问题
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,
(-(0x80000000u)) = 0x100000000u-0x80000000u = 0x80000000u
But I can not find the answer to the question: what will be unary minus for signed 0x80000000? Is it UB, implementation defined, or ... ?
The question is mostly about run-time calculation.
Say
signed int my_minus(signed int i) { return -i;}
....
int main() {
signed int a = -0x7FFFFFFF; // a looks like 0x80000001
signed int b = a - 1; // b looks like 0x80000000
std::cout << my_minus(b);
....
}
Still, your comments on other 2 cases are welcome:
Compile-time constant folding, say,
-(INT_MIN)Compile-time calculation of
constexpr(if there is a difference with compile-time constant folding).
( Please look at https://meta.stackexchange.com/questions/123713/is-splitting-a-question-a-good-practice before voting for duplicate. )
回答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]
回答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.
来源：https://stackoverflow.com/questions/9476036/c-only-unary-minus-for-0x80000000