Question

即使只是尝试开始，我也会收到此代码的错误：

note: candidate template ignored: could not match 'double' against 'long'

::

#include <numeric>
#include <chrono>
using namespace std::chrono_literals;

// how to write a function that will take any duration and turn it 
//   into a float representation of seconds?
template <class T>
void go(std::chrono::duration<double, T> d) {

    // what I want to do (that may not work because I haven't gotten this far):
    float seconds = std::chrono::duration_cast<std::chrono::seconds>(d);
}

int main()
{
  go(1ms);
  go(1s);
}

Answer 1

我只能猜测你想要完成什么，这是我最好的猜测：

#include <chrono>
#include <iostream>
using namespace std::chrono_literals;

void go(std::chrono::duration<float> d) {
    std::cout << d.count() << '\n';
}

int main()
{
  go(1ms);
  go(1s);
}

输出：

0.001
1

Answer 2

转换为float并调用count（）：

#include <iostream>
#include <numeric>
#include <chrono>

template< class T, class P >
float to_secs(std::chrono::duration< T, P > t)
{
    std::chrono::duration< float > f = t;
    return f.count();
}

int main(int argc, char*argv[])
{
    std::cout << to_secs(std::chrono::milliseconds(1)) << std::endl;
    std::cout << to_secs(std::chrono::minutes(1)) << std::endl; 
    std::cout << to_secs(std::chrono::hours(1)) << std::endl;
    // output:
    // 0.001
    // 60
    // 3600
    return 0;
}

Answer 3

template<typename duration_t> float seconds(const duration_t &d);

template<class Rep, class Period>
float seconds(const std::chrono::duration<Rep, Period> &d)
{
    typedef std::chrono::duration<Rep, Period> duration_t;

    auto one_second=std::chrono::duration_cast<duration_t>
                  (std::chrono::seconds(1)).count();

    if (one_second == 0)
        return d.count() *
          std::chrono::duration_cast<std::chrono::seconds>
                (duration_t(1)).count();
    else
        return float(d.count())/one_second;
}

如何编写一个函数，它接受任何std :: chrono :: duration（1ms，4s，7h）并将秒作为浮点数？

3 个答案: