chrono：如何从两个高分辨率时间点计算毫秒持续时间？

Question

我正在尝试从两个计时time_points计算一个计时duration

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

int main() {
    time_point<high_resolution_clock> start = high_resolution_clock::now();
    time_point<high_resolution_clock> end = high_resolution_clock::now();
    duration<milli> difference = duration_cast<milli>(end - start);
    return 0;
}

http://ideone.com/wUKb9S

我对编译器错误消息完全感到困惑：

Compilation error   time: 0 memory: 0 signal:0

prog.cpp: In function 'int main()':
prog.cpp:9:63: error: no matching function for call to 'duration_cast(std::__success_type<std::chrono::duration<long long int, std::ratio<1ll, 1000000000ll> > >::type)'
  duration<milli> difference = duration_cast<milli>(end - start);
                                                               ^
prog.cpp:9:63: note: candidate is:
In file included from prog.cpp:2:0:
/usr/include/c++/4.9/chrono:194:7: note: template<class _ToDur, class _Rep, class _Period> constexpr typename std::enable_if<std::chrono::__is_duration<_Tp>::value, _ToDur>::type std::chrono::duration_cast(const std::chrono::duration<_Rep, _Period>&)
       duration_cast(const duration<_Rep, _Period>& __d)
       ^
/usr/include/c++/4.9/chrono:194:7: note:   template argument deduction/substitution failed:
/usr/include/c++/4.9/chrono: In substitution of 'template<class _ToDur, class _Rep, class _Period> constexpr typename std::enable_if<std::chrono::__is_duration<_Tp>::value, _ToDur>::type std::chrono::duration_cast(const std::chrono::duration<_Rep, _Period>&) [with _ToDur = std::ratio<1ll, 1000ll>; _Rep = <missing>; _Period = <missing>]':
prog.cpp:9:63:   required from here
/usr/include/c++/4.9/chrono:194:7: error: no type named 'type' in 'struct std::enable_if<false, std::ratio<1ll, 1000ll> >'

两个time_point之间的差异不是持续时间吗？ 我不能在持续时间类型之间进行转换以获得毫秒的持续时间表示吗？

Answer 1

milli是std::ratio<1, 1000>的typedef，对于duration不是有效的表示形式，应该类似于duration<long, milli>或只是std::chrono::milliseconds

auto difference = duration_cast<milliseconds>(end - start);