How do I pass a function as a parameter to another function? - php

Question

I have a list of functions a(), b(), c()
I have a main() function.
Depending on case, I need to pass a different function to main() to use.
In javascript it would be:
var a = function(){}
var b = function(){}
var c = function(){}
var func = (some logic) a, b or c;
main(func);
How do I do that in php5.3?
I am trying to avoid using
$func_name = "a";
main($func_name){
$func_name();
}
Or may be it is the best way, and I should not use closures of any type?

Answer 1

Same idea in PHP 5.3, as you can create anonymous functions:
$sayHello = function($var)
{
echo "Hello ", $var;
};
// Logic here to determine what $func is set to
$func = $sayHello;
function callCustom($function)
{
if(!is_callback($function))
{
// throw exception
}
$function("World");
}
callCustom($func); // Hello World

Answer 2

Try this:
$a = function(){};
$b = function(){};
$c = function(){};
switch (rand(0, 2)) {
case 0: $func = $a; break;
case 1: $func = $b; break;
case 2: $func = $c; break;
}
var_dump($func);
You can see a working example here http://codepad.viper-7.com/Ut5yGQ
Note: I used var_dump instead of main as main is undefined

Answer 3

In PHP, anything that is "callable" (is_callable­Docs) can be called (invoked). The manual often names these parameters as Callback, a pseudo-type in PHP­Docs.
You can use any of the different callbacks and pass them as a function parameter:
function main($callback)
{
call_user_func($callback);
}
Then your function will work with any valid PHP callback.
Next to that, functions can be called with variables, some examples:
Variable Function Call:
function name() {};
$function = 'name'; # function name as string
$function(); # invoke
Anonymous Function as Variable Function Call:
$function = function() {}; # define function
$function(); # invoke
See as well Variable Functions­Docs.