Question

Is the behavior of the following code well defined?

int32_t i = -0x80000000;
uint32_t u = abs(i);

There may be an overflow/underflow on the first line. But on platforms when that is ok, does the second line have defined behavior?

Answer 1

Do you mean -0x80000000 with seven 0s? The behavior of abs(-0x8000000) is certainly well-defined (both uint32_t and int32_t contain ±0x8000000 in their valid range if those types exist.

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When the standard says one behavior is undefined, it is undefined even if some particular platform work in some defined way.

The C++’s abs function is taken from C, and the C standard says (§7.20.6.1/2):

The abs, labs, and llabs functions compute the absolute value of an integer j. If the
result cannot be represented, the behavior is undefined.²⁵⁹⁾

^{259) The absolute value of the most negative number cannot be represented in two’s complement.}

So the result of abs is undefined. Pratically it will return 0x80000000, but that’s not the behavior defined by the standard.