c++ - 当二元运算符两边的符号不同时，提升规则如何工作？

Question

考虑以下程序:

// http://ideone.com/4I0dT
#include <limits>
#include <iostream>

int main()
{
    int max = std::numeric_limits<int>::max();
    unsigned int one = 1;
    unsigned int result = max + one;
    std::cout << result;
}

和

// http://ideone.com/UBuFZ
#include <limits>
#include <iostream>

int main()
{
    unsigned int us = 42;
    int neg = -43;
    int result = us + neg;
    std::cout << result;
}

+ 运算符如何“知道”哪个是要返回的正确类型？一般规则是将所有参数转换为最宽的类型，但这里 int 和 unsigned int 之间没有明确的“赢家”。在第一种情况下，必须选择 unsigned int 作为 operator+ 的结果，因为我得到了 2147483648 的结果。在第二种情况下，它必须选择 int，因为我得到了 -1 的结果。然而，在一般情况下，我看不出这是如何确定的。这是我看到的未定义行为还是其他？

Answer 1

第 5/9 节明确概述了这一点:

Many binary operators that expect operands of arithmetic or enumeration type cause conversions and yield result types in a similar way. The purpose is to yield a common type, which is also the type of the result. This pattern is called the usual arithmetic conversions, which are defined as follows:

• If either operand is of type long double, the other shall be converted to long double.
• Otherwise, if either operand is double, the other shall be converted to double.
• Otherwise, if either operand is float, the other shall be converted to float.
• Otherwise, the integral promotions shall be performed on both operands.
• Then, if either operand is unsigned long the other shall be converted to unsigned long.
• Otherwise, if one operand is a long int and the other unsigned int, then if a long int can represent all the values of an unsigned int, the unsigned int shall be converted to a long int; otherwise both operands shall be converted to unsigned long int.
• Otherwise, if either operand is long, the other shall be converted to long.
• Otherwise, if either operand is unsigned, the other shall be converted to unsigned.

[Note: otherwise, the only remaining case is that both operands are int]

在您的两种情况下，operator+ 的结果都是 unsigned。因此，第二种情况实际上是:

int result = static_cast<int>(us + static_cast<unsigned>(neg));

因为在这种情况下 us + neg 的值不能由 int 表示，所以 result 的值是实现定义的 - § 4.7/3:

If the destination type is signed, the value is unchanged if it can be represented in the destination type (and bit-field width); otherwise, the value is implementation-defined.