How to use PHP closure with predefined arguments / parameters

Question

What I am trying to achieve is this:

$x = 5;
$b = function ($x) {
    echo 'This should be 5 :' . $x;
};
function a($fn){
    echo 'In a ';
    $fn();
}
a($b);

That when you run this code, we get

In a

This should be 5 :5

What we get instead is

Warning: Missing argument 1 for {closure}(), called in writecodeonline.com/php on line 10 and defined on line 3 This should be 5

I don't want to redefine the argument I already defined

What I don't want neither is hide the $x. I don't want to change its visibility.

Is there a way for this?

fejese · Accepted Answer · 2014-12-19 23:45:18Z

3

Read the docs: you can use the use expression for it:

$x = 5;

$b = function () use ($x) {
    echo 'This is x: ' . $x . "\n";
};

$b();

$c = function ($fn) {
    echo 'In c: ';
    $fn();
};

$c($b);
$x = 10;
$c($b);

Output:

This is x: 5
In c: This is x: 5
In c: This is x: 5

Note that despite the change of $x later the assigned value does not get changed. You can achieve that if you pass the variable by reference:

$x = 5;

$b = function () use (&$x) { // << Note the difference here
    echo 'This is x: ' . $x . "\n";
};

$b();

$c = function ($fn) {
    echo 'In c: ';
    $fn();
};

$c($b);
$x = 10;
$c($b);

Output:

This is x: 5
In c: This is x: 5
In c: This is x: 10

answered Dec 19, 2014 at 23:45

fejese

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Comments

Mark Baker · Accepted Answer · 2014-12-19 23:37:17Z

1

Read about variable scope, $x cannot be automagically visible inside function a() unless you pass the argument into function a()

$x = 5;

$b = function ($x) {
    echo 'This should be 5 :' . $x;
};

function a($fn, $value) {
    echo 'In a ';
    $fn($value);
}

a($b, $x);

answered Dec 19, 2014 at 23:37

Mark Baker

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2 Comments

Toskan Over a year ago

this would certainly work, but you make the microing of $value through your whole application to get it to work.

Mark Baker Over a year ago

microing? If you need to pass an argument to your function $b(), then it must be passed to $b() at the point where $b() is called, not simply where it is defined. You're calling $b() from within function a(), so the argument must exist within the scope of a()

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