c++ - std::chrono::duration缺少立即刻度计数操纵的原因是什么？

假设我们有

#include <chrono>
#include <iostream>
#include <ctime>

namespace Ratios { typedef std::ratio<60*60*24,1> Days; }

typedef std::chrono::system_clock Clock;
typedef Clock::time_point TimePoint;

我们的main看起来像

int main(int argc, char *argv[])
{
    // argc check left out for brevity
    const Clock::rep d = static_cast<Clock::rep>(std::atoi(argv[1]));
    // Right now
    TimePoint now = Clock::now();
    // Start with zero days
    auto days = std::chrono::duration<Clock::rep, Ratios::Days>::zero();

    // Now we'd like to add d to the days
    days += d; // Error!
    days.count() = d; // Error!
    days = days + d; // Error!
    days += std::chrono::duration<Clock::rep, Ratios::Days>(d); // Okay
    days = days + std::chrono::duration<Clock::rep, Ratios::Days>(d); // Okay

    days *= d; // Why is this okay?
    days %= d; // And this too?

    TimePoint later = now + days;

    return 0;
}

禁止用户直接操作duration的背后原因是什么？

最佳答案

这样做是为了让您坚持使用强类型值而不是任意值。

Bjarne Stroustrup在“C++编程语言”(第4版，35.2.1，第1011页)中提供了有关此行为的示例:

duration<long long, milli> d1{7}; // 7 milliseconds
d1 += 5; // error
[...]

d1 += duration<long long, milli>{5}; //OK: milliseconds"