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I have a PHP class like so:

<?php
class MyClass {
    public $my_variable = array();

    public function func1() {
        $var1 = $this->my_variable;
        array_push($var1, 'var1');
        return $this->my_variable;
    }

    public function func2() {
        $var2 = $this->my_variable;
        array_push($var2, 'var2');
        return $this->my_variable;
    }
}

$my_class = new MyClass;

print_r($my_class->func1());
print_r($my_class->func2());
?>

The two print_r functions return an empty array, and there are no errors displayed.

How can I get the "var1" and "var2" strings added to the $my_variable array? I'm not sure where I am going wrong here...!

Thanks.

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2 Answers

up vote 2 down vote accepted

$var1 = $this->my_variable actually creates a copy of the array, which you then push a value onto.

Instead, you can do this: $var1 = &$this->my_variable to create a reference instead, but it would just be better to not have the pointless variable at all:

public function func1() {
    $this->my_variable[] = 'var1';
    return $this->my_variable;
}
public function func2() {
    $this->my_variable[] = 'var2';
    return $this->my_variable;
}

Or, more appropriately:

public function add($value) {
    $this->my_variable[] = $value;
    return $this->my_variable;
}
// call with `$my_class->add('var1'); $my_class->add('var2');
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People use such pointless variables to shorten the $var. – bwoebi Apr 21 at 20:46
1  
@bwoebi: People interested in short code don't use array_push... :P – rynah Apr 21 at 20:48
@minitech People who don't know about [] use array_push ^^ – bwoebi Apr 21 at 20:54
I dont think I've ever used array_pop, ever, in 10ish years of writing php. But I also find that people who dont even know about array_push (regardless if they use it) also never know about array_pop and array_shift and array_unshift – Uberfuzzy Apr 21 at 21:24
up vote 0 down vote

You have to assign the $var's by reference. You copy the array and then add to the copy some entry and then return the initial array.

$var2 = &$this->my_variable;

would be right. The & is marking here a reference.

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