在函数中使用默认参数

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.

Answer 1

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
    } else if (is_string($params)) {
        // single argument given as string
    } else if (is_array($params)) {
        // params could be an array of properties like array('x' => 'x1', 'y' => 'y1')
    } else if (func_num_args() == 3) {
        $args = func_get_args();

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