将函数的引用作为参数传递

Question

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

<?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);

Answer 1

Change your $product = line to:

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

And change your calling line to:

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