Is overflow abs good?

Question

Is the behavior of the following code correct?

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

The first line may be overflow / underflow. But on platforms, when this is normal, does the second line make 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 these types exist.

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When a standard says that one behavior is undefined, it is undefined, even if some particular platform works in a certain way.

The C ++ 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 the integer j . If the result cannot be presented, the behavior is undefined. ²⁵⁹⁾

^{259) The absolute value of the negative number itself cannot be represented in a double complement.}

So the result of abs is undefined. Pratically returns 0x80000000 , but this is not the behavior defined by the standard.