I was reading an article on usage of size_t and ptrdiff_t data types here, when I came across this example:
The code:
int A = -2;
unsigned B = 1;
int array[5] = { 1, 2, 3, 4, 5 };
int *ptr = array + 3;
ptr = ptr + (A + B); //Error
printf("%i\n", *ptr);
I am unable to understand a couple of things. First, how can adding a signed and an unsigned number cast the enter result into unsigned type? If the result is indeed 0xFFFFFFFF of unsigned type, why in a 32 bit system, while adding it with ptr, will it be interpreted as ptr-1, given that the number is actually unsigned type and the leading 1 should not signify sign?
Second, why is the result different in 64 bit system?
Can anyone explain this please?
1. I am unable to understand a couple of things. First, how can adding a signed and an unsigned number cast the enter result into unsigned type?
This is defined by integer promotions and integer conversion rank.
6.3.1.8 p1: 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.
In this case unsigned has a higher rank than int, therefore int is promoted to unsigned.
The conversion of int ( -2 ) to unsigned is performed as described:
6.3.1.3 p2: Otherwise, if the new type is unsigned, the value is converted by repeatedly adding or
subtracting one more than the maximum value that can be represented in the new type
until the value is in the range of the new type
2. If the result is indeed 0xFFFFFFFF of unsigned type, why in a 32 bit system, while adding it with ptr, will it be interpreted as ptr-1, given that the number is actually unsigned type and the leading 1 should not signify sign?
This is undefined behavior and should not be relied on, since C doesn't define pointer arithmetic overflow.
6.5.6 p8: If both the pointer
operand and the result point to elements of the same array object, or one past the last
element of the array object, the evaluation shall not produce an overflow; otherwise, the
behavior is undefined.
3. Second, why is the result different in 64 bit system?
( This assumes( as does the picture ) that int and unsigned are 4 bytes. )
The result of A and B is the same as described in 1., then that result is added to the pointer. Since the pointer is 8 bytes and assuming the addition doesn't overflow( it still could if ptr had a large address, giving the same undefined behavior as in 2. ) the result is an address.
This is undefined behavior because the pointer points way outside of the bounds of the array.
The operands of the expression A + B are subject to usual arithmetic conversion, covered in C11 (n1570) 6.3.1.8 p1:
[...]
Otherwise, the integer promotions [which leave int and unsigned int unchanged] are performed on both operands. Then the following rules are applied to the promoted operands:
If both operands have the same type, [...]
Otherwise, if both operands have signed integer types or both have unsigned integer types, [...]
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.
[...]
The types int and unsigned int have the same rank (ibid. 6.3.1.1 p1, 4th bullet); the result of the addition has type unsigned int.
On 32-bit systems, int and pointers usually have the same size (32 bit). From a hardware-centric point of view (and assuming 2's complement), subtracting 1 and adding -1u is the same (addition for signed and unsigned types is the same!), so the access to the array element appears to work.
However, this is undefined behaviour, as array doesn't contain a 0x100000003rd element.
On 64-bit, int usually has still 32 bit, but pointers have 64 bit. Thus, there is no wraparound and no equivalence to subtracting 1 (from a hardware-centric point of view, the behaviour is undefined in both cases).
To illustrate, say ptr is 0xabcd0123, adding 0xffffffff yields
abcd0123
+ ffffffff
1abcd0122
^-- The 1 is truncated for a 32-bit calculation, but not for 64-bit.
On most 64bits, int are 32bits, but on 32bits systems, pointers are also 32bits.
Recall that in 32bits arithmetic - on two's complement based hardware, adding 0xFFFFFFFF is almost the same as subtracting 1: it overflows and you get that same number minus 1 (it's the same phenomenon when you add 9 to a number between 0 and 9, you get than number minus 1 and a carry). On that type of hardware, the encoding of -1 is actually that same value 0xFFFFFFFF, only the operation is different (signed add versus unsigned add), and so a carry will be produced in the unsigned case.
On 64bits pointers are... 64bits. Adding a 32bits value to a 64bits one requires to extend that 32bit value to 64. unsigned values are zero extended (ie. the missing bits are just filled with zeros), while signed values are sign extended (ie. missing bits are filled with the sign bit value).
In this case, adding an unsigned value (which will therefore not be sign extended) will not overflow, thus yielding a very different value from the original.
Related
Suppose I have the following C code.
unsigned int u = 1234;
int i = -5678;
unsigned int result = u + i;
What implicit conversions are going on here, and is this code safe for all values of u and i? (Safe, in the sense that even though result in this example will overflow to some huge positive number, I could cast it back to an int and get the real result.)
Short Answer
Your i will be converted to an unsigned integer by adding UINT_MAX + 1, then the addition will be carried out with the unsigned values, resulting in a large result (depending on the values of u and i).
Long Answer
According to the C99 Standard:
6.3.1.8 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.
In your case, we have one unsigned int (u) and signed int (i). Referring to (3) above, since both operands have the same rank, your i will need to be converted to an unsigned integer.
6.3.1.3 Signed and unsigned integers
When a value with integer type is converted to another integer type other than _Bool, if the value can be represented by the new type, it is unchanged.
Otherwise, if the new type is unsigned, the value is converted by repeatedly adding or subtracting one more than the maximum value that can be represented in the new type until the value is in the range of the new type.
Otherwise, the new type is signed and the value cannot be represented in it; either the result is implementation-defined or an implementation-defined signal is raised.
Now we need to refer to (2) above. Your i will be converted to an unsigned value by adding UINT_MAX + 1. So the result will depend on how UINT_MAX is defined on your implementation. It will be large, but it will not overflow, because:
6.2.5 (9)
A computation involving unsigned operands can never overflow, because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the largest value that can be represented by the resulting type.
Bonus: Arithmetic Conversion Semi-WTF
#include <stdio.h>
int main(void)
{
unsigned int plus_one = 1;
int minus_one = -1;
if(plus_one < minus_one)
printf("1 < -1");
else
printf("boring");
return 0;
}
You can use this link to try this online: https://repl.it/repls/QuickWhimsicalBytes
Bonus: Arithmetic Conversion Side Effect
Arithmetic conversion rules can be used to get the value of UINT_MAX by initializing an unsigned value to -1, ie:
unsigned int umax = -1; // umax set to UINT_MAX
This is guaranteed to be portable regardless of the signed number representation of the system because of the conversion rules described above. See this SO question for more information: Is it safe to use -1 to set all bits to true?
Conversion from signed to unsigned does not necessarily just copy or reinterpret the representation of the signed value. Quoting the C standard (C99 6.3.1.3):
When a value with integer type is converted to another integer type other than _Bool, if
the value can be represented by the new type, it is unchanged.
Otherwise, if the new type is unsigned, the value is converted by repeatedly adding or
subtracting one more than the maximum value that can be represented in the new type
until the value is in the range of the new type.
Otherwise, the new type is signed and the value cannot be represented in it; either the
result is implementation-defined or an implementation-defined signal is raised.
For the two's complement representation that's nearly universal these days, the rules do correspond to reinterpreting the bits. But for other representations (sign-and-magnitude or ones' complement), the C implementation must still arrange for the same result, which means that the conversion can't just copy the bits. For example, (unsigned)-1 == UINT_MAX, regardless of the representation.
In general, conversions in C are defined to operate on values, not on representations.
To answer the original question:
unsigned int u = 1234;
int i = -5678;
unsigned int result = u + i;
The value of i is converted to unsigned int, yielding UINT_MAX + 1 - 5678. This value is then added to the unsigned value 1234, yielding UINT_MAX + 1 - 4444.
(Unlike unsigned overflow, signed overflow invokes undefined behavior. Wraparound is common, but is not guaranteed by the C standard -- and compiler optimizations can wreak havoc on code that makes unwarranted assumptions.)
Referring to The C Programming Language, Second Edition (ISBN 0131103628),
Your addition operation causes the int to be converted to an unsigned int.
Assuming two's complement representation and equally sized types, the bit pattern does not change.
Conversion from unsigned int to signed int is implementation dependent. (But it probably works the way you expect on most platforms these days.)
The rules are a little more complicated in the case of combining signed and unsigned of differing sizes.
When converting from signed to unsigned there are two possibilities. Numbers that were originally positive remain (or are interpreted as) the same value. Number that were originally negative will now be interpreted as larger positive numbers.
When one unsigned and one signed variable are added (or any binary operation) both are implicitly converted to unsigned, which would in this case result in a huge result.
So it is safe in the sense of that the result might be huge and wrong, but it will never crash.
As was previously answered, you can cast back and forth between signed and unsigned without a problem. The border case for signed integers is -1 (0xFFFFFFFF). Try adding and subtracting from that and you'll find that you can cast back and have it be correct.
However, if you are going to be casting back and forth, I would strongly advise naming your variables such that it is clear what type they are, eg:
int iValue, iResult;
unsigned int uValue, uResult;
It is far too easy to get distracted by more important issues and forget which variable is what type if they are named without a hint. You don't want to cast to an unsigned and then use that as an array index.
What implicit conversions are going on here,
i will be converted to an unsigned integer.
and is this code safe for all values of u and i?
Safe in the sense of being well-defined yes (see https://stackoverflow.com/a/50632/5083516 ).
The rules are written in typically hard to read standards-speak but essentially whatever representation was used in the signed integer the unsigned integer will contain a 2's complement representation of the number.
Addition, subtraction and multiplication will work correctly on these numbers resulting in another unsigned integer containing a twos complement number representing the "real result".
division and casting to larger unsigned integer types will have well-defined results but those results will not be 2's complement representations of the "real result".
(Safe, in the sense that even though result in this example will overflow to some huge positive number, I could cast it back to an int and get the real result.)
While conversions from signed to unsigned are defined by the standard the reverse is implementation-defined both gcc and msvc define the conversion such that you will get the "real result" when converting a 2's complement number stored in an unsigned integer back to a signed integer. I expect you will only find any other behaviour on obscure systems that don't use 2's complement for signed integers.
https://gcc.gnu.org/onlinedocs/gcc/Integers-implementation.html#Integers-implementation
https://msdn.microsoft.com/en-us/library/0eex498h.aspx
Horrible Answers Galore
Ozgur Ozcitak
When you cast from signed to unsigned
(and vice versa) the internal
representation of the number does not
change. What changes is how the
compiler interprets the sign bit.
This is completely wrong.
Mats Fredriksson
When one unsigned and one signed
variable are added (or any binary
operation) both are implicitly
converted to unsigned, which would in
this case result in a huge result.
This is also wrong. Unsigned ints may be promoted to ints should they have equal precision due to padding bits in the unsigned type.
smh
Your addition operation causes the int
to be converted to an unsigned int.
Wrong. Maybe it does and maybe it doesn't.
Conversion from unsigned int to signed
int is implementation dependent. (But
it probably works the way you expect
on most platforms these days.)
Wrong. It is either undefined behavior if it causes overflow or the value is preserved.
Anonymous
The value of i is converted to
unsigned int ...
Wrong. Depends on the precision of an int relative to an unsigned int.
Taylor Price
As was previously answered, you can
cast back and forth between signed and
unsigned without a problem.
Wrong. Trying to store a value outside the range of a signed integer results in undefined behavior.
Now I can finally answer the question.
Should the precision of int be equal to unsigned int, u will be promoted to a signed int and you will get the value -4444 from the expression (u+i). Now, should u and i have other values, you may get overflow and undefined behavior but with those exact numbers you will get -4444 [1]. This value will have type int. But you are trying to store that value into an unsigned int so that will then be cast to an unsigned int and the value that result will end up having would be (UINT_MAX+1) - 4444.
Should the precision of unsigned int be greater than that of an int, the signed int will be promoted to an unsigned int yielding the value (UINT_MAX+1) - 5678 which will be added to the other unsigned int 1234. Should u and i have other values, which make the expression fall outside the range {0..UINT_MAX} the value (UINT_MAX+1) will either be added or subtracted until the result DOES fall inside the range {0..UINT_MAX) and no undefined behavior will occur.
What is precision?
Integers have padding bits, sign bits, and value bits. Unsigned integers do not have a sign bit obviously. Unsigned char is further guaranteed to not have padding bits. The number of values bits an integer has is how much precision it has.
[Gotchas]
The macro sizeof macro alone cannot be used to determine precision of an integer if padding bits are present. And the size of a byte does not have to be an octet (eight bits) as defined by C99.
[1] The overflow may occur at one of two points. Either before the addition (during promotion) - when you have an unsigned int which is too large to fit inside an int. The overflow may also occur after the addition even if the unsigned int was within the range of an int, after the addition the result may still overflow.
Consider following example:
#include <stdio.h>
int main(void)
{
unsigned char a = 15; /* one byte */
unsigned short b = 15; /* two bytes */
unsigned int c = 15; /* four bytes */
long x = -a; /* eight bytes */
printf("%ld\n", x);
x = -b;
printf("%ld\n", x);
x = -c;
printf("%ld\n", x);
return 0;
}
To compile I am using GCC 4.4.7 (and it gave me no warnings):
gcc -g -std=c99 -pedantic-errors -Wall -W check.c
My result is:
-15
-15
4294967281
The question is why both unsigned char and unsigned short values are "propagated" correctly to (signed) long, while unsigned int is not ? Is there any reference or rule on this ?
Here are results from gdb (words are in little-endian order) accordingly:
(gdb) x/2w &x
0x7fffffffe168: 11111111111111111111111111110001 11111111111111111111111111111111
(gdb) x/2w &x
0x7fffffffe168: 11111111111111111111111111110001 00000000000000000000000000000000
This is due to how the integer promotions applied to the operand and the requirement that the result of unary minus have the same type. This is covered in section 6.5.3.3 Unary arithmetic operators and says (emphasis mine going forward):
The result of the unary - operator is the negative of its (promoted) operand. The integer promotions are performed on the operand, and the result has the promoted type.
and integer promotion which is covered in the draft c99 standard section 6.3 Conversions and says:
if an int can represent all values of the original type, the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions.48) All other types are unchanged by the integer promotions.
In the first two cases, the promotion will be to int and the result will be int. In the case of unsigned int no promotion is required but the result will require a conversion back to unsigned int.
The -15 is converted to unsigned int using the rules set out in section 6.3.1.3 Signed and unsigned integers which says:
Otherwise, if the new type is unsigned, the value is converted by repeatedly adding or subtracting one more than the maximum value that can be represented in the new type until the value is in the range of the new type.49)
So we end up with -15 + (UMAX + 1) which results in UMAX - 14 which results in a large unsigned value. This is sometimes why you will see code use -1 converted to to an unsigned value to obtain the max unsigned value of a type since it will always end up being -1 + UMAX + 1 which is UMAX.
int is special. Everything smaller than int gets promoted to int in arithmetic operations.
Thus -a and -b are applications of unary minus to int values of 15, which just work and produce -15. This value is then converted to long.
-c is different. c is not promoted to an int as it is not smaller than int. The result of unary minus applied to an unsigned int value of k is again an unsigned int, computed as 2N-k (N is the number of bits).
Now this unsigned int value is converted to long normally.
This behavior is correct. Quotes are from C 9899:TC2.
6.5.3.3/3:
The result of the unary - operator is the negative of its (promoted) operand. The integer promotions are performed on the operand, and the result has the promoted type.
6.2.5/9:
A computation involving unsigned operands can never overflow, because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the largest value that can be represented by the resulting type.
6.3.1.1/2:
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 whose integer conversion rank is less than or equal to the rank of int and unsigned int.
A bit-field of type _Bool, int, signed int, or unsigned int.
If an int can represent all values of the original type, 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.
So for long x = -a;, since the operand a, an unsigned char, has conversion rank less than the rank of int and unsigned int, and all unsigned char values can be represented as int (on your platform), we first promote to type int. The negative of that is simple: the int with value -15.
Same logic for unsigned short (on your platform).
The unsigned int c is not changed by promotion. So the value of -c is calculated using modular arithmetic, giving the result UINT_MAX-14.
C's integer promotion rules are what they are because standards-writers wanted to allow a wide variety of existing implementations that did different things, in some cases because they were created before there were "standards", to keep on doing what they were doing, while defining rules for new implementations that were more specific than "do whatever you feel like". Unfortunately, the rules as written make it extremely difficult to write code which doesn't depend upon a compiler's integer size. Even if future processors would be able to perform 64-bit operations faster than 32-bit ones, the rules dictated by the standards would cause a lot of code to break if int ever grew beyond 32 bits.
It would probably in retrospect have been better to have handled "weird" compilers by explicitly recognizing the existence of multiple dialects of C, and recommending that compilers implement a dialect that handles various things in consistent ways, but providing that they may also implement dialects which do them differently. Such an approach may end up ultimately being the only way that int can grow beyond 32 bits, but I've not heard of anyone even considering such a thing.
I think the root of the problem with unsigned integer types stems from the fact that they are sometimes used to represent numerical quantities, and are sometimes used to represent members of a wrapping abstract algebraic ring. Unsigned types behave in a manner consistent with an abstract algebraic ring in circumstances which do not involve type promotion. Applying a unary minus to a member of a ring should (and does) yield a member of that same ring which, when added to the original, will yield zero [i.e. the additive inverse]. There is exactly one way to map integer quantities to ring elements, but multiple ways exist to map ring elements back to integer quantities. Thus, adding a ring element to an integer quantity should yield an element of the same ring regardless of the size of the integer, and conversion from rings to integer quantities should require that code specify how the conversion should be performed. Unfortunately, C implicitly converts rings to integers in cases where either the size of the ring is smaller than the default integer type, or when an operation uses a ring member with an integer of a larger type.
The proper solution to solve this problem would be to allow code to specify that certain variables, return values, etc. should be regarded as ring types rather than numbers; an expression like -(ring16_t)2 should yield 65534 regardless of the size of int, rather than yielding 65534 on systems where int is 16 bits, and -2 on systems where it's larger. Likewise, (ring32)0xC0000001 * (ring32)0xC0000001 should yield (ring32)0x80000001 even if int happens to be 64 bits [note that if int is 64 bits, the compiler could legally do anything it likes if code tries to multiply two unsigned 32-bit values which equal 0xC0000001, since the result would be too large to represent in a 64-bit signed integer.
Negatives are tricky. Especially when it comes to unsigned values. If you look at the c-documentation, you'll notice that (contrary to what you'd expect) unsigned chars and shorts are promoted to signed ints for computing, while an unsigned int will be computed as an unsigned int.
When you compute the -c, the c is treated as an int, it becomes -15, then is stored in x, (which still believes it is an UNSIGNED int) and is stored as such.
For clarification - No ACTUAL promotion is done when "negativeing" an unsigned. When you assign a negative to any type of int (or take a negative) the 2's compliment of the number is instead used. Since the only practical difference between unsigned and signed values is that the MSB acts as a sign flag, it is taken as a very large positive number instead of a negative one.
I'm a little confused about how the C language treats different sized ints when you do basic arithmetic and bitwise operation. what would happen if in this case:
int8_t even = 0xBB;
int16_t twice = even << 8;
twice = twice + even;
What If i went the other way and attempted to add a 16 bit int to an 8 bit? is the normal int declaration dynamic? Why would i want to designate a size? What happens when I add to 8 bit ints that are both 0xFF?
As I mentioned in my comment, everything gets promoted to at least an int during arithmetic.
Here's an example program to demonstrate that:
main(){
struct {
char x : 1;
}t;
t.x = 1;
int i = t.x << (sizeof i * 8 - 1);
printf("i = %x\n",i);
}
t.x is only one bit, but in this operation it is promoted all the way to an integer to give the output:
i = 80000000
On the other hand, if we add the line
long long j = t.x << (sizeof j * 8 - 1)
gcc gave me the warning:
warning: left shift count >= width of type [enabled by default]
What If i went the other way and attempted to add a 16 bit int to an 8 bit?
All the arithmetic would be done at integer precision (probably 32 bits) and then the bottom 8 bits of the result would be stored in the 8 bit number.
Is the normal int declaration dynamic?
No. Its a fixed width on an implementation (probably 32 bits on yours).
Why would i want to designate a size?
Maybe you have space constraints. Maybe your algorithm was made to work with a specific size integer. Maybe you really want to work on some modulo field provided by uint16_t (Z_65536).
What happens when I add to 8 bit ints that are both 0xFF?
The promotion doesn't matter here, the 8 bit result will be 0xFE. If you were to store the result in a 16 bit number, then the result will be 0x1FE (Unless int's in your implementation are only 8 bits. This is highly unlikely except for some esoteric embedded applications).
Edit
I wrote about the unsigned convention here, because you representing the number as 0xFF seemed to refer to unsigned numbers (in K&R C 0xFF is actually an unsigned literal). If you were actually referring to the signed 8-bit value 0xFF that is equivalent to -1, and your problem becomes sort of trivial. No matter how big the integer, it should always be able to represent -1, as well as -1 + (-1) = -2 (you only actually need two bits to represent these numbers).
Normal "int" is not dynamic, but is not common across compilers; it's the size that is most convenient for the CPU you're compiling for. You would want to designate a size if you have some reason (network protocol, file format, etc) why you actually care about the representation.
Answers to your exercises below the answer to your question:
Your answer can be found in the C11 final draft spec (was ratified in October 2011, but the "final" document is not free): http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1570.pdf
The first relevant section is 6.3.1.1 Conversions : Arithmetic Operands : Boolean, characters and integers, beginning on page 50
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.
The integer promotions preserve value including sign. As discussed earlier, whether a "plain" char is treated as signed is implementation-defined.
Essentially, the integer types are all promoted to at least int, and possibly to larger sizes than that if any of the operands had a rank greater than that. They generally become signed, but remain unsigned if either operand's type has greater range than int.
There's a bunch of stuff specifically defining how the conversions are performed, and codifying the rules of which types are "higher rank" (bigger) than others, and then we reach the real meat in 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:
... some stuff about floats ...
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.
int8_t even = 0xBB;
0xBB formed as a constant of type int (0x000000BB), then truncated down to int8_t
int16_t twice = even << 8;
the value of even is sign extended from int8_t to int, becoming 0xFFFFFFBB (if I assume int is 32-bits on this machine)
The value of 8 is formed as type int
These types match, so no further conversions occur
The operator << is performed (now that types match), yielding 0xFFFFBB00
This value is truncated to fit in int16_t, yielding 0xBB00
twice = twice + even;
The value of twice is sign extended from int16_t to int, yielding 0xFFFFBB00 (-17664)
The value of even is sign extended from int8_t to int, yielding 0xFFFFFFBB (-69)
These values are added, yielding 0xFFFFBABB (-17733)
This value is then truncated to fit in int16_t, yielding 0xBABB
What happens when I add to 8 bit ints that are both 0xFF?
They are both sign extended to int yielding 0xFFFFFFFF, then added yielding 0xFFFFFFFE. This may or may not be truncated back down depending on where you store it.
The compiler is of course free to notice that much of this extending/truncating stuff is worthless busywork because the output bits are not kept, and optimize it out. But the result is required to be "as if" if had done things the long way.
Suppose I have the following C code.
unsigned int u = 1234;
int i = -5678;
unsigned int result = u + i;
What implicit conversions are going on here, and is this code safe for all values of u and i? (Safe, in the sense that even though result in this example will overflow to some huge positive number, I could cast it back to an int and get the real result.)
Short Answer
Your i will be converted to an unsigned integer by adding UINT_MAX + 1, then the addition will be carried out with the unsigned values, resulting in a large result (depending on the values of u and i).
Long Answer
According to the C99 Standard:
6.3.1.8 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.
In your case, we have one unsigned int (u) and signed int (i). Referring to (3) above, since both operands have the same rank, your i will need to be converted to an unsigned integer.
6.3.1.3 Signed and unsigned integers
When a value with integer type is converted to another integer type other than _Bool, if the value can be represented by the new type, it is unchanged.
Otherwise, if the new type is unsigned, the value is converted by repeatedly adding or subtracting one more than the maximum value that can be represented in the new type until the value is in the range of the new type.
Otherwise, the new type is signed and the value cannot be represented in it; either the result is implementation-defined or an implementation-defined signal is raised.
Now we need to refer to (2) above. Your i will be converted to an unsigned value by adding UINT_MAX + 1. So the result will depend on how UINT_MAX is defined on your implementation. It will be large, but it will not overflow, because:
6.2.5 (9)
A computation involving unsigned operands can never overflow, because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the largest value that can be represented by the resulting type.
Bonus: Arithmetic Conversion Semi-WTF
#include <stdio.h>
int main(void)
{
unsigned int plus_one = 1;
int minus_one = -1;
if(plus_one < minus_one)
printf("1 < -1");
else
printf("boring");
return 0;
}
You can use this link to try this online: https://repl.it/repls/QuickWhimsicalBytes
Bonus: Arithmetic Conversion Side Effect
Arithmetic conversion rules can be used to get the value of UINT_MAX by initializing an unsigned value to -1, ie:
unsigned int umax = -1; // umax set to UINT_MAX
This is guaranteed to be portable regardless of the signed number representation of the system because of the conversion rules described above. See this SO question for more information: Is it safe to use -1 to set all bits to true?
Conversion from signed to unsigned does not necessarily just copy or reinterpret the representation of the signed value. Quoting the C standard (C99 6.3.1.3):
When a value with integer type is converted to another integer type other than _Bool, if
the value can be represented by the new type, it is unchanged.
Otherwise, if the new type is unsigned, the value is converted by repeatedly adding or
subtracting one more than the maximum value that can be represented in the new type
until the value is in the range of the new type.
Otherwise, the new type is signed and the value cannot be represented in it; either the
result is implementation-defined or an implementation-defined signal is raised.
For the two's complement representation that's nearly universal these days, the rules do correspond to reinterpreting the bits. But for other representations (sign-and-magnitude or ones' complement), the C implementation must still arrange for the same result, which means that the conversion can't just copy the bits. For example, (unsigned)-1 == UINT_MAX, regardless of the representation.
In general, conversions in C are defined to operate on values, not on representations.
To answer the original question:
unsigned int u = 1234;
int i = -5678;
unsigned int result = u + i;
The value of i is converted to unsigned int, yielding UINT_MAX + 1 - 5678. This value is then added to the unsigned value 1234, yielding UINT_MAX + 1 - 4444.
(Unlike unsigned overflow, signed overflow invokes undefined behavior. Wraparound is common, but is not guaranteed by the C standard -- and compiler optimizations can wreak havoc on code that makes unwarranted assumptions.)
Referring to The C Programming Language, Second Edition (ISBN 0131103628),
Your addition operation causes the int to be converted to an unsigned int.
Assuming two's complement representation and equally sized types, the bit pattern does not change.
Conversion from unsigned int to signed int is implementation dependent. (But it probably works the way you expect on most platforms these days.)
The rules are a little more complicated in the case of combining signed and unsigned of differing sizes.
When converting from signed to unsigned there are two possibilities. Numbers that were originally positive remain (or are interpreted as) the same value. Number that were originally negative will now be interpreted as larger positive numbers.
When one unsigned and one signed variable are added (or any binary operation) both are implicitly converted to unsigned, which would in this case result in a huge result.
So it is safe in the sense of that the result might be huge and wrong, but it will never crash.
As was previously answered, you can cast back and forth between signed and unsigned without a problem. The border case for signed integers is -1 (0xFFFFFFFF). Try adding and subtracting from that and you'll find that you can cast back and have it be correct.
However, if you are going to be casting back and forth, I would strongly advise naming your variables such that it is clear what type they are, eg:
int iValue, iResult;
unsigned int uValue, uResult;
It is far too easy to get distracted by more important issues and forget which variable is what type if they are named without a hint. You don't want to cast to an unsigned and then use that as an array index.
What implicit conversions are going on here,
i will be converted to an unsigned integer.
and is this code safe for all values of u and i?
Safe in the sense of being well-defined yes (see https://stackoverflow.com/a/50632/5083516 ).
The rules are written in typically hard to read standards-speak but essentially whatever representation was used in the signed integer the unsigned integer will contain a 2's complement representation of the number.
Addition, subtraction and multiplication will work correctly on these numbers resulting in another unsigned integer containing a twos complement number representing the "real result".
division and casting to larger unsigned integer types will have well-defined results but those results will not be 2's complement representations of the "real result".
(Safe, in the sense that even though result in this example will overflow to some huge positive number, I could cast it back to an int and get the real result.)
While conversions from signed to unsigned are defined by the standard the reverse is implementation-defined both gcc and msvc define the conversion such that you will get the "real result" when converting a 2's complement number stored in an unsigned integer back to a signed integer. I expect you will only find any other behaviour on obscure systems that don't use 2's complement for signed integers.
https://gcc.gnu.org/onlinedocs/gcc/Integers-implementation.html#Integers-implementation
https://msdn.microsoft.com/en-us/library/0eex498h.aspx
Horrible Answers Galore
Ozgur Ozcitak
When you cast from signed to unsigned
(and vice versa) the internal
representation of the number does not
change. What changes is how the
compiler interprets the sign bit.
This is completely wrong.
Mats Fredriksson
When one unsigned and one signed
variable are added (or any binary
operation) both are implicitly
converted to unsigned, which would in
this case result in a huge result.
This is also wrong. Unsigned ints may be promoted to ints should they have equal precision due to padding bits in the unsigned type.
smh
Your addition operation causes the int
to be converted to an unsigned int.
Wrong. Maybe it does and maybe it doesn't.
Conversion from unsigned int to signed
int is implementation dependent. (But
it probably works the way you expect
on most platforms these days.)
Wrong. It is either undefined behavior if it causes overflow or the value is preserved.
Anonymous
The value of i is converted to
unsigned int ...
Wrong. Depends on the precision of an int relative to an unsigned int.
Taylor Price
As was previously answered, you can
cast back and forth between signed and
unsigned without a problem.
Wrong. Trying to store a value outside the range of a signed integer results in undefined behavior.
Now I can finally answer the question.
Should the precision of int be equal to unsigned int, u will be promoted to a signed int and you will get the value -4444 from the expression (u+i). Now, should u and i have other values, you may get overflow and undefined behavior but with those exact numbers you will get -4444 [1]. This value will have type int. But you are trying to store that value into an unsigned int so that will then be cast to an unsigned int and the value that result will end up having would be (UINT_MAX+1) - 4444.
Should the precision of unsigned int be greater than that of an int, the signed int will be promoted to an unsigned int yielding the value (UINT_MAX+1) - 5678 which will be added to the other unsigned int 1234. Should u and i have other values, which make the expression fall outside the range {0..UINT_MAX} the value (UINT_MAX+1) will either be added or subtracted until the result DOES fall inside the range {0..UINT_MAX) and no undefined behavior will occur.
What is precision?
Integers have padding bits, sign bits, and value bits. Unsigned integers do not have a sign bit obviously. Unsigned char is further guaranteed to not have padding bits. The number of values bits an integer has is how much precision it has.
[Gotchas]
The macro sizeof macro alone cannot be used to determine precision of an integer if padding bits are present. And the size of a byte does not have to be an octet (eight bits) as defined by C99.
[1] The overflow may occur at one of two points. Either before the addition (during promotion) - when you have an unsigned int which is too large to fit inside an int. The overflow may also occur after the addition even if the unsigned int was within the range of an int, after the addition the result may still overflow.
Suppose I have the following C code.
unsigned int u = 1234;
int i = -5678;
unsigned int result = u + i;
What implicit conversions are going on here, and is this code safe for all values of u and i? (Safe, in the sense that even though result in this example will overflow to some huge positive number, I could cast it back to an int and get the real result.)
Short Answer
Your i will be converted to an unsigned integer by adding UINT_MAX + 1, then the addition will be carried out with the unsigned values, resulting in a large result (depending on the values of u and i).
Long Answer
According to the C99 Standard:
6.3.1.8 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.
In your case, we have one unsigned int (u) and signed int (i). Referring to (3) above, since both operands have the same rank, your i will need to be converted to an unsigned integer.
6.3.1.3 Signed and unsigned integers
When a value with integer type is converted to another integer type other than _Bool, if the value can be represented by the new type, it is unchanged.
Otherwise, if the new type is unsigned, the value is converted by repeatedly adding or subtracting one more than the maximum value that can be represented in the new type until the value is in the range of the new type.
Otherwise, the new type is signed and the value cannot be represented in it; either the result is implementation-defined or an implementation-defined signal is raised.
Now we need to refer to (2) above. Your i will be converted to an unsigned value by adding UINT_MAX + 1. So the result will depend on how UINT_MAX is defined on your implementation. It will be large, but it will not overflow, because:
6.2.5 (9)
A computation involving unsigned operands can never overflow, because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the largest value that can be represented by the resulting type.
Bonus: Arithmetic Conversion Semi-WTF
#include <stdio.h>
int main(void)
{
unsigned int plus_one = 1;
int minus_one = -1;
if(plus_one < minus_one)
printf("1 < -1");
else
printf("boring");
return 0;
}
You can use this link to try this online: https://repl.it/repls/QuickWhimsicalBytes
Bonus: Arithmetic Conversion Side Effect
Arithmetic conversion rules can be used to get the value of UINT_MAX by initializing an unsigned value to -1, ie:
unsigned int umax = -1; // umax set to UINT_MAX
This is guaranteed to be portable regardless of the signed number representation of the system because of the conversion rules described above. See this SO question for more information: Is it safe to use -1 to set all bits to true?
Conversion from signed to unsigned does not necessarily just copy or reinterpret the representation of the signed value. Quoting the C standard (C99 6.3.1.3):
When a value with integer type is converted to another integer type other than _Bool, if
the value can be represented by the new type, it is unchanged.
Otherwise, if the new type is unsigned, the value is converted by repeatedly adding or
subtracting one more than the maximum value that can be represented in the new type
until the value is in the range of the new type.
Otherwise, the new type is signed and the value cannot be represented in it; either the
result is implementation-defined or an implementation-defined signal is raised.
For the two's complement representation that's nearly universal these days, the rules do correspond to reinterpreting the bits. But for other representations (sign-and-magnitude or ones' complement), the C implementation must still arrange for the same result, which means that the conversion can't just copy the bits. For example, (unsigned)-1 == UINT_MAX, regardless of the representation.
In general, conversions in C are defined to operate on values, not on representations.
To answer the original question:
unsigned int u = 1234;
int i = -5678;
unsigned int result = u + i;
The value of i is converted to unsigned int, yielding UINT_MAX + 1 - 5678. This value is then added to the unsigned value 1234, yielding UINT_MAX + 1 - 4444.
(Unlike unsigned overflow, signed overflow invokes undefined behavior. Wraparound is common, but is not guaranteed by the C standard -- and compiler optimizations can wreak havoc on code that makes unwarranted assumptions.)
Referring to The C Programming Language, Second Edition (ISBN 0131103628),
Your addition operation causes the int to be converted to an unsigned int.
Assuming two's complement representation and equally sized types, the bit pattern does not change.
Conversion from unsigned int to signed int is implementation dependent. (But it probably works the way you expect on most platforms these days.)
The rules are a little more complicated in the case of combining signed and unsigned of differing sizes.
When converting from signed to unsigned there are two possibilities. Numbers that were originally positive remain (or are interpreted as) the same value. Number that were originally negative will now be interpreted as larger positive numbers.
When one unsigned and one signed variable are added (or any binary operation) both are implicitly converted to unsigned, which would in this case result in a huge result.
So it is safe in the sense of that the result might be huge and wrong, but it will never crash.
As was previously answered, you can cast back and forth between signed and unsigned without a problem. The border case for signed integers is -1 (0xFFFFFFFF). Try adding and subtracting from that and you'll find that you can cast back and have it be correct.
However, if you are going to be casting back and forth, I would strongly advise naming your variables such that it is clear what type they are, eg:
int iValue, iResult;
unsigned int uValue, uResult;
It is far too easy to get distracted by more important issues and forget which variable is what type if they are named without a hint. You don't want to cast to an unsigned and then use that as an array index.
What implicit conversions are going on here,
i will be converted to an unsigned integer.
and is this code safe for all values of u and i?
Safe in the sense of being well-defined yes (see https://stackoverflow.com/a/50632/5083516 ).
The rules are written in typically hard to read standards-speak but essentially whatever representation was used in the signed integer the unsigned integer will contain a 2's complement representation of the number.
Addition, subtraction and multiplication will work correctly on these numbers resulting in another unsigned integer containing a twos complement number representing the "real result".
division and casting to larger unsigned integer types will have well-defined results but those results will not be 2's complement representations of the "real result".
(Safe, in the sense that even though result in this example will overflow to some huge positive number, I could cast it back to an int and get the real result.)
While conversions from signed to unsigned are defined by the standard the reverse is implementation-defined both gcc and msvc define the conversion such that you will get the "real result" when converting a 2's complement number stored in an unsigned integer back to a signed integer. I expect you will only find any other behaviour on obscure systems that don't use 2's complement for signed integers.
https://gcc.gnu.org/onlinedocs/gcc/Integers-implementation.html#Integers-implementation
https://msdn.microsoft.com/en-us/library/0eex498h.aspx
Horrible Answers Galore
Ozgur Ozcitak
When you cast from signed to unsigned
(and vice versa) the internal
representation of the number does not
change. What changes is how the
compiler interprets the sign bit.
This is completely wrong.
Mats Fredriksson
When one unsigned and one signed
variable are added (or any binary
operation) both are implicitly
converted to unsigned, which would in
this case result in a huge result.
This is also wrong. Unsigned ints may be promoted to ints should they have equal precision due to padding bits in the unsigned type.
smh
Your addition operation causes the int
to be converted to an unsigned int.
Wrong. Maybe it does and maybe it doesn't.
Conversion from unsigned int to signed
int is implementation dependent. (But
it probably works the way you expect
on most platforms these days.)
Wrong. It is either undefined behavior if it causes overflow or the value is preserved.
Anonymous
The value of i is converted to
unsigned int ...
Wrong. Depends on the precision of an int relative to an unsigned int.
Taylor Price
As was previously answered, you can
cast back and forth between signed and
unsigned without a problem.
Wrong. Trying to store a value outside the range of a signed integer results in undefined behavior.
Now I can finally answer the question.
Should the precision of int be equal to unsigned int, u will be promoted to a signed int and you will get the value -4444 from the expression (u+i). Now, should u and i have other values, you may get overflow and undefined behavior but with those exact numbers you will get -4444 [1]. This value will have type int. But you are trying to store that value into an unsigned int so that will then be cast to an unsigned int and the value that result will end up having would be (UINT_MAX+1) - 4444.
Should the precision of unsigned int be greater than that of an int, the signed int will be promoted to an unsigned int yielding the value (UINT_MAX+1) - 5678 which will be added to the other unsigned int 1234. Should u and i have other values, which make the expression fall outside the range {0..UINT_MAX} the value (UINT_MAX+1) will either be added or subtracted until the result DOES fall inside the range {0..UINT_MAX) and no undefined behavior will occur.
What is precision?
Integers have padding bits, sign bits, and value bits. Unsigned integers do not have a sign bit obviously. Unsigned char is further guaranteed to not have padding bits. The number of values bits an integer has is how much precision it has.
[Gotchas]
The macro sizeof macro alone cannot be used to determine precision of an integer if padding bits are present. And the size of a byte does not have to be an octet (eight bits) as defined by C99.
[1] The overflow may occur at one of two points. Either before the addition (during promotion) - when you have an unsigned int which is too large to fit inside an int. The overflow may also occur after the addition even if the unsigned int was within the range of an int, after the addition the result may still overflow.