c++ - 为什么固定宽度无符号整数的输出为负，而无符号整数输出按预期回绕？

Question

#include <iostream>

#define TRY_INT

void testRun() 
{
    #ifdef TRY_INT          //test with unsigned
    unsigned int value1{1}; //define some unsigned variables
    unsigned int value2{1};
    unsigned int value3{2};
    #else                   //test with fixed width
    uint16_t value1{1};     //define fixed width unsigned variables
    uint16_t value2{1};
    uint16_t value3{2};

    #endif

    if ( value1 > value2 - value3 )
    {
        std::cout << value1 << " is bigger than: " << value2 - value3 << "\n";
    }
    else
    {
        std::cout << value1 << " is smaller than: " << value2 - value3 << "\n";
    }

}

int main()
{
    testRun();

    return 0;
}

使用无符号整数我得到:

1 is smaller than: 4294967295

使用固定宽度的无符号整数，输出为:

1 is smaller than: -1

我的期望是它也会环绕，这与 std::cout 有关系吗？

Answer 1

我猜是积分推广造成的。引用表格cppreference :

...arithmetic operators do not accept types smaller than int as arguments, and integral promotions are automatically applied after lvalue-to-rvalue conversion, if applicable.

unsigned char, char8_t (since C++20) or unsigned short can be converted to int if it can hold its entire value range...

因此，如果 uint16_t 只是您实现中 unsigned Short 的别名，则 value2 - value3 会使用 int< 进行计算 类型，结果也是 int，这就是显示 -1 的原因。

使用unsigned int，不会应用任何提升，并且整个计算都以此类型执行。

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在最新的在线 C++ 草案中，请参阅 [conv.prom/1] :

A prvalue of an integer type other than bool, char16_­t, char32_­t, or wchar_­t whose integer conversion rank is less than the rank of int can be converted to a prvalue of type int if int can represent all the values of the source type; otherwise, the source prvalue can be converted to a prvalue of type unsigned int.