Make a function which executes others in C++?

Question

I need a function which as the first parameter it takes a pointer to a function, and the other ones are just parameters for the function(the function can have any number of parameters).

For example, if I define a function
int add(int a, int b)
I could execute it with the wanted function with:
wantedF(address of function "add", x, y) and would return an int.

Is this possible for arbitrary function?

Thanks.

Answer 1

C++03 and C++11 have undergone a lot of changes, and now, we have variadic templates.

Thus, the wantedF() function to execute a function can now be declared for arbitrary functions :

template<typename Fx, Fx fn, typename... Args>
typename std::result_of<Fn(Args...)>::type
wantedF (Args&&... args)
{
return fn(std::forward<Args>(args)...);
}

This is called as :
wantedF<decltype(&add),&add>(2,3);

Now, to shorten it for use, you may use a define:
#define exec(fx) wantedF<decltype(&fx),&fx>

Then use it like this :

int add(int a,int b){ return a+b;}
int sub(int a,int b,int c){ return a-b-c;}

int main()
{
int sum = exec (add)(2,3);
int diff = exec(sub)(2,3,4);
cout << sum << " " << diff ;
// Prints 5 and -5...
}

NOTE - You may require <type_traits> for the function result_of.

Answer 2

There are two ways of doing it:

#include <iostream>
#include <functional>

int add(int a, int b)
{
return a + b;
}

int wantedF(const std::function<int(int, int)>& function, int x, int y)
{
return function(x, y);
}

template<typename t>
int wantedF2(t* function, int x, int y)
{
return function(x, y);
}

int main()
{
std::cout << wantedF(&add, 2, 2) << "\n";
std::cout << wantedF2(&add, 2, 2) << "\n";

std::cin.get();
}

wantedF is the safest and recommended way to do it.

Answer 3

@alkex
Will any of those ways work for the following function?
int add3 (int a, int b, int c) {return a+b+c;}

Answer 4

Of course, and if he wants it to be arbitrary, you can use std::function with template args