PHP: How to pass instance variable to closure?

Question

If you want to use a closure from within a class, how do you pass in an instance variable from that class?

class Example {
    private $myVar;
    public function test() {
        $this->myVar = 5;
        $func = function() use ($this->myVar) { echo 'myVar is: ' . $this->myVar; };

        // The next line is for example purposes only if you want to run this code.
        // $func is actually passed as a callback to a library, so I don't have
        // control over the actual call.
        $func();
    }
}
$e = new Example();
$e->test();

PHP doesn't like this syntax:

PHP Fatal error:  Cannot use $this as lexical variable in example.php on line 5

If you take off $this-> then it can't find the variable:

PHP Notice:  Undefined variable: myVar in example.php on line 5

If you use use (xxx as $blah) as suggested in some places, it seems invalid syntax whether you have $this or not:

PHP Parse error:  syntax error, unexpected 'as' (T_AS), expecting ',' or ')' in example.php on line 5

Is there a way to do this? The only way I can get it to work is with a dodgy workaround:

$x = $this->myVar;
... function() use ($x) { ...

Why do you think it's dodgy? As it says, you can't use $this as a lexical variable, so you need to use another variable. That's what $x is. — Barmar
– Barmar, Commented May 23, 2014 at 1:25
It's dodgy because it adds an extra copy operation. I have to copy the variable into $x, then $x gets copied a second time into the closure. If I could do something like function () use ($this->myVar as $x) then it would avoid that extra copy and make the code neater. — Malvineous
– Malvineous, Commented May 23, 2014 at 1:27
PHP uses lazy copying, it initially just uses references, copying only when one of them is modified. — Barmar
– Barmar, Commented May 23, 2014 at 1:33

Felix Kling · Accepted Answer · 2014-05-23 02:30:21Z

13

If you are using PHP 5.4 or later, then you can use $this directly inside the closure:

$func = function() {
  echo 'myVar is: ' . $this->myVar;
};

answered May 23, 2014 at 2:30

Felix Kling

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1 Comment

Tom Jenkinson Over a year ago

It looks like this also includes the parent:: variable in addition to $this, to access the parent class of the object $this (if there is one).

Barmar · Accepted Answer · 2014-05-23 02:07:54Z

13

You can use your workaround. You can also be more general:

$self = $this;
$func = function() use ($self) {
    echo "myVar = " . $self->myVar;
};

Within the closure you can access any public properties or methods using $self instead of $this.

But it won't work for the original question, because the variable in question is private.

edited May 23, 2014 at 2:07

answered May 23, 2014 at 1:28

Barmar

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2 Comments

Malvineous Over a year ago

voodoo417 makes a good point - this will only work with public variables, not private ones.

Barmar Over a year ago

Didn't think of that. I've updated my answer to reflect this limitation.

voodoo417 · Accepted Answer · 2014-05-23 02:21:23Z

2

Why not ?:

    $func = function($param){ echo 'myVar is: ' . $param; };
    $func($this->myVar);

Update:

@Bramar right. But, if only $myVar will be public, it will work. Closures have no associated scope , so they cannot access private and protected members. In you specific case, you can do:

    $this->myVar = 5;   
    $_var = $this->myVar;
    $func = function() use ($_var) { echo 'myVar is: ' . $_var; };
    $func();

edited May 23, 2014 at 2:21

answered May 23, 2014 at 1:22

voodoo417

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1 Comment

Malvineous Over a year ago

Sorry, $func() was just for example purposes. I am really passing this function to a library as a callback, so I have no control over the actual calling of my function.

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