Converting pointer to member function to std::function

Question

I have a slightly convoluted use case of passing a member function pointer to an outside function which is then called again by a member function (Don't ask!). I'm learning about std::function and std::mem_fn but I can't seem to be able to convert my old school function pointer

void (T::*func)(int) to a std::function<void (T::*)(int) func>

in the code below, I'd like to be able to pass a std::function to memFuncTaker in the call from anotherMember

#include "class2.hpp" 
#include <iostream> 

class outer{ 
public: 
  void aMember(int a){ 
    std::cout << a <<std::endl; 
  } 
  void anotherMember(double){ 
    memFuncTaker(this, &outer::aMember); 
  } 

}; 


template<class T> 
void memFuncTaker(T* obj , void (T::*func)(int) ){ 
  (obj->*func)(7); 
}

I don't see any attempts to use std::function in your code. — AnT stands with Russia
– AnT stands with Russia, Commented Nov 1, 2017 at 3:13
Your question title says "Converting pointer to member function to std::function", but the question sounds like "Converting std::function to pointer to member function". — Christian Hackl
– Christian Hackl, Commented Nov 1, 2017 at 8:16

AnT stands with Russia · Accepted Answer · 2017-11-01 03:28:53Z

2

When you bind std::function to a non-static member function pointer, it "reveals" the hidden this parameter, making it the first explicit parameter of the resultant functor. So in your case for outer::aMember you'd use std::function<void(outer *, int)> and end up with a two-parameter functor

#include <functional>
#include <iostream> 

template<class T> 
void memFuncTaker(T *obj , std::function<void(T *, int)> func){ 
  func(obj, 7);
} 

class outer{ 
public: 
  void aMember(int a){ 
    std::cout << a <<std::endl; 
  } 
  void anotherMember(double){ 
    memFuncTaker(this, std::function<void(outer *, int)>{&outer::aMember}); 
  } 
}; 

int main() {
  outer o;
  o.anotherMember(0);
}

http://coliru.stacked-crooked.com/a/5e9d2486c4c45138

Of course, if you prefer, you can bind the first argument of that functor (by using std::bind or lambda) and thus "hide" it again

#include <functional>
#include <iostream> 

using namespace std::placeholders;

void memFuncTaker(std::function<void(int)> func){ 
  func(7);
} 

class outer{ 
public: 
  void aMember(int a){ 
    std::cout << a <<std::endl; 
  } 
  void anotherMember(double){ 
    memFuncTaker(std::function<void(int)>(std::bind(&outer::aMember, this, _1))); 
  } 
}; 

int main() {
  outer o;
  o.anotherMember(0);
}

Note that in this version memFuncTaker no longer has to be a template (which happens to be one of the primary purposes of std::function - employ type erasure techniques to "de-templatize" the code).

edited Nov 1, 2017 at 3:28

answered Nov 1, 2017 at 3:21

AnT stands with Russia

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3 Comments

cppprogrammer Over a year ago

How come the return type of std::bind(&outer::aMember, this, _1) is std::function<void(int)>. I know aMember is a member function but how come this doesn't show up the function signature bound to std::function. I expected std::bind(&outer::aMember, this, _1) to return std::function<void(outer*, int)>

AnT stands with Russia Over a year ago

@creationist: Why? The whole purpose of that std::bind is to eliminate (bind) the first (outer *) parameter. Before std::bind the functor had two parameters - outer * and int, - but after std::bind only one was left: int.

AnT stands with Russia Over a year ago

For another example, if you do std::mem_fn(&outer::aMember) you will get a two-parameter functior (just like you described). But if after that you do std::bind(std::mem_fn(&outer::aMember), this, _1) only one parameter will be left. What you see in my answer above is the same thing, since std::mem_fn is optional in this context and can be omitted. std::bind is smart enough to realize what is meant without explicit application of std::mem_fn.

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