php cannot pass function as argument?

Question

Works: returns [false,null]

array_filter([1, 2, 3, false, null], function($value) {
   return empty($value);
});

Does not Work:

array_filter([1, 2, 3, false, null], empty); // syntax error, unexpected ')',expecting '('

Why cant i just pass in the function as an argument?

Answer 1

What is empty meant to be in your second example?

Note that php's "empty" is not a function, but a language construct. Therefore you cannot use it as a function. That is explained in the documentation: http://php.net/manual/en/function.empty.php (Scroll down to the "Notes" section...)

That explains why the first variant does work, whilst the second results in a syntax error.

Answer 2

Functions can be passed into any function/method that takes a callable argument. However, empty (), in spite of how it's written in code, is not a function. It's a language construct.

If you want to pass empty as a callable, you have no choice but to wrap it in a user-defined function.

$func = function ($param) { return empty ($param); };

print_r (array_filter ([1,2, NULL, 3, 4, FALSE, 5, 6, 0, "", 7, 8, 9], $func));

or

print_r array_filter (([1,2, NULL, 3, 4, FALSE, 5, 6, 0, "", 7, 8, 9], function ($param) { 
    return empty ($param); 
}));

Other language constructs are affected as well, so you might want to consult the manual for a list of language constructs (all the ones that terminate with () like functions are things you can't use with functions that take a callable argument).

http://php.net/manual/en/reserved.keywords.php

Answer 3

empty() is not a function, it's a language construct. array_filter expects a callable as the second argument. You will need to use a wrapper function exactly as you have in your first example:

array_filter([1, 2, 3, false, null], function($value) {
   return empty($value);
});