion to y. Then we get (unsigned int)x-(int)y. Since they still don't have the same type, we apply the usual arithmetic converions, we get (unsigned int)x-(unsigned int)y = 4294967295.
The output from the above code:(same as what we expected):
4294967295
-1
Similarly, the following code gets the same result:
int main(){
unsigned char x = 0;
unsigned int y = 1;
printf("%u\n", x - y);
printf("%d\n", x - y);
}
change both of them to unsigned int
int main(){
unsigned int x = 0;
unsigned int y = 1;
printf("%u\n", x - y);
printf("%d\n", x - y);
}
Since both of them are unsigned int, no integer promotion is needed. By the usual arithmetic converison(have the same type), (unsigned int)x-(unsigned int)y = 4294967295.
The output from the above code:(same as what we expected):
4294967295
-1
One of possible ways to fix the code:(add a type cast in the end)
int main(){
unsigned char x = 0;
unsigned char y = 1;
printf("%u\n", x - y);
printf("%d\n", x - y);
unsigned char z = x-y;
printf("%u\n", z);
}
The output from the above code:
4294967295
-1
255
Example 2)
int main(){
unsigned int a = 1;
signed int b = -2;
if(a + b > 0)
puts("-1 is larger than 0");
printf("%u\n", a+b);
}
Since both of them are integers, no integer promotion is needed. By the usual arithmetic conversion, we get (unsigned int)a+(unsigned int)b = 1+4294967294 = 4294967295.
The output from the above code:(same as what we expected)
-1 is larger than 0
4294967295
How to fix it?
int main(){
unsigned int a = 1;
signed int b = -2;
signed int c = a+b;
if(c < 0)
puts("-1 is smaller than 0");
printf("%d\n", c);
}
The output from the above code:
-1 is smaller than 0
-1
Example 3)
int main(){
unsigned short a = 1;
signed short b = -2;
if(a + b < 0)
puts("-1 is smaller than 0");
printf("%d\n", a+b);
}
The last example fixed the problem since a and b both converted to int due to the integer promotion.
The output from the above code:
-1 is smaller than 0
-1
If I got some concepts mixed up, please let me know. Thanks~

Integer and floating point rank and promotion rules in C and C++
I'd like to take a stab at this to summarize the rules so I can quickly reference them. I've fully studied the question and both of the other two answers here, including the main one by #Lundin. If you want more examples beyond the ones below, go study that answer in detail as well, while referencing my "rules" and "promotion flow" summaries below.
I've also written my own example and demo code here: integer_promotion_overflow_underflow_undefined_behavior.c.
Despite normally being incredibly verbose myself, I'm going to try to keep this a short summary, since the other two answers plus my test code already have sufficient detail via their necessary verbosity.
Integer and variable promotion quick reference guide and summary
3 simple rules
For any operation where multiple operands (input variables) are involved (ex: mathematical operations, comparisons, or ternary), the variables are promoted as required to the required variable type before the operation is performed.
Therefore, you must manually, explicitly cast the output to any desired type you desire if you do not want it to be implicitly chosen for you. See the example below.
All types smaller than int (int32_t on my 64-bit Linux system) are "small types". They cannot be used in ANY operation. So, if all input variables are "small types", they are ALL first promoted to int (int32_t on my 64-bit Linux system) before performing the operation.
Otherwise, if at least one of the input types is int or larger, the other, smaller input type or types are promoted to this largest-input-type's type.
Example
Example: with this code:
uint8_t x = 0;
uint8_t y = 1;
...if you do x - y, they first get implicitly promoted to int (which is int32_t on my 64-bit
system), and you end up with this: (int)x - (int)y, which results in an int type with value
-1, rather than a uint8_t type of value 255. To get the desired 255 result, manually
cast the result back to uint8_t, by doing this: (uint8_t)(x - y).
Promotion flow
The promotion rules are as follows. Promotion from smallest to largest types is as follows.
Read "-->" as "gets promoted to".
The types in square brackets (ex: [int8_t]) are the typical "fixed-width integer types" for the given standard type on a typical 64-bit Unix (Linux or Mac) architecture. See, for example:
https://www.cs.yale.edu/homes/aspnes/pinewiki/C(2f)IntegerTypes.html
https://www.ibm.com/docs/en/ibm-mq/7.5?topic=platforms-standard-data-types
And even better, test it for yourself on your machine by running my code here!: stdint_sizes.c from my eRCaGuy_hello_world repo.
1. For integer types
Note: "small types" = bool (_Bool), char [int8_t], unsigned char [uint8_t], short [int16_t], unsigned short [uint16_t].
SMALL TYPES: bool (_Bool), char [int8_t], unsigned char [uint8_t], short [int16_t], unsigned short [uint16_t]
--> int [int32_t]
--> unsigned int [uint32_t]
--> long int [int64_t]
--> unsigned long int [uint64_t]
--> long long int [int64_t]
--> unsigned long long int [uint64_t]
Pointers (ex: void*) and size_t are both 64-bits, so I imagine they fit into the uint64_t category above.
2. For floating point types
float [32-bits] --> double [64-bits] --> long double [128-bits]

I would like to add two clarifications to #Lundin's otherwise excellent answer, regarding example 1, where there are two operands of identical integer type, but are "small types" that require integer promotion.
I'm using the N1256 draft since I don't have access to a paid copy of the C standard.
First: (normative)
6.3.1.1's definition of integer promotion isn't the triggering clause of actually doing integer promotion. In reality it is 6.3.1.8 Usual arithmetic conversions.
Most of the time, the "usual arithmetic conversions" apply when the operands are of different types, in which case at least one operand must be promoted. But the catch is that for integer types, integer promotion is required in all cases.
[clauses of floating-point types come first]
Otherwise, the integer promotions are performed on both operands. Then the
following rules are applied to the promoted operands:
If both operands have the same type, then no further conversion is needed.
Otherwise, if both operands have signed integer types or both have unsigned
integer types, the operand with the type of lesser integer conversion rank is
converted to the type of the operand with greater rank.
Otherwise, if the operand that has unsigned integer type has rank greater or
equal to the rank of the type of the other operand, then the operand with
signed integer type is converted to the type of the operand with unsigned
integer type.
Otherwise, if the type of the operand with signed integer type can represent
all of the values of the type of the operand with unsigned integer type, then
the operand with unsigned integer type is converted to the type of the
operand with signed integer type.
Otherwise, both operands are converted to the unsigned integer type
corresponding to the type of the operand with signed integer type.
Second: (non-normative)
There is an explicit example cited by the standard to demonstrate this:
EXAMPLE 2 In executing the fragment
char c1, c2;
/* ... */
c1 = c1 + c2;
the "integer promotions" require that the abstract machine promote the value of each variable to int size
and then add the two ints and truncate the sum. Provided the addition of two chars can be done without
overflow, or with overflow wrapping silently to produce the correct result, the actual execution need only
produce the same result, possibly omitting the promotions.

Implicit type promotion rules

This post is meant to be used as a FAQ regarding implicit integer promotion in C, particularly implicit promotion caused by the usual arithmetic conversions and/or the integer promotions.
Example 1)
Why does this give a strange, large integer number and not 255?
unsigned char x = 0;
unsigned char y = 1;
printf("%u\n", x - y);
Example 2)
Why does this give "-1 is larger than 0"?
unsigned int a = 1;
signed int b = -2;
if(a + b > 0)
puts("-1 is larger than 0");
Example 3)
Why does changing the type in the above example to short fix the problem?
unsigned short a = 1;
signed short b = -2;
if(a + b > 0)
puts("-1 is larger than 0"); // will not print
(These examples were intended for a 32 or 64 bit computer with 16 bit short.)

C was designed to implicitly and silently change the integer types of the operands used in expressions. There exist several cases where the language forces the compiler to either change the operands to a larger type, or to change their signedness.
The rationale behind this is to prevent accidental overflows during arithmetic, but also to allow operands with different signedness to co-exist in the same expression.
Unfortunately, the rules for implicit type promotion cause much more harm than good, to the point where they might be one of the biggest flaws in the C language. These rules are often not even known by the average C programmer and therefore cause all manner of very subtle bugs.
Typically you see scenarios where the programmer says "just cast to type x and it works" - but they don't know why. Or such bugs manifest themselves as rare, intermittent phenomena striking from within seemingly simple and straight-forward code. Implicit promotion is particularly troublesome in code doing bit manipulations, since most bit-wise operators in C come with poorly-defined behavior when given a signed operand.
Integer types and conversion rank
The integer types in C are char, short, int, long, long long and enum.
_Bool/bool is also treated as an integer type when it comes to type promotions.
All integers have a specified conversion rank. C11 6.3.1.1, emphasis mine on the most important parts:
Every integer type has an integer conversion rank defined as follows:
— No two signed integer types shall have the same rank, even if they have the same representation.
— The rank of a signed integer type shall be greater than the rank of any signed integer type with less precision.
— The rank of long long int shall be greater than the rank of long int, which shall be greater than the rank of int, which shall be greater than the rank of short int, which shall be greater than the rank of signed char.
— The rank of any unsigned integer type shall equal the rank of the corresponding signed integer type, if any.
— The rank of any standard integer type shall be greater than the rank of any extended integer type with the same width.
— The rank of char shall equal the rank of signed char and unsigned char.
— The rank of _Bool shall be less than the rank of all other standard integer types.
— The rank of any enumerated type shall equal the rank of the compatible integer type (see 6.7.2.2).
The types from stdint.h sort in here too, with the same rank as whatever type they happen to correspond to on the given system. For example, int32_t has the same rank as int on a 32 bit system.
Further, C11 6.3.1.1 specifies which types are regarded as the small integer types (not a formal term):
The following may be used in an expression wherever an int or unsigned int may
be used:
— An object or expression with an integer type (other than int or unsigned int) whose integer conversion rank is less than or equal to the rank of int and unsigned int.
What this somewhat cryptic text means in practice, is that _Bool, char and short (and also int8_t, uint8_t etc) are the "small integer types". These are treated in special ways and subject to implicit promotion, as explained below.
The integer promotions
Whenever a small integer type is used in an expression, it is implicitly converted to int which is always signed. This is known as the integer promotions or the integer promotion rule.
Formally, the rule says (C11 6.3.1.1):
If an int can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions.
This means that all small integer types, no matter signedness, get implicitly converted to (signed) int when used in most expressions.
This text is often misunderstood as: "all small signed integer types are converted to signed int and all small, unsigned integer types are converted to unsigned int". This is incorrect. The unsigned part here only means that if we have for example an unsigned short operand, and int happens to have the same size as short on the given system, then the unsigned short operand is converted to unsigned int. As in, nothing of note really happens. But in case short is a smaller type than int, it is always converted to (signed) int, regardless of it the short was signed or unsigned!
The harsh reality caused by the integer promotions means that almost no operation in C can be carried out on small types like char or short. Operations are always carried out on int or larger types.
This might sound like nonsense, but luckily the compiler is allowed to optimize the code. For example, an expression containing two unsigned char operands would get the operands promoted to int and the operation carried out as int. But the compiler is allowed to optimize the expression to actually get carried out as an 8-bit operation, as would be expected. However, here comes the problem: the compiler is not allowed to optimize out the implicit change of signedness caused by the integer promotion because there is no way for the compiler to tell if the programmer is purposely relying on implicit promotion to happen, or if it is unintentional.
This is why example 1 in the question fails. Both unsigned char operands are promoted to type int, the operation is carried out on type int, and the result of x - y is of type int. Meaning that we get -1 instead of 255 which might have been expected. The compiler may generate machine code that executes the code with 8 bit instructions instead of int, but it may not optimize out the change of signedness. Meaning that we end up with a negative result, that in turn results in a weird number when printf("%u is invoked. Example 1 could be fixed by casting the result of the operation back to type unsigned char.
With the exception of a few special cases like ++ and sizeof operators, the integer promotions apply to almost all operations in C, no matter if unary, binary (or ternary) operators are used.
The usual arithmetic conversions
Whenever a binary operation (an operation with 2 operands) is done in C, both operands of the operator have to be of the same type. Therefore, in case the operands are of different types, C enforces an implicit conversion of one operand to the type of the other operand. The rules for how this is done are named the usual artihmetic conversions (sometimes informally referred to as "balancing"). These are specified in C11 6.3.18:
(Think of this rule as a long, nested if-else if statement and it might be easier to read :) )
6.3.1.8 Usual arithmetic conversions
Many operators that expect operands of arithmetic type cause conversions and yield result
types in a similar way. The purpose is to determine a common real type for the operands
and result. For the specified operands, each operand is converted, without change of type
domain, to a type whose corresponding real type is the common real type. Unless
explicitly stated otherwise, the common real type is also the corresponding real type of
the result, whose type domain is the type domain of the operands if they are the same,
and complex otherwise. This pattern is called the usual arithmetic conversions:
First, if the corresponding real type of either operand is long double, the other operand is converted, without change of type domain, to a type whose corresponding real type is long double.
Otherwise, if the corresponding real type of either operand is double, the other operand is converted, without change of type domain, to a type whose corresponding real type is double.
Otherwise, if the corresponding real type of either operand is float, the other operand is converted, without change of type domain, to a type whose corresponding real type is float.
Otherwise, the integer promotions are performed on both operands. Then the
following rules are applied to the promoted operands:
If both operands have the same type, then no further conversion is needed.
Otherwise, if both operands have signed integer types or both have unsigned
integer types, the operand with the type of lesser integer conversion rank is
converted to the type of the operand with greater rank.
Otherwise, if the operand that has unsigned integer type has rank greater or
equal to the rank of the type of the other operand, then the operand with
signed integer type is converted to the type of the operand with unsigned
integer type.
Otherwise, if the type of the operand with signed integer type can represent
all of the values of the type of the operand with unsigned integer type, then
the operand with unsigned integer type is converted to the type of the
operand with signed integer type.
Otherwise, both operands are converted to the unsigned integer type
corresponding to the type of the operand with signed integer type.
Notable here is that the usual arithmetic conversions apply to both floating point and integer variables. In the case of integers, we can also note that the integer promotions are invoked from within the usual arithmetic conversions. And after that, when both operands have at least the rank of int, the operators are balanced to the same type, with the same signedness.
This is the reason why a + b in example 2 gives a strange result. Both operands are integers and they are at least of rank int, so the integer promotions do not apply. The operands are not of the same type - a is unsigned int and b is signed int. Therefore the operator b is temporarily converted to type unsigned int. During this conversion, it loses the sign information and ends up as a large value.
The reason why changing type to short in example 3 fixes the problem, is because short is a small integer type. Meaning that both operands are integer promoted to type int which is signed. After integer promotion, both operands have the same type (int), no further conversion is needed. And then the operation can be carried out on a signed type as expected.

According to the previous post, I want to give more information about each example.
Example 1)
int main(){
unsigned char x = 0;
unsigned char y = 1;
printf("%u\n", x - y);
printf("%d\n", x - y);
}
Since unsigned char is smaller than int, we apply the integer promotion on them, then we have (int)x-(int)y = (int)(-1) and unsigned int (-1) = 4294967295.
The output from the above code:(same as what we expected)
4294967295
-1
How to fix it?
I tried what the previous post recommended, but it doesn't really work.
Here is the code based on the previous post:
change one of them to unsigned int
int main(){
unsigned int x = 0;
unsigned char y = 1;
printf("%u\n", x - y);
printf("%d\n", x - y);
}
Since x is already an unsigned integer, we only apply the integer promotion to y. Then we get (unsigned int)x-(int)y. Since they still don't have the same type, we apply the usual arithmetic converions, we get (unsigned int)x-(unsigned int)y = 4294967295.
The output from the above code:(same as what we expected):
4294967295
-1
Similarly, the following code gets the same result:
int main(){
unsigned char x = 0;
unsigned int y = 1;
printf("%u\n", x - y);
printf("%d\n", x - y);
}
change both of them to unsigned int
int main(){
unsigned int x = 0;
unsigned int y = 1;
printf("%u\n", x - y);
printf("%d\n", x - y);
}
Since both of them are unsigned int, no integer promotion is needed. By the usual arithmetic converison(have the same type), (unsigned int)x-(unsigned int)y = 4294967295.
The output from the above code:(same as what we expected):
4294967295
-1
One of possible ways to fix the code:(add a type cast in the end)
int main(){
unsigned char x = 0;
unsigned char y = 1;
printf("%u\n", x - y);
printf("%d\n", x - y);
unsigned char z = x-y;
printf("%u\n", z);
}
The output from the above code:
4294967295
-1
255
Example 2)
int main(){
unsigned int a = 1;
signed int b = -2;
if(a + b > 0)
puts("-1 is larger than 0");
printf("%u\n", a+b);
}
Since both of them are integers, no integer promotion is needed. By the usual arithmetic conversion, we get (unsigned int)a+(unsigned int)b = 1+4294967294 = 4294967295.
The output from the above code:(same as what we expected)
-1 is larger than 0
4294967295
How to fix it?
int main(){
unsigned int a = 1;
signed int b = -2;
signed int c = a+b;
if(c < 0)
puts("-1 is smaller than 0");
printf("%d\n", c);
}
The output from the above code:
-1 is smaller than 0
-1
Example 3)
int main(){
unsigned short a = 1;
signed short b = -2;
if(a + b < 0)
puts("-1 is smaller than 0");
printf("%d\n", a+b);
}
The last example fixed the problem since a and b both converted to int due to the integer promotion.
The output from the above code:
-1 is smaller than 0
-1
If I got some concepts mixed up, please let me know. Thanks~

Integer and floating point rank and promotion rules in C and C++
I'd like to take a stab at this to summarize the rules so I can quickly reference them. I've fully studied the question and both of the other two answers here, including the main one by #Lundin. If you want more examples beyond the ones below, go study that answer in detail as well, while referencing my "rules" and "promotion flow" summaries below.
I've also written my own example and demo code here: integer_promotion_overflow_underflow_undefined_behavior.c.
Despite normally being incredibly verbose myself, I'm going to try to keep this a short summary, since the other two answers plus my test code already have sufficient detail via their necessary verbosity.
Integer and variable promotion quick reference guide and summary
3 simple rules
For any operation where multiple operands (input variables) are involved (ex: mathematical operations, comparisons, or ternary), the variables are promoted as required to the required variable type before the operation is performed.
Therefore, you must manually, explicitly cast the output to any desired type you desire if you do not want it to be implicitly chosen for you. See the example below.
All types smaller than int (int32_t on my 64-bit Linux system) are "small types". They cannot be used in ANY operation. So, if all input variables are "small types", they are ALL first promoted to int (int32_t on my 64-bit Linux system) before performing the operation.
Otherwise, if at least one of the input types is int or larger, the other, smaller input type or types are promoted to this largest-input-type's type.
Example
Example: with this code:
uint8_t x = 0;
uint8_t y = 1;
...if you do x - y, they first get implicitly promoted to int (which is int32_t on my 64-bit
system), and you end up with this: (int)x - (int)y, which results in an int type with value
-1, rather than a uint8_t type of value 255. To get the desired 255 result, manually
cast the result back to uint8_t, by doing this: (uint8_t)(x - y).
Promotion flow
The promotion rules are as follows. Promotion from smallest to largest types is as follows.
Read "-->" as "gets promoted to".
The types in square brackets (ex: [int8_t]) are the typical "fixed-width integer types" for the given standard type on a typical 64-bit Unix (Linux or Mac) architecture. See, for example:
https://www.cs.yale.edu/homes/aspnes/pinewiki/C(2f)IntegerTypes.html
https://www.ibm.com/docs/en/ibm-mq/7.5?topic=platforms-standard-data-types
And even better, test it for yourself on your machine by running my code here!: stdint_sizes.c from my eRCaGuy_hello_world repo.
1. For integer types
Note: "small types" = bool (_Bool), char [int8_t], unsigned char [uint8_t], short [int16_t], unsigned short [uint16_t].
SMALL TYPES: bool (_Bool), char [int8_t], unsigned char [uint8_t], short [int16_t], unsigned short [uint16_t]
--> int [int32_t]
--> unsigned int [uint32_t]
--> long int [int64_t]
--> unsigned long int [uint64_t]
--> long long int [int64_t]
--> unsigned long long int [uint64_t]
Pointers (ex: void*) and size_t are both 64-bits, so I imagine they fit into the uint64_t category above.
2. For floating point types
float [32-bits] --> double [64-bits] --> long double [128-bits]

I would like to add two clarifications to #Lundin's otherwise excellent answer, regarding example 1, where there are two operands of identical integer type, but are "small types" that require integer promotion.
I'm using the N1256 draft since I don't have access to a paid copy of the C standard.
First: (normative)
6.3.1.1's definition of integer promotion isn't the triggering clause of actually doing integer promotion. In reality it is 6.3.1.8 Usual arithmetic conversions.
Most of the time, the "usual arithmetic conversions" apply when the operands are of different types, in which case at least one operand must be promoted. But the catch is that for integer types, integer promotion is required in all cases.
[clauses of floating-point types come first]
Otherwise, the integer promotions are performed on both operands. Then the
following rules are applied to the promoted operands:
If both operands have the same type, then no further conversion is needed.
Otherwise, if both operands have signed integer types or both have unsigned
integer types, the operand with the type of lesser integer conversion rank is
converted to the type of the operand with greater rank.
Otherwise, if the operand that has unsigned integer type has rank greater or
equal to the rank of the type of the other operand, then the operand with
signed integer type is converted to the type of the operand with unsigned
integer type.
Otherwise, if the type of the operand with signed integer type can represent
all of the values of the type of the operand with unsigned integer type, then
the operand with unsigned integer type is converted to the type of the
operand with signed integer type.
Otherwise, both operands are converted to the unsigned integer type
corresponding to the type of the operand with signed integer type.
Second: (non-normative)
There is an explicit example cited by the standard to demonstrate this:
EXAMPLE 2 In executing the fragment
char c1, c2;
/* ... */
c1 = c1 + c2;
the "integer promotions" require that the abstract machine promote the value of each variable to int size
and then add the two ints and truncate the sum. Provided the addition of two chars can be done without
overflow, or with overflow wrapping silently to produce the correct result, the actual execution need only
produce the same result, possibly omitting the promotions.

C usual arithmetic conversions

I was reading in the C99 standard about the usual arithmetic conversions.
If both operands have the same type, then no further conversion is
needed.
Otherwise, if both operands have signed integer types or both have
unsigned integer types, the operand with the type of lesser integer
conversion rank is converted to the type of the operand with greater
rank.
Otherwise, if the operand that has unsigned integer type has rank
greater or equal to the rank of the type of the other operand, then
the operand with signed integer type is converted to the type of the
operand with unsigned integer type.
Otherwise, if the type of the operand with signed integer type can
represent all of the values of the type of the operand with unsigned
integer type, then the operand with unsigned integer type is converted
to the type of the operand with signed integer type.
Otherwise, both operands are converted to the unsigned integer type
corresponding to the type of the operand with signed integer type.
So let's say I have the following code:
#include <stdio.h>
int main()
{
unsigned int a = 10;
signed int b = -5;
printf("%d\n", a + b); /* 5 */
printf("%u\n", a + b); /* 5 */
return 0;
}
I thought the bolded paragraph applies (since unsigned int and signed int have the same rank. Why isn't b converted to unsigned ? Or perhaps it is converted to unsigned but there is something I don't understand ?
Thank you for your time :-)

Indeed b is converted to unsigned. However what you observed is that b converted to unsigned and then added to 10 gives as value 5.
On x86 32bit this is what happens
b, coverted to unsigned, becomes 4294967291 (i.e. 2**32 - 5)
adding 10 becomes 5 because of wrap-around at 2**32 (2**32 - 5 + 10 = 2**32 + 5 = 5)

0x0000000a plus 0xfffffffb will always be 0x00000005 regardless of whether you are dealing with signed or unsigned types, as long as only 32 bits are used.

Repeating the relevant portion of the code from the question:
unsigned int a = 10;
signed int b = -5;
printf("%d\n", a + b); /* 5 */
printf("%u\n", a + b); /* 5 */
In a + b, b is converted to unsigned int, (yielding UINT_MAX + 1 - 5 by the rule for unsigned-to-signed conversion). The result of adding 10 to this value is 5, by the rules of unsigned arithmetic, and its type is unsigned int. In most cases, the type of a C expression is independent of the context in which it appears. (Note that none of this depends on the representation; conversion and arithmetic are defined purely in terms of numeric values.)
For the second printf call, the result is straightforward: "%u" expects an argument of type unsigned int, and you've given it one. It prints "5\n".
The first printf is a little more complicated. "%d" expects an argument of type int, but you're giving it an argument of type unsigned int. In most cases, a type mismatch like this results in undefined behavior, but there's a special-case rule that corresponding signed and unsigned types are interchangeable as function arguments -- as long as the value is representable in both types (as it is here). So the first printf also prints "5\n".
Again, all this behavior is defined in terms of values, not representations (except for the requirement that a given value has the same representation in corresponding signed and unsigned types). You'd get the same result on a system where signed int and unsigned int are both 37 bits, signed int has 7 padding bits, unsigned int has 11 padding bits, and signed int uses a 1s'-complement or sign-and-magnitude representation. (No such system exists in real life, as far as I know.)

It is converted to unsigned, the unsigned arithmetic just happens to give the result you see.
The result of unsigned arithmetic is equivalent to doing signed arithmetic with two's complement and no out of range exception.

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