PHP closure scope problem

Question

Apparently $pid is out of scope here. Shouldn't it be "closed" in with the function? I'm fairly sure that is how closures work in javascript for example.

According to some articles php closures are broken, so I cannot access this?

So how can $pid be accessed from this closure function?

class MyClass {
  static function getHdvdsCol($pid) {
    $col = new PointColumn();
    $col->key = $pid;
    $col->parser = function($row) {
        print $pid; // Undefined variable: pid
    };
    return $col;
  }
}

$func = MyClass::getHdvdsCol(45);
call_user_func($func, $row);

Edit I have gotten around it with use: $col->parser = function($row) use($pid). However I feel this is ugly.

It is not ugly, it is how closures work in php

zerkms
– zerkms

2011-07-01 03:56:26 +00:00
Commented Jul 1, 2011 at 3:56 — zerkms
– zerkms, Commented Jul 1, 2011 at 3:56
that's just how it is in php so..

tradyblix
– tradyblix

2011-07-01 03:57:47 +00:00
Commented Jul 1, 2011 at 3:57 — tradyblix
– tradyblix, Commented Jul 1, 2011 at 3:57
It's an ugly language, but I'm stuck with it. :(

Benbob
– Benbob

2011-07-01 04:00:08 +00:00
Commented Jul 1, 2011 at 4:00 — Benbob
– Benbob, Commented Jul 1, 2011 at 4:00
It becomes less and less ugly every day. (:

rsk82
– rsk82

2011-08-07 06:09:46 +00:00
Commented Aug 7, 2011 at 6:09 — rsk82
– rsk82, Commented Aug 7, 2011 at 6:09
From PHP 5.4 you can access $this within the closure.

jgivoni
– jgivoni

2012-06-28 21:41:04 +00:00
Commented Jun 28, 2012 at 21:41 — jgivoni
– jgivoni, Commented Jun 28, 2012 at 21:41

zerkms · Accepted Answer · 2011-07-01 03:53:32Z

24

You need to specify which variables should be closed in this way:

function($row) use ($pid) { ... }

answered Jul 1, 2011 at 3:53

zerkms

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4 Comments

Josh Nankin Over a year ago

If this closure is saved in a variable and then passed to another scope where $pid does not exist, will pid be wrapped in the closure somehow, or not accessible?

zerkms Over a year ago

@Josh Nankin: actually use($var,...) is how closures are implemented in php. So if you pass the anonymous function to another variable scope - the $pid variable will contain the last defined value.

Matjaz Muhic Over a year ago

Does this "use" thingy feel just weird or is it just me? I just can't get over it after using closures in other languages. :/

zerkms Over a year ago

@Matjaz Muhic: yep, it's weird for everyone not just you

Knarf · Accepted Answer · 2015-04-28 14:20:33Z

0

You can use the bindTo method.

class MyClass {
  static function getHdvdsCol($pid) {
    $col = new PointColumn();
    $col->key = $pid;
    $parser = function($row) {
        print $this->key;
    };
    $col->parser = $parser->bindTo($parser, $parser);
    return $col;
  }
}

$func = MyClass::getHdvdsCol(45);
call_user_func($func, $row);

answered Apr 28, 2015 at 14:20

Knarf

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Comments

Savas Alp · Accepted Answer · 2015-04-29 15:06:54Z

-4

I think PHP is very consistent in scoping of variables. The rule is, if a variable is defined outside a function, you must specify it explicitly. For lexical scope 'use' is used, for globals 'global' is used.

For example, you can't also use a global variable directly:

$n = 5;

function f()
{
    echo $n; // Undefined variable
}

You must use the global keyword:

$n = 5;

function f()
{
    global $n;
    echo $n;
}

answered Apr 29, 2015 at 15:06

Savas Alp

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