使用时间单位模板返回 chrono::duration 的函数 [英] Function to return a chrono::duration using templates for the time unit

问题描述

我是 C++ 模板的新手，我正在尝试编写一个函数，该函数返回具有指定时间单位和类型的 chrono::duration.例如，这一行给了我两倍的时间差(以秒为单位):

I'm new to C++ templates and I'm trying to write a function which returns a chrono::duration with the specified time unit and type. For instance, this line gives me the time difference in seconds as double:

std::chrono::duration<double> secd =
     std::chrono::duration_cast<std::chrono::duration<double,std::ratio<1>>>(end - start);

我有一个类函数，它给了我一个持续时间，我想使用模板来指示返回值的类型和单位(在前面的例子中，这将是 double 和比率＜1>).我想要的是类似于这个伪代码的东西:

I have a class function which gives me a time duration, and I would like to use templates to indicate the type and unit for the return value (in the previous example, that would be double and ratio<1>). What I would like to have is something similar to this pseudocode:

template typename<class T, class R> std::chrono::duration<T, R> getStepTime() {
    return std::chrono::duration_cast<std::chrono::duration<T, R>>(_time);
}

其中 _time 是具有持续时间的类成员.到目前为止，我所有的尝试都没有编译通过.

where _time is a class member with the duration. All my attempts so far didn't even compile.

如果有更好的方法可以在不使用模板的情况下实现这一目标，我会全力以赴.

In case there is a better way to achieve this without using templates, I'm all ears.

推荐答案

typename 的错误使用，并且模板中缺少结束 >.这是一个测试编译的调整示例:

Bad usage of typename and there's a missing closing > in your template. Here is a tweaked example to test compilation :

template <typename T, typename R> 
std::chrono::duration<T, R> getStepTime() 
{
    std::chrono::duration<T, R> duration;
    return std::chrono::duration_cast<std::chrono::duration<T, R>>(duration);
}

http://ideone.com/QGYm8u

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