In PHP, when passing a parameter by reference and then assigning it to NULL, the reference is

While working with passing parameters by reference to an object method, I noticed a peculiar behavior:

class Test
{
    private $value;
    public function Set($value)
    {
        $this->value = $value;
    }

    public function Get(&$ref)
    {
        $ref = &$this->value; //Assigning reference parameter to the value of this object
    }
}

$test = new Test();
$test->Set('test');
$test->Get($value1);

var_dump($value1); //Returns NULL instead of 'test'!

*Update: For clarification purposes, changing the function name from GetByRef(...) to Get(...)

*Update 2: Forgot to mention a real test case scenario where I encountered some difficulties:

$test->Get($value1);
$test->Get($value2);

$value1 = 'Another test value';
echo $value2; //Expected output is 'Another test value';

The issue arises because $value2 does not know whether $value1 has been initialized or not. Hence, standard assignment like $value2 = &$value1 will not work in this situation.

Answer №1

When you assign to a reference by reference, you end up with null. To avoid this issue, simply assign the value normally:

public function RetrieveDataByRef(&$ref) {
    $ref = $this->data;
}

By having &$ref in the method signature and then calling the method, a variable is created in the calling scope with an initial value of null. This variable is then referenced as $ref within the method. When you do $ref = &$this->data, you are essentially creating a new reference that replaces the existing one. Using =& always creates a new reference variable; if you want to update its value instead, you must use = to assign to it. Therefore, the variable in the calling scope remains unchanged at its original value of null, breaking its reference inside the method.

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