PHP: passing a function with parameters as parameter
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by Oden
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Published on 2010-06-05T10:57:57Z
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2010/06/05
11:02 UTC
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Hey,
I'm not sure that silly question, but I ask:
So, if there is an anonymous function I can give it as another anonymous functions parameter, if it has been already stored a variable.
But, whats in that case, if I have stored only one function in a variable, and add the second directly as a parameter into it? Can I add parameters to the non-stored function?
Fist example (thats what i understand :) ):
$func = function($str){ return $str; };
$func2 = function($str){ return $str; };
$var = $func($func2('asd'));
var_dump($var);
// prints out string(3) "asd"
That makes sense for me, but what is with the following one?
$func = function($str){ return $str; };
$var = $func(function($str = "asd"){ return $str; });
var_dump($var);
/** This prints out:
object(Closure)#1 (1) {
["parameter"]=>
array(1) {
["$str"]=>
string(10) ""
}
}
But why?
*/
And at the end, can someone recommend me a book or an article, from what i can learn this lambda coding feature of php?
Thank you in advance for your answers :)
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