Closure with parameter inside array

Question

I have problem with closure in PHP 5.4

I have array

public function check(){
return ['int'=>['filter'=>2],
'min'=>function($val){
return ['int'=>2,'min'=>$val];
}
]
}

When I use

(new Obj())->check()['int'];

it works. But I don't know how use min with parameter for example 3

I tryed

(new Obj())->check()['min'](3);
(new Obj())->check()['min'(3)];
(new Obj())->check()['min(3)'];

don't work.

The most valid way is (new Obj())->check()['min'](3); - what is your result ? Any error ? — hsz
– hsz, Commented Jul 15, 2013 at 9:16
@hsz: That would be valid and correct in a language with a formal grammar. PHP is not that language. — Jon
– Jon, Commented Jul 15, 2013 at 9:20
Parse error: syntax error, unexpected '(' Buy in real enviroment I use (new Valid())->checkVal()['min'](3); — 110precent
– 110precent, Commented Jul 15, 2013 at 9:21
@110precent Even error message says that this is wrong syntax for php. Also, I hope you don't use stuff like this in real-world projects. 1. This really hurts eyes (and might hurt brain). 2. Syntax may change from version to version and the first thing that may become invalid is this kind of structure. — Leri
– Leri, Commented Jul 15, 2013 at 9:25

Jon · Accepted Answer · 2013-07-15 09:18:37Z

3

PHP's parser is simply not up to the task, so you can't write this expression as you could have in other languages. You will have to use a workaround, for example:

call_user_func((new Obj())->check()['min'], 3));

Or alternatively:

$f = (new Obj())->check()['min'];
$f(3);

answered Jul 15, 2013 at 9:18

Jon

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2 Comments

hsz Over a year ago

What about ((new Obj())->check()['min'])(3) ?

Leri Over a year ago

@hsz still invalid syntax.

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