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By passing a reference to a function as an argument

How can I pass a function reference to another function as an argument? I was trying to do a callback and I need to pass the reference to the function returnProduct

before. How to do it?

<?php
class Tester {
    public function calculate($var_1,$var_2,$var_3) {
        $product = var_3($var_1,$var_2);
        echo $product;
    }

    public function returnProduct($var_1,$var_2) {
        return $var_1*$var_2;   
    }
}

$obj = new Tester();
$obj->calculate(100,2,$obj->returnProduct);
+3


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2 answers


Change the line $product =

to:

$product = call_user_func($var_3,$var_1,$var_2);


And change your phone line to:

$obj->calculate(100,2,array($obj,'returnProduct'));
+3


source


If you only wanted to use a method Tester

, you can pass the method name as a string, for example

public function calculate($var_1, $var_2, $var_3) {
    $product = $this->$var_3($var_1, $var_2);
    echo $product;
}

Then call it



$obj->calculate(100, 2, 'returnProduct');

To err on the side of caution, you can check if a method exists using the aptly named method_exists()

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