如何用'auto'字推导出哪种类型的参数来编写函数？

Question

我正在搜索一个干净的c ++ 11(最多c ++ 17)方法来编写一个函数,该函数只需将fps写入输出流,并给定'start'和'stop'次(例如给定一个间隔时间).所以我有这个代码,例如:

#include <iostream>
int main(int argc, char** argv) {
    typedef std::chrono::high_resolution_clock time_t;
    while (1) {
        auto start = time_t::now();

        // here is the call of function that do something
        // in this example will be printing
        std::cout << "Hello world!"

        auto stop = time_t::now();
        fsec_t duration = stop - start;

        double seconds = duration.count();
        double fps = (1.0 / seconds);

        std::stringstream s;
        s << "FPS: " << fps;

        std::cout << s.str();
    }
}

我想做的事情如下:

#include <iostream>

std::ostream & printFPS(std::ostream &stream, auto start);

int main(int argc, char** argv) {

    while (1) {
        auto start = std::chrono::high_resolution_clock::now();

        // here is the call of function that do something
        // in this example will be printing
        std::cout << "Hello world!"

        printFPS(std::cout, start);
    }
}

std::ostream & printFPS(std::ostream &stream, auto start){

    auto stop = std::chrono::high_resolution_clock::now();
    std::chrono::duration<float> duration = stop - start;

    double seconds = duration.count();
    double fps = (1.0 / seconds);

    std::stringstream s;
    s << "FPS: " << fps;

    return stream << s.str();
}

GCC给了我一些提示,推断出'start'的类型std::chrono::time_point<std::chrono::_V2::system_clock, std::chrono::duration<long int, std::ratio<1, 1000000000> > >,但我不想写函数这个类型(可能是演绎会改变(？),也很长,需要typedef),是否可以写更优雅的功能,因为不允许自动参数？谢谢!

Answer 1

您可以使用它decltype来推断时间类型并将其用作参数的类型.

using time_type = decltype(std::chrono::high_resolution_clock::now());

std::ostream & printFPS(std::ostream &stream, time_type start);

Answer 2

表达式std::chrono::high_resolution_clock::now()返回type的值std::chrono::high_resolution_clock::time_point.所以你可以直接这样做:

std::ostream& printFPS(std::ostream&, std::chrono::high_resolution_clock::time_point );

但是你的打印功能可能并不关心你从哪个时钟开始time_point.它只关心它有一个time_point,所以你可以更普遍地做到这一点:

template <typename Clock, typename Duration>
std::ostream& printFPS(std::ostream&, std::chrono::time_point<Clock, Duration> );