c++ only: unary minus for 0x80000000

Question

This question is supposedly for language-lawyers.

Suppose that signed and unsigned int are both 32 bits wide. As stated in the n3337.pdf draft, 5.3.1.8,

(-

Answer 1

Signed integer overflow is always undefined, as far as I know. From the C++ spec section 5 Expressions, paragraph 4:

If during the evaluation of an expression, the result is not mathematically defined or not in the range of representable values for its type, the behavior is undefined. [Note: most existing implementations of C++ ignore integer overflows. Treatment of division by zero, forming a remainder using a zero divisor, and all floating point exceptions vary among machines, and is usually adjustable by a library function. —endnote]

Answer 2

Signed integral types obey the rules of mathematical integers without added computer bullshit. So -std::numeric_limits< signed_type >::min() is going to be undefined behavior, if the given type cannot represent the resulting number.

In a constexpr, the implementation is required to reject that expression, as anything causing undefined behavior renders a constant expression invalid, as a diagnosable rule. In this case the rule is one of the forbidden items in §5.19,

— a result that is not mathematically defined or not in the range of representable values for its type;

In constant folding, the compiler is most likely to insert the overflowed value.