Function pointer to member function

Question

I'd like to set up a function pointer as a member of a class that is a pointer to another function in the same class. The reasons why I'm doing this are complicated.

In this example, I would like the output to be "1"

class A {
public:
 int f();
 int (*x)();
}

int A::f() {
 return 1;
}


int main() {
 A a;
 a.x = a.f;
 printf("%d\n",a.x())
}

But this fails at compiling. Why?

Answer 1

The syntax is wrong. A member pointer is a different type category from a ordinary pointer. The member pointer will have to be used together with an object of its class:

class A {
public:
 int f();
 int (A::*x)(); // <- declare by saying what class it is a pointer to
};

int A::f() {
 return 1;
}


int main() {
 A a;
 a.x = &A::f; // use the :: syntax
 printf("%d\n",(a.*(a.x))()); // use together with an object of its class
}

a.x does not yet say on what object the function is to be called on. It just says that you want to use the pointer stored in the object a. Prepending a another time as the left operand to the .* operator will tell the compiler on what object to call the function on.

Answer 2

int (*x)() is not a pointer to member function. A pointer to member function is written like this: int (A::*x)(void) = &A::f;.

Answer 3

Call member function on string command

#include <iostream>
#include <string>


class A 
{
public: 
    void call();
private:
    void printH();
    void command(std::string a, std::string b, void (A::*func)());
};

void A::printH()
{
    std::cout<< "H\n";
}

void A::call()
{
    command("a","a", &A::printH);
}

void A::command(std::string a, std::string b, void (A::*func)())
{
    if(a == b)
    {
        (this->*func)();
    }
}

int main()
{
    A a;
    a.call();
    return 0;
}

Pay attention to (this->*func)(); and the way to declare the function pointer with class name void (A::*func)()

Answer 4

You need to use a pointer to a member function, not just a pointer to a function.

class A { 
    int f() { return 1; }
public:
    int (A::*x)();

    A() : x(&A::f) {}
};

int main() { 
   A a;
   std::cout << (a.*a.x)();
   return 0;
}

Answer 5

While this is based on the sterling answers elsewhere on this page, I had a use case which wasn't completely solved by them; for a vector of pointers to functions do the following:

#include <iostream>
#include <vector>
#include <stdio.h>
#include <stdlib.h>

class A{
public:
  typedef vector<int> (A::*AFunc)(int I1,int I2);
  vector<AFunc> FuncList;
  inline int Subtract(int I1,int I2){return I1-I2;};
  inline int Add(int I1,int I2){return I1+I2;};
  ...
  void Populate();
  void ExecuteAll();
};

void A::Populate(){
    FuncList.push_back(&A::Subtract);
    FuncList.push_back(&A::Add);
    ...
}

void A::ExecuteAll(){
  int In1=1,In2=2,Out=0;
  for(size_t FuncId=0;FuncId<FuncList.size();FuncId++){
    Out=(this->*FuncList[FuncId])(In1,In2);
    printf("Function %ld output %d\n",FuncId,Out);
  }
}

int main(){
  A Demo;
  Demo.Populate();
  Demo.ExecuteAll();
  return 0;
}

Something like this is useful if you are writing a command interpreter with indexed functions that need to be married up with parameter syntax and help tips etc. Possibly also useful in menus.

Answer 6

While you unfortunately cannot convert an existing member function pointer to a plain function pointer, you can create an adapter function template in a fairly straightforward way that wraps a member function pointer known at compile-time in a normal function like this:

template <class Type>
struct member_function;

template <class Type, class Ret, class... Args>
struct member_function<Ret(Type::*)(Args...)>
{
    template <Ret(Type::*Func)(Args...)>
    static Ret adapter(Type &obj, Args&&... args)
    {
        return (obj.*Func)(std::forward<Args>(args)...);
    }
};

template <class Type, class Ret, class... Args>
struct member_function<Ret(Type::*)(Args...) const>
{
    template <Ret(Type::*Func)(Args...) const>
    static Ret adapter(const Type &obj, Args&&... args)
    {
        return (obj.*Func)(std::forward<Args>(args)...);
    }
};

 

int (*func)(A&) = &member_function<decltype(&A::f)>::adapter<&A::f>;

Note that in order to call the member function, an instance of A must be provided.