std::common_type(std::chrono::duration) (3) - Linux Manuals

std::common_type(std::chrono::duration): std::common_type(std::chrono::duration)

NAME

std::common_type(std::chrono::duration) - std::common_type(std::chrono::duration)

Synopsis

template <class Rep1, class Period1, class Rep2, class Period2>
struct common_type<std::chrono::duration<Rep1, Period1>,
std::chrono::duration<Rep2, Period2>> { (since C++11)
typedef std::chrono::duration<
typename std::common_type<Rep1, Rep2>::type, /*see note*/> type;
};

Exposes the type named type, which is the common type of two std::chrono::durations.

Note

The period of the resulting duration is the greatest common divisor of Period1 and Period2.

Example

// Run this code

  #include <iostream>
  #include <chrono>

  // std::chrono already finds the greatest common divisor,
  // likely using std::common_type<>. We make the type
  // deduction externally.

  template <typename T,typename S>
  auto durationDiff(const T& t, const S& s) -> typename std::common_type<T,S>::type
  {
      typedef typename std::common_type<T,S>::type Common;
      return Common(t) - Common(s);
  }

  int main()
  {
      typedef std::chrono::milliseconds milliseconds;
      typedef std::chrono::microseconds microseconds;

      auto ms = milliseconds(30);
      auto us = microseconds(1100);

      std::cout << ms.count() << "ms - " << us.count() << "us = "
                << durationDiff(ms,us).count() << "\n";
  }

Output:

  30ms - 1100us = 28900