Arsthbt

I have a function Mult that takes two integers and returns the product of its parameters. And a function Double that takes an integer and returns a pointer to function that returns an integer and takes two integer parameters like Mult.

• 
  Mult's second parameter is default So when I call Double, Double returns the address of Mult thus I can pass only one argument.

But It doesn't work with pointer to function:

int Mult(int x, int y = 2) // y is default
 return x * y;


using pFn = int(*)(int, int);


pFn Double(int x) 
 return Mult;


int main(int argc, char* argv[])

 pFn func = Double(0);
 cout << func(7, 4) << endl; // ok
 //cout << func(7) << endl; // error: Too few arguments
 cout << Mult(4) << endl; // ok. the second argument is default

Above if I call Mult with a single argument it works fine because the second argument is default but calling it through the pointer func it fails. func is pointer to function that takes two integers and returns an int.

Defaulted arguments are a bit of C++ syntactic sugar; when calling the function directly with insufficient arguments, the compiler inserts the default as if the caller had passed it explicitly, so the function is still called with the full complement of arguments (Mult(4) is compiled into the same code as Mult(4, 2) in this case).

The default isn't actually part of the function type though, so you can't use the default for an indirect call; the syntactic sugar breaks down there, since as soon as you are calling through a pointer, the information about the defaults is lost.

For the "why not" I refer you to this answer. If you want to somehow keep the ability to use a default, you need to provide something more than a function pointer, eg a lamdba will do:

#include <iostream>

int Mult(int x, int y = 2) // y is default
 return x * y;


auto Double() 
 return [](int x,int y=2) return Mult(x,y); ;


int main(int argc, char* argv[]) 
 auto func = Double();
 std::cout << func(7, 4) << 'n'; // ok
 std::cout << func(7) << 'n'; // ok
 std::cout << Mult(4) << 'n'; // ok

Live demo

Take the example with some grain of salt. It is merely meant to demonstrate that it is possible to do something like this. The usefulness of this approach is rather limited. You have to repeat the default parameters in Double to make it work.