Wie schreibe ich eine Funktion, die eine Std :: Chrono :: Dauer (1ms, 4s, 7h) dauert und die Sekunden als Float bekommt?

Question

Auch nur zu beginnen versucht, bekomme ich einen Fehler mit diesem Code:

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

ich nur raten kann, was Sie versuchen, zu erreichen und hier ist meine beste Vermutung:

#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); 
} 

Diese Ausgänge:

0.001 
1 

Answer 2

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; 
} 

Answer 3

Guss zu schweben und Call-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; 
}