CODEWITHSUNDEEP
phpreferencecallback
Steven Oxley
Steven Oxley

Reputation: 6723

Is it possible to pass parameters by reference using call_user_func_array()?

When using call_user_func_array() I want to pass a parameter by reference. How would I do this. For example

function toBeCalled( &$parameter ) {
    //...Do Something...
}

$changingVar = 'passThis';
$parameters = array( $changingVar );
call_user_func_array( 'toBeCalled', $parameters );

Upvotes: 50

Views: 20679

Answers (5)

Dharman
Dharman

Reputation: 33305

Yes, it is possible but you have to call the function directly. Don't use call_user_func_....

<?php

function toBeCalled( &$parameter ) {
    //...Do Something...
}

$changingVar = 'passThis';

toBeCalled($changingVar);

And if you have a list of parameters you can use the spread operator (available as of PHP 5.6):

$parameters = array( $changingVar );
toBeCalled(...$parameters);

If you still need to execute code on an older version of PHP, you must use the hacks from the other answers.

Upvotes: 0

Danny
Danny

Reputation: 1

This works by double referencing,the original variable is modified when the $parameter variable is modified.

$a = 2;
$a = toBeCalled($a);
echo $a //50

function toBeCalled( &$par_ref ) {
    $parameter = &$par_ref;
    $parameter = $parameter*25;
}

Upvotes: -1

Sophia_ES
Sophia_ES

Reputation: 1361

Directly, it may be impossible -- however, if you have control both over the function you are implementing and of the code that calls it - then there is one work-around that you might find suitable.

Would you be okay with having to embed the variable in question into an object? The code would look (somewhat) like this if you did so.

function toBeCalled( $par_ref ) {
    $parameter = $par_ref->parameter;
    //...Do Something...
    $par_ref->parameter = $parameter;
}

$changingVar = 'passThis';
$parembed = new stdClass; // Creates an empty object
$parembed->parameter = array( $changingVar );
call_user_func_array( 'toBeCalled', $parembed );

You see, an object variable in PHP is merely a reference to the contents of the object --- so if you pass an object to a function, any changes that the function makes to the content of the object will be reflected in what the calling function has access to as well.

Just make sure that the calling function never does an assignment to the object variable itself - or that will cause the function to, basically, lose the reference. Any assignment statement the function makes must be strictly to the contents of the object.

Upvotes: 1

voidstate
voidstate

Reputation: 7990

Except you are using deprecated functionality here. You'll generate a warning in PHP5 making it less than perfect.

Warning: Call-time pass-by-reference has been deprecated; If you would like to pass it by reference, modify the declaration of runtime function name. If you would like to enable call-time pass-by-reference, you can set allow_call_time_pass_reference to true in your INI file in ...

Unfortunately, there doesn't appear to be any other option as far as I can discover.

Upvotes: -3

Steven Oxley
Steven Oxley

Reputation: 6723

To pass by reference using call_user_func_array(), the parameter in the array must be a reference - it does not depend on the function definition whether or not it is passed by reference. For example, this would work:

function toBeCalled( &$parameter ) {
    //...Do Something...
}

$changingVar = 'passThis';
$parameters = array( &$changingVar );
call_user_func_array( 'toBeCalled', $parameters );

See the notes on the call_user_func_array() function documentation for more information.

Upvotes: 57

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