clarity of -1L < 1U and -1L > 1UL

clarity of -1L < 1U and -1L > 1UL - c

I have been reading the book The C Programming Language by K&R and came across this.
Conversion rules are more complicated when unsigned operands are involved. The problem
is that comparisons between signed and unsigned values are machine-dependent, because
they depend on the sizes of the various integer types. For example, suppose that int is 16 bits and long is 32 bits. Then -1L < 1U, because 1U, which is an unsigned int, is promoted to a signed long. But -1L > 1UL because -1L is promoted to unsigned long and thus appears
to be a large positive number.
From my beginning of C, i have little difficult regarding this Unsigned and Signed values. Can any one Explain how an unsigned value converted to signed value and viceversa with an explample.

You have touched the surface of usual arithmetic conversions with integral promotions (newer standards use term of integer promotions there).
What are these usual arithmetic conversions ?
In general when you have expression, that involves arithmetic or logical operator and operands don't match, then they need to be unified (as we say promoted) into common form. In C90 rules between integer operands (I am skipping intentionally rules for types "lower" that int here for simplicity) are in following gradation (note that there is no long long int type in C90):
int → unsigned int → long int → unsigned long int
However, there is one exception between unsigned int and long int. If these have the same size (in bits), then both operands of such types are promoted to common type unsigned long int. In any other case common type is that, which is on right side (e.g. when you have int and unsigned int operands, then first one is promoted to unsigned int). If both operands have the same type, then no promotion is made for types listed above.
How does this relate to -1L < 1U and -1L > 1UL ?
In your case it's assumed that sizeof(int) == 16 and sizeof(long) = 32, so for:
-1L < 1U second operand is promoted to long int
-1L > 1UL first operand is promoted to unsigned long
In former example value of expression is 1 as -1L < 1L. In latter case -1L is promoted to unsigned type by repeatedly adding or substracting n+1, when n is the largest value of type unsigned long (in your case n+1 == 2^32), which yields into large number (i.e. 2^32-1), thus value of whole expression is 1 (of type int) as well.

Hum... there is a longer (detailed) answer, and a shorter (approximate) answer...
...starting with the longer answer:
The standard (C99, 6.3.1) specifies an integer conversion rank, and where two integer operands have different rank, one will be converted up. The rank is related to the size of the integer; larger sizes have greater rank than smaller sizes. Where the sizes are the same there may be differences in rank -- but that's not important just now. However, a signed integer and an unsigned integer of the same size have the same rank.
It's worth remembering that signed int and unsigned int have the same size, by definition. They are examples of what the standard calls corresponding integer types (C99, 6.2.5). Each signed integer type has a corresponding unsigned one, and vice versa -- except for _Bool which doesn't have a sign (and, FWIW, has the lowest rank). As above, corresponding signed and unsigned types have the same rank.
C is very fond of int and unsigned int. It will promote integers of lesser rank to int (or to unsigned int, if the promoted integer is unsigned and the same size as int). This is called integer promotion (which is a subset of all the possible conversions). For many operators, C will perform integer promotion on both operands before doing anything else. It will also perform integer promotion on arguments for things like printf().
Now we get to what are known as the usual arithmetic conversions (C99, 6.3.1.8), which for two integer operands are:
both arguments are promoted as required, so will be at least int or unsigned int
if they are now not the same type (type, not rank):
if both types are signed, or both types are unsigned, the lower rank is converted to the higher.
This is straightforward, and there is clearly no loss of signed-ness.
...otherwise, for different types, of different signed-ness, size matters:
if the signed operand has the greater size (and hence rank), the unsigned operand is converted to the type of the signed one.
This is also straightforward, the unsigned value can be represented, without loss, in the signed type.
if the unsigned operand has the greater size (and hence rank), the signed operand is converted to the type of the unsigned one.
So, for (say) unsigned long long and int (which one assumes are different sizes), the int is converted up, by adding ULLONG_MAX + 1 to it.
This is not so straightforward, the signed-ness is lost.
...otherwise, for different types, of different signed-ness, of the same size, the result will be two unsigned operands of their current size. The rank kicks in to decide what type the operands will be:
if the operands have the same rank, then the signed one is converted to the unsigned operand's type.
This is what happens when you have int and unsigned int operands, the int is converted (by adding UINT_MAX + 1 to it).
if the unsigned operand has the greater rank, then the signed operand is converted to the unsigned operand's type (as above).
otherwise, the signed operand has the greater rank, and both operands are converted to the unsigned type corresponding to the signed operand's type (ie to the unsigned type with the greater rank).
This all looks terrifyingly complicated :-( But, keep in mind that the size of an integer is the major component of its rank -- what makes it look complicated dealing with integers of the same size, but different signedness and type.
(For the picky: I have glossed over the wrinkle in the standard which allows for integer representations to include "padding" bits -- which is deeply exotic -- so, for size read width, where required.)
...but the essence of it is, in short:
After promotion to int or unsigned int:
if the operands have different sizes, the smaller is converted up, and if larger is unsigned but the smaller is not, the smaller loses its signed-ness.
if the operands have the same size, but different signed-ness, the signed one is converted to unsigned, losing signed-ness.
The conversion to unsigned (the loss of signed-ness) requires the addition of MAX+1, where MAX is the maximum value of the unsigned type being converted to.
Hence: -1L > 0UL... indeed -1L == ULONG_MAX.

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.)

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]

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.)

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]

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.)

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]

Implicit conversion confusion between signed and unsigned when reading K&R book

I am learning the c language using the K&R book. In the second chapter book, the author talks about implicit conversion. There book says this:
Conversion rules are more complicated when unsigned operands are involved. The problem is that
comparisons between signed and unsigned values are machine-dependent, because they depend on the sizes of the various integer types. For example, suppose that int is 16 bits and long is 32 bits. Then -1L < 1U, because 1U, which is an unsigned int, is promoted to a signed long. But -1L >
1UL because -1L is promoted to unsigned long and thus appears to be a large positive number.
I tried the code below in two different scenarios:
compiled on an x86 64bits platform and executed. Where sizeof(-1L) -> 8byte and sizeof(1U) -> 4 bytes
compiled on an x86 32bits platform and executed. Where sizeof(-1L) -> 4byte and sizeof(1U) -> 4 bytes
The code:
int main() {
if(-1L > 1U)
printf("true");
else
printf("false");
return 0;
}
The results:
x86 64bits: false
x86 32bits: true
so I'm getting two different OP in each case.
As author says, for 2 different data sizes one being 16 and the other 32, it holds good in my x86-64 case.
But im not able to understand why in the second case for 32 bits, I'm getting true.
As author says unsigned int is promoted to signed long int, if this is true then both
should be 4 bytes wide, then why is it printing true instead of false? As now both should be signed long.
As the author says it is machine dependent, then both long and int should have same byte size, so how the implicit conversion is happening here?
My understanding is that -1 is stored as two's complement i.e 0xFFFFFFFF > 0x1 so in the second case it should be true.
But this explanation contradicts the 1st case.
Please correct me if what I think is wrong, as I am new to implicit conversion.
Can anyone please explain this behaviour?

lets explain the rank system first
6.3.1 Arithmetic operand(c99 standard)
A) The rank of a signed integer type shall be greater than the rank of any signed integer
type with less precision(more bytes higher precision higher rank)
B) 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.
C) The rank of any unsigned integer type shall equal the rank of the corresponding signed
integer type, if any.
(in other words if your system unsigned int is 32bits and your int is 32bits then the
ranks of these are the same.)
the above explains the rank.
now coming to arithmetic conversions.
6.3.1.8 Usual arithmetic conversions (c99 standard)
1)If both operands have the same type, then no further conversion is needed.
2)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.(similar to 1)
3)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.
4)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
5)Otherwise, both operands are converted to the unsigned integer type corresponding to the
type of the operand with signed integer type.
2) compiled on an x86 32bits platform and executed. Where sizeof(-1L) -> 4byte and sizeof(1U) -> 4 bytes
in your case look at statement 3 & C. the unsigned value(4bytes) has rank equal to the signed value(4btyes) therefore the singed value is converted to an unsigned value, when this happens the, the sign bit makes this look like a extremely large value. -1L > 1U therefore is true
1) compiled on an x86 64bits platform and executed. Where sizeof(-1L) -> 8byte and sizeof(1U) -> 4 bytes
in this case, the unsigned value rank is less than the rank of the singed value. look at 4).
the signed integer(8bytes) can represent any 4byte unsigned value. therefore the unsigned 4byte value is converted to a signed value.(this will preserve the sign bit, sign bit is 0)
therefore -1L > 1U is false

But im not able to understand why in second case in 32 bit its OP-->true. As author says unsigned int is promoted to signed long int if so then both are 4 byte wide, why its printing true instead of false.? since now both are signed long.
The auther says, that if int and long have different size, then unsigned int is promoted to signed long.
If int and long have the same size, then long is too small to hold all values of unsigned int and therefore both are converted to unsigned long.

For binary arithmetic and relational operators:
If either operand has type long double, the other operand is converted to long double. Otherwise, if either operand has type double, the other operand is converted to double. Otherwise, if either operand has type float, the other operand is converted to float. Otherwise the integral promotions are performed on both operands.
(Integral promotion: A char, a short int, or an int bit-field, or their signed or unsigned varieties, or an enumeration type, may be used in an expression wherever an int or unsigned int may be used. If an int can represent all the values of the original type, the value is converted to an int; otherwise it is converted to an unsigned int.)
Then if either operand has type unsigned long int, the other operand is converted to unsigned long int. Otherwise, if one operand has type long int and the other has type unsigned int, if a long int can represent all values of an unsigned int the operand of type unsigned int is converted to long int; if a long int cannot represent all the values of an unsigned int, both operands are converted to unsigned long int. Otherwise, if either operand has type long int, the other operand is converted to long int. Otherwise, if either operand has type unsigned int, the other operand is converted to unsigned int. Otherwise, both operands have type int.
The sentence in bold explains your second case, where long int has the same width as unsigned int thus cannot hold all values of unsigned int.
(The above description lacks the type unsigned long long int and long long it, but the rules are basically the same.)

As all who answered above are correct, Just to add more clarity and my understanding writing here to get more clarity.
-->if one operand has type long int and the other has type unsigned int,
-->if a long int can represent all values of an unsigned int the operand of type unsigned int is converted to long int;
-->if a long int cannot represent all the values of an unsigned int, both operands are converted to unsigned long int.
So from above one operand has type long int i.e -1L and the other has type unsigned int i.e 1U
suppose sizeof -1L is --->8byte and sizeof 1U is 4 byte
0X0000-0XFFFFF values can be represented using in long int whose sizeof is 8 byte
so in this case long int can represent all values of an unsigned int i.e using 8byte ---> it can represent all the values unsigned int 1U.
so----> here operand of type unsigned int is converted to long int ---> -1L > 1U --> is false
coming 2nd case
if a long int cannot represent all the values of an unsigned int
i.e sizeof -1L -->4byte and sizeof 1U -->4byte
here long int cannot represent all the values i.e using 4 bytes--> it cannot represent all the values of unsigned int 1U. so both operands are converted to unsigned long int
-1L appears to large value since its unsigned now when compared to 1U.
i.e---->0xFFFFFFFF > 0x1 ---> its true

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