c++ - 为什么(-i)的类型，其中i是unsigned int，还是unsigned int？

Question

int main() {
    unsigned int i = 1;
    typeof(-i) j = -i;
}

当我在 gdb 中查看 j 的类型时，它显示它是 unsigned int 类型。为什么类型没有移动到相应的签名类型？选择这种行为背后有什么原因吗？我发现该类型是否会更直观地转换为相应的带符号类型。

Answer 1

一元否定运算符 - 不会改变其参数的类型，除了应用整数提升。 C standard 的第 6.5.3.3p3 节状态:

The result of the unary - operator is the negative of its (promoted) operand. The integer promotions are performed on the operand, and the result has the promoted type

整数提升的细节在6.3.1.1p2节有详细介绍:

The following may be used in an expression wherever an int or unsigned int may be used:

• An object or expression with an integer type (other than int or unsigned int) whose integer conversion rank is less than or equal to the rank of int and unsigned int.
• A bit-field of type _Bool, int, signed int, or unsigned int.

If an int can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions. All other types are unchanged by the integer promotions.

i 的类型是 unsigned int，根据上面的段落，它不受整数提升的影响(实际上，它是整数提升的目标类型)。因此不执行任何提升，-i 的类型与 i 的类型相同。