Question

Is it possible to pass functions by reference?

Something like this:

function call($func){
    $func();
}

function test(){
    echo "hello world!";
}

call(test);

I know that you could do 'test', but I don’t really want that, as I need to pass the function by reference.

Is the only way to do so via anonymous functions?

Clarification: If you recall from C++, you could pass a function via pointers:

void call(void (*func)(void)){
    func();
}

Or in Python:

def call(func):
    func()

That’s what i’m trying to accomplish.

Answer 1

For what it’s worth, how about giving something like this a shot? (Yes, I know it’s an anonymous function which was mentioned in the post, but I was disgruntled at the abundance of replies that did not mention closures/function-objects at all so this is mostly a note for people running across this post.)

I don’t use PHP, but using a closure appears to work in PHP 5.3 (but not PHP 5.2) as demonstrated here. I am not sure what the limitations, if any, there are. (For all I know the closure will eat your children. You have been warned.)

function doIt ($fn) {
  echo "doIt\n";
  return $fn();
}

function doMe () {
  echo "doMe\n";
}

// I am using a closure here.
// There may be a more clever way to "get the function-object" representing a given
// named function, but I do not know what it is. Again, I *don't use PHP* :-)
echo doIt(function () { doMe(); });

Happy coding.