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How can I use std::chrono::duration as a template parameter?

I have a template class, something like:

template < typename T, size_t Seconds > class MyClass {}

Now, I would like to change Seconds to be a duration, so the class can be parametrized with std::chrono::duration. For example, I'd like to be able to do this:

MyClass < std::string, std::chrono::seconds(30) > object;

Also, in the template, I'd like to specify a default value, something like std::chrono::seconds(30).

like image 520
Amy Avatar asked Mar 05 '15 10:03

Amy


1 Answers

You can design your template in a clever way:

template < typename T, typename Duration = std::chrono::seconds, int duration_value = 30 > 
class MyClass 
{
    // Now you can use duration here:
    // auto duration = Duration(duration_value);
};

And then you can instantiate your template as

MyClass < std::string, std::chrono::seconds, 30 > object;

Or, actually having these values as defaults, simply

MyClass < std::string > object;

Edit:

Taking into account PaperBirdMaster's request, you can limit template's Duration type, to be std::chrono::duration only, this way:

template <typename T>
struct is_chrono_duration
{
    static constexpr bool value = false;
};

template <typename Rep, typename Period>
struct is_chrono_duration<std::chrono::duration<Rep, Period>>
{
    static constexpr bool value = true;
};

template < typename T, typename Duration = std::chrono::seconds, int duration_value = 30 >
class MyClass
{
    static_assert(is_chrono_duration<Duration>::value, "duration must be a std::chrono::duration");
    // Now you can use duration here:
    // auto duration = Duration(duration_value);
};

int main(int argc, char ** argv)
{
    MyClass < std::string, std::chrono::seconds, 1> obj1;       // Ok
    MyClass < std::string, std::chrono::milliseconds, 1> obj2;  // Ok
    MyClass < std::string, int, 1> obj3;                        // Error
    return 0;
}
like image 54
GreenScape Avatar answered Sep 23 '22 03:09

GreenScape