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PHP PHP Functions Function Returns and More PHP Closures

Posted by Jordan Olson
Jordan Olson
6,155 Points

Why use 'use' when we can just pass the variable into the actual function?

<?php

$name = 'Mike';

$greet = function($name){ echo "Hello, $name!"; };

$greet();

VS

<?php

$name = 'Mike';

$greet = function() use($name){ echo "Hello, $name!"; };

$greet();

1 Answer

Kevin Korte
Kevin Korte
28,149 Points
Kevin Korte
Kevin Korte
28,149 Points

Because you can't always pass in all the variables you need into a function.

First, there is a small error to correct.

It should be

$name = 'Mike';

$greet = function($name){ echo "Hello, $name!"; };

$greet($name);

Otherwise, $greet doesn't know what you're passing to it.

With that side, this is using the concept of closure. The second example, is passing a copy of the variable from outside it's scope into the variable.

Take for instance, you wanted to use array_walk(). http://php.net/manual/en/function.array-walk.php The callback typically only allows 2 parameters. If you need more than two parameters passed into the function, you could use the use closure to pass a copy of those variables in so you could do whatever you needed to do.

I think this is helpful: http://stackoverflow.com/questions/10692817/whats-the-difference-between-closure-parameters-and-the-use-keyword