How to pass reference-to-function into another function

Question

I have been reading about function pointers and about using them as parameters for other functions.

My question is how would you pass a function by reference without using pointers? I have been trying to find the answer on the Internet but I haven't found a good answer.

I know that you can pass variables by reference like this: void funct(int& anInt);. How would you do something similar to this, but instead of a reference to a variable, a reference to a function was the parameter?

Also, how would you use a reference to the function in a function body?

Note: You can't pass functions. You can only pass references or pointers to functions. — Sebastian Mach
– Sebastian Mach, Commented Aug 22, 2011 at 12:08

Cheers and hth. - Alf · Accepted Answer · 2021-11-14 20:51:20Z

35

#include <iostream>
using namespace std;

void doCall( void (&f)(int) )
{
    f( 42 );
}

void foo( int x )
{
    cout << "The answer might be " << x << "." << endl;
}

int main()
{
    doCall( foo );
}

edited Nov 14, 2021 at 20:51

user17242583

answered Aug 22, 2011 at 3:32

Cheers and hth. - Alf

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4 Comments

fredoverflow Over a year ago

Just out of curiosity, why would I want to use a reference to a function instead of a pointer to a function?

Cheers and hth. - Alf Over a year ago

@FredOverflow: Because a pointer can be null or invalid. A reference guarantees that it's valid and not null (there is no null-reference), or else that there is an error in the program.

newacct Over a year ago

@Alf P. Steinbach: "or else that there is an error in the program" That sounds great in theory but if you have something like sometype &ref = *ptr; the address in the pointer will be simply copied into the address in the reference, which may be non-valid or null. You might say that the statement above is undefined behavior when ptr is non-valid or null, but there won't be any indication of this at compile- or run-time, so it is still up to the programmer to make sure it is valid and non-null, same as if you used a pointer. So I say there isn't any difference to the programmer.

Mark Vincze Over a year ago

I think there is a great difference. If you build up you whole codebase with using references in every API where null is not a sensible value, then the sometype &ref = *ptr; situation won't come up that often because you will have references everywhere. Also there is a semantical difference between accepting a reference and a pointer in an API. Accepting a reference clearly states that the function can't be called with null. If it is a reference, then it's the caller's responsibility, if it's a pointer, then it's the callee's.

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