什么是std :: chrono :: duration缺乏立即计数操作的背后的原因？ [英] What is the reason behind std::chrono::duration's lack of immediate tick count manipulation?

问题描述

假设我们有

  #include< chrono> 
 #include< iostream> 
 #include< ctime> 
 
 namespace比例{typedef std :: ratio< 60 * 60 * 24,1>天; } 
 
 typedef std :: chrono :: system_clock时钟; 
 typedef Clock :: time_point TimePoint; 
  

我们的 main 看起来像

  int main（int argc，char * argv []）
 {
 // argc 
 const Clock :: rep d = static_cast< Clock :: rep>（std :: atoi（argv [1]））; 
 //现在
 TimePoint now = Clock :: now（）; 
 //从0开始
 auto days = std :: chrono :: duration< Clock :: rep，Ratios :: Days> :: zero（）; 
 
 //现在我们要添加d到天
天+ = d; //错误！ 
 days.count（）= d; //错误！ 
天=天+ d; //错误！ 
 days + = std :: chrono :: duration< Clock :: rep，Ratios :: Days>（d）; // Okay 
 days = days + std :: chrono :: duration< Clock :: rep，Ratios :: Days>（d）; // Okay 
 
 days * = d; //为什么这么好？ 
 days％= d; //这也是吗？ 
 
时间点后=现在+天; 
 
 return 0; 
} 
  

禁止用户操作 duration 直接？

解决方案

这样做是为了强制你坚持强类型值，任何值。

Bjarne Stroustrup在C ++编程语言（第4版，35.2.1，第1011页）中提供了关于此行为的示例：

---

期间是一个单位系统，因此没有 = 或 + = 采用一个简单的值。允许添加 5






  duration< long long，milli> d1 {7}; // 7 milliseconds 
 d1 + = 5; // error 
 [...] 
  / pre> 
 
        ，例如：
 
 
 
 
 
  d1 + = duration< long long，milli> {5}; // OK：milliseconds
  
 
 
 
 
Suppose we have
#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;
And our main looks like
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;
}
What is the reason behind prohibiting the user to manipulate a duration directly?
解决方案
It is done to enforce you to stick to strongly typed values rather than arbitrary values.

Bjarne Stroustrup has examples regarding this behaviour in "The C++ Programming Language" (4th Ed., 35.2.1, pp. 1011):




  
"The period is a unit system, so there is no = or += taking a plain value. Allowing that would be like allowing the addition of 5 of an unknown SI unit to a length in meters. Consider:


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



  
What would 5 mean here? 5 seconds? 5 milliseconds? [...] If you know what you mean, be explicit about it. For example:


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




                        
这篇关于什么是std :: chrono :: duration缺乏立即计数操作的背后的原因？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！