CODEWITHSUNDEEP
php
Dan
Dan

Reputation: 454

Problem with optional arguments in PHP

I have a function which takes four optional arguments:

public function doSomething($x = null, $y = null, $a = null, $b = null) {  }

However when I try to call this function and specify only $y for instance:

$object->doSomething($y=3)

It seems to ignore that I am setting $y as 3, and instead sets $x to be 3. Is there any reason why this might happen with PHP? I never used to have this issue before...

Thanks,

Dan

Upvotes: 2

Views: 164

Answers (2)

BoltClock
BoltClock

Reputation: 724542

You must pass arguments in the order that you declare in your method signature, whether they're optional or not. That means you must specify $x before $y, no matter what.

If you don't need to pass any other value of $x, you'll have to pass null. You can still skip the remaining optional arguments, of course:

$object->doSomething(NULL, 3)

Additionally, PHP does not support named arguments. Therefore, you cannot explicitly write $y in the calling code because in PHP that actually sets $y within the scope of the calling code, and not the scope of doSomething()'s method body.

EDIT: per dpk's suggestion, as an alternative route you can change the method to accept a hash (associative array) instead, and optionally overwrite some defaults and extract() the values of it into scope:

public function doSomething(array $args = array()) {
    $defaults = array('x' => NULL, 'y' => NULL, 'a' => NULL, 'b' => NULL);
    extract(array_merge($defaults, $args));

    // Use $x et al as normal...
}

In your calling code:

$object->doSomething(array('y' => 3));

Upvotes: 6

Ben
Ben

Reputation: 2992

Even though $x is optional, the position is not so $y needs to be the second parameter. Try:

$object->doSomething(null,3);

Upvotes: 1

Related Questions