Passing an Array as Arguments, not an Array, in PHP


I seem to remember that in PHP there is a way to pass an array as a list of arguments for a function, dereferencing the array into the standard func($arg1, $arg2) manner. But now I'm lost on how to do it. I recall the manner of passing by reference, how to "glob" incoming parameters ... but not how to de-list the array into a list of arguments.

It may be as simple as func(&$myArgs), but I'm pretty sure that isn't it. But, sadly, the manual hasn't divulged anything so far. Not that I've had to use this particular feature for the last year or so.

6/23/2012 12:26:05 PM

Accepted Answer

4/13/2009 3:02:05 PM

As has been mentioned, as of PHP 5.6+ you can (should!) use the ... token (aka "splat operator", part of the variadic functions functionality) to easily call a function with an array of arguments:

function variadic($arg1, $arg2)
    // Do stuff
    echo $arg1.' '.$arg2;

$array = ['Hello', 'World'];

// 'Splat' the $array in the function call

// 'Hello World'

Note: array items are mapped to arguments by their position in the array, not their keys.

As per CarlosCarucce's comment, this form of argument unpacking is the fastest method by far in all cases. In some comparisons, it's over 5x faster than call_user_func_array.


Because I think this is really useful (though not directly related to the question): you can type-hint the splat operator parameter in your function definition to make sure all of the passed values match a specific type.

(Just remember that doing this it MUST be the last parameter you define and that it bundles all parameters passed to the function into the array.)

This is great for making sure an array contains items of a specific type:


// Define the function...

function variadic($var, SomeClass ...$items)
    // $items will be an array of objects of type `SomeClass`

// Then you can call...

variadic('Hello', new SomeClass, new SomeClass);

// or even splat both ways

$items = [
    new SomeClass,
    new SomeClass,

variadic('Hello', ...$items);

Licensed under: CC-BY-SA with attribution
Not affiliated with: Stack Overflow