What happens when I return a variable reference, then try to modify it?

return a reference

<?php

// what happens when I return a reference to a variable, then try to modify it?

class ReferenceOrValue
{
    private $variable;
    
    public function __construct($value)
    {
        $this->variable = $value;
    }

    public function method()
    {
        $variable =& $this->variable;
        
        return $variable;
    }
    
    public function getValue()
    {
        return $this->variable;
    }
}

$value = 'Hello World!';
$object = new ReferenceOrValue($value);
$return = $object->method();

echo "Return should be '{$value}'.<br />"; // Return should be 'Hello World!'.
echo "Return is '{$return}'.<br />"; // Return is 'Hello World!'.

// if return is reference . . .
$before = $return;
$return = 'Hello Mars!';
$after = $object->getValue();
echo "Is '{$before}' the same as '{$after}'?"; // Is 'Hello World!' the same as 'Hello World!'?

// answer is nothing.

Hello, you left ##php before I could reply to your query, I suggest you look at this man page and modify your code accordingly:

http://php.net/manual/en/language.references.return.php

Actually, the sample did exactly what I wanted it to do. Within the method, I wanted the variable to be a reference of a class variable, but after returning, I did not want the variable to reference the class variable. The reason I create a variable reference is that I wanted a shorthand inside the method. But I don't want a reference after the method returns because I want to protect class variables. Thanks though! That is a good reference from the php manual!

…

On Sat, Oct 15, 2011 at 11:11 PM, Yousif Masoud < ***@***.***>wrote: Hello, you left ##php before I could reply to your query, I suggest you look at this man page and modify your code accordingly: http://php.net/manual/en/language.references.return.php ## Reply to this email directly or view it on GitHub: https://gist.github.com/1290560