Passing member function pointer as template parameter

Question

I want to create a wrapper class which is able to call member functions (of any type) of the wrapped class with help of templates. This is what I have so far:

template <typename T>
class wrapper {
  public:    
    template<typename R, R (T::*func)()>
    void call_func() {
      (wrapped.*func)();
    }

  private:
    T wrapped;  
};

class some_class {
  private:
    int i = 2;
  public:
    void some_func() {
      std::cout << i << std::endl;
    }
};

int main() {
    wrapper<some_class> wr;
    // How I need to call at the moment:
    wr.call_func<void, &some_class::some_func>();
    // How I want call:
    wr.call_func<&some_class::some_func>();
    return 0;
}

As you can see in the comments of the main function, I want to call the wrapper-function without explicitly specifying the return type of the wrapped member function. (How) Can this be done in C++11?

petersohn · Accepted Answer · 2018-02-24 16:22:21Z

10

template<typename F>
void call_func(F func) {
  (wrapped.*func)();
}

then call like this:

wr.call_func(&some_class::some_func);

If you want to use the return value too, you'll need this:

template<typename F>
auto call_func(F func) -> decltype((std::declval<T>().*func)()) {
  return (wrapped.*func)();
}

If you have C++14, you can omit the -> decltype(...) part and use decltype(auto) as the return value.

If you also want to pass functions, you can use variadic templates and forwarding for that.

template<typename F, typename... Args>
decltype(auto) call_func(F func, Args&&... args) {
  return (wrapped.*func)(std::forward<Args>(args)...);
}

edited Feb 24, 2018 at 16:22

answered Feb 23, 2018 at 13:50

petersohn

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5 Comments

Mike van Dyke Over a year ago

Isn't it possible to pass the member function as template parameter instead of function parameter?

petersohn Over a year ago

@MikevanDyke There is, as it is in the question. But why do you want it that way?

StoryTeller - Unslander Monica Over a year ago

If you have C++14 and omit the decltype, you should substitute the return type for decltype(auto) to conserve the same semantics for the template.

Mike van Dyke Over a year ago

@petersohn Because I want to pass some arguments too and that way, I could seperate the function from the arguments. It would just look kind of "cleaner" to me.

petersohn Over a year ago

@MikevanDyke I added the usage of arguments to the answer.

NathanOliver · Accepted Answer · 2018-02-23 13:57:00Z

0

You can directly use std::function

For more information on std::function see: http://en.cppreference.com/w/cpp/utility/functional/function

edited Feb 23, 2018 at 13:57

NathanOliver

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answered Feb 23, 2018 at 13:52

Navie

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1 Comment

petersohn Over a year ago

That one works, but not very efficient. std::function is useful if you want to save the call for later.

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