涉及输出计时持续时间的 C++ 错误

Question

这是我正在使用的类(class)(某些部分被剪掉了)

const int records = 7;

    class Timed {
    public:

        friend std::ostream& operator<<(std::ostream& os, Timed right) {
            for (int i = 0; i < records; i++) {
                os << right.eR[i].vName << " " << right.eR[i].duration << " " << right.eR[i].seconds << std::endl;
            }
            return os;
        }

    private:

        std::chrono::time_point<std::chrono::steady_clock> start, end;

        struct {
            std::string vName;
            std::string seconds;
            std::chrono::duration<float> duration;
        } eR[records];

    };

基本上，我正在尝试输出匿名结构的值。但是，我收到错误:

binary '<<': no operator found which takes a right-hand operand of type 'std::chrono::duration<float,std::ratio>1,1>>' (or there is no acceptable conversion)

我想知道如何在持续时间内打印此值？提前致谢

Answer 1

在 C++11/14/17 中，chrono::duration 没有流式运算符类型。你有两个选择:

提取.count() :

|

 os << right.eR[i].duration.count() << "s ";

使用这个open-source, header-only date/time library :

|

#include "date/date.h"
// ...
using date::operator<<;
os << right.eR[i].duration << " ";

上面的 datelib 现在是 C++20 的一部分，但尚未发布。供应商正在为此努力。当您移植到它时，您只需删除 #include "date/date.h"和 using date::operator<<; .