How to make a function return a pointer to a function? (C++)

Solution 1:

int f(char) {
    return 0;
}

int (*return_f())(char) {
    return f;
}

No, seriously, use a typedef :)

Solution 2:

#include <iostream>
using namespace std;

int f1() {
    return 1;
}

int f2() {
    return 2;
}

typedef int (*fptr)();


fptr f( char c ) {
    if ( c == '1' ) {
        return f1;
    }
    else {
        return f2;
    }
}

int main() {
    char c = '1';
    fptr fp = f( c );
    cout << fp() << endl;
}

Solution 3:

Create a typedef for the function signature:

typedef void (* FuncSig)(int param);

Then declare your function as returning FuncSig:

FuncSig GetFunction();

Solution 4:

Assuming int f(char) and ret_f which returns &f.

C++98/C++03 compatible ways:

• Ugly way:

  int (*ret_f()) (char) { return &f; }
  

• With typedef:

  typedef int (sig)(char);
  
  sig* ret_f() { return &f; }
  

  or:

  typedef int (*sig_ptr)(char);
  
  sig_ptr ret_f() { return &f; }
  

Since C++11, in addition we have:

• with decltype:

  decltype(&f) ret_f() { return &f; }
  

• trailing return type:

  auto ret_f() -> int(*)(char) { return &f; }
  

  or:

  auto ret_f() -> decltype(&f) { return &f; }
  

• typedef with using:

  using sig = int(char);
  
  sig* ret_f() { return &f; }
  

  or:

  using sig_ptr = int (*)(char);
  
  sig_ptr ret_f() { return &f; }
  

C++14 adds:

• auto deduction:

  auto ret_f() { return &f; }