Truncating an int to char - is it defined?

Question

unsigned char a, b;
b = something();
a = ~b;

A static analyzer complained of truncation in the last line, presumably because b is promoted to int before its bits are flipped and the result will be of type int.

I am only interested in the last byte of the promoted int - if b was 0x55, I need a to be 0xAA. My question is, does the C spec say anything about how the truncation happens, or is it implementation defined/undefined? Is it guaranteed that a will always get assigned the value I expect or could it go wrong on a conforming platform?

Of course, casting the result before assigning will silence the static analyzer, but I want to know if it is safe to ignore this warning in the first place.

Answer 1

The C standard specifies this for unsigned types:

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.

In this case, if your unsigned char is 8 bits, it means that the result will be reduced modulo 256, which means that if b was 0x55, a will indeed end up as 0xAA.

But note that if unsigned char is wider than 8 bits (which is perfectly legal), you will get a different result. To ensure that you will portably get 0xAA as the result, you can use:

a = ~b & 0xff;

(The bitwise and should be optimised out on platforms where unsigned char is 8 bits).

Note also that if you use a signed type, the result is implementation-defined.