Using Default Arguments in a Function

Question

I am confused about default values for PHP functions. Say I have a function like this:

function foo($blah, $x = "some value", $y = "some other value") {
    // code here!
}

What if I want to use the default argument for $x and set a different argument for $y?

I have been experimenting with different ways and I am just getting more confused. For example, I tried these two:

foo("blah", null, "test");
foo("blah", "", "test");

But both of those do not result in a proper default argument for $x. I have also tried to set it by variable name.

foo("blah", $x, $y = "test");

I fully expected something like this to work. But it doesn't work as I expected at all. It seems like no matter what I do, I am going to have to end up typing in the default arguments anyway, every time I invoke the function. And I must be missing something obvious.

You're probably not missing anything, the PHP developers probably missed this. — Matti Virkkunen
That's an interesting question. I've never actually had the need to do this as you've stated since I only use default args to expand functions inherited by other programmers when I am not sure what depends on the original. I'm curious about an answer. — Kai Qing
One of the best questions I've seen in a while! Have you tried passing nothing? foo("blah", , "test");? — Madara's Ghost
The rule is that default arguments can only follow arguments. That is you could have foo("blah") and x,y are default, or foo("Blah","xyz") and y is the default. — Mouse Food
Expected behavior. See "Default argument values" section in: php.net/manual/en/functions.arguments.php -- null value and empty string are accepted as actual arguments to the function — jmdavalos

drew010 drew010 · Accepted Answer · 2012-02-06T20:40:12

I would propose changing the function declaration as follows so you can do what you want:

function foo($blah, $x = null, $y = null) {
    if (null === $x) {
        $x = "some value";
    }

    if (null === $y) {
        $y = "some other value";
    }

    code here!

}

This way, you can make a call like foo('blah', null, 'non-default y value'); and have it work as you want, where the second parameter $x still gets its default value.

With this method, passing a null value means you want the default value for one parameter when you want to override the default value for a parameter that comes after it.

As stated in other answers,

default parameters only work as the last arguments to the function. If you want to declare the default values in the function definition, there is no way to omit one parameter and override one following it.

If I have a method that can accept varying numbers of parameters, and parameters of varying types, I often declare the function similar to the answer shown by Ryan P.

Here is another example (this doesn't answer your question, but is hopefully informative:

public function __construct($params = null)
{
    if ($params instanceof SOMETHING) {
        // single parameter, of object type SOMETHING
    } elseif (is_string($params)) {
        // single argument given as string
    } elseif (is_array($params)) {
        // params could be an array of properties like array('x' => 'x1', 'y' => 'y1')
    } elseif (func_num_args() == 3) {
        $args = func_get_args();

        // 3 parameters passed
    } elseif (func_num_args() == 5) {
        $args = func_get_args();
        // 5 parameters passed
    } else {
        throw new \InvalidArgumentException("Could not figure out parameters!");
    }
}

Using Default Arguments in a Function

14 Answers

UPDATE: