c++ - 为什么在 C++ 中混合 size_t 和 unsigned int 时模除法会出错

Question

给定一个程序

#include <iostream>
using namespace std;

int main()
{
     const size_t DoW = 7;
     const unsigned int DAYS_OF_WEEK = static_cast<unsigned int> (DoW);
     unsigned int dayOfFirstDay = 0;
     unsigned int _firstDayOfWeek = 1;
     unsigned int diff = (DAYS_OF_WEEK+ (dayOfFirstDay - _firstDayOfWeek) ) % DAYS_OF_WEEK;
     cout << "diff = ("  << DAYS_OF_WEEK << " + (" << dayOfFirstDay << " - " << _firstDayOfWeek << ")) %" << DAYS_OF_WEEK
         << " = " << diff << endl;
     return 0;
}

该程序的输出是

diff = (7 + (0 - 1)) %7 = 6

这是预期的。但是一个没有 static_cast 的修改过的程序

#include <iostream>
using namespace std;

int main()
{
     const size_t DAYS_OF_WEEK = 7;
     unsigned int dayOfFirstDay = 0;
     unsigned int _firstDayOfWeek = 1;
     unsigned int diff = (DAYS_OF_WEEK+ (dayOfFirstDay - _firstDayOfWeek) ) % DAYS_OF_WEEK;
     cout << "diff = ("  << DAYS_OF_WEEK << " + (" << dayOfFirstDay << " - " << _firstDayOfWeek << ")) %" << DAYS_OF_WEEK
         << " = " << diff << endl;
     return 0;
}

产出

diff = (7 + (0 - 1)) %7 = 3

这是意料之中的。为什么？
(这两个程序都是在 Ubuntu 64 位上使用 g++ 9.3.0 编译的)

Answer 1

在您的平台上，size_t 似乎是 64 位，而 unsigned int 是 32 位。
64位1没有integral promotion。这是在表达式中混合 64 位操作数的危险。
因此，当转换为 64 位时，-1 的 32 位环绕保留为 4294967295。
我们得到 7 + 4294967295(以 64 位执行)= 4294967302(无环绕)。
4294967302 % 7 = 3

1 除了 ( unsigned ) int 本身是 64 位的系统，这目前不太可能。