PHP Functions, null default parameters

Question

I have a PHP function signature that looks like this, where I want the 4th param to be NULL by default if nothing is supplied.

testfunction($param1, $param2, $param3, $param4=NULL) {
    //do stuff
}

I can also call using 4 params as such

x = testfunction('100', 'abc', 'xxx', null);

Why is it that I can also call the function using 3 parameters without any errors?

x = testfunction('100', 'abc', 'xxx');

Is what i'm doing even correct (using 3 params)? Any thoughts about how this relates to traditional method overloading where separate function signatures are defined?

My question is general in nature. Hope someone can shed some light on and around it.

You have not to pass parameters which have a default value. So what you're doing is correct. — Jérôme
– Jérôme, Commented May 5, 2014 at 9:51
because the 4th parameter is optional. you may be interested in this SO question stackoverflow.com/questions/4697705/php-function-overloading — bansi
– bansi, Commented May 5, 2014 at 9:52

Zander Rootman · Accepted Answer · 2014-05-05 10:02:08Z

9

When you declare a function like this:

testfunction($param1, $param2, $param3, $param4=NULL) {
    //do stuff
}

You're telling PHP that $param4 already has a value, so when that function is called, a value is already assigned to it. It doesn't expect you to send an argument for that parameter because a default one has already been assigned.

It therefore knows the function will be able to operate (to a certain extent) as it should. It's useful for making optional parameters.

However when you declare a function as this:

testfunction($param1, $param2, $param3, $param4) {
    //do stuff
}

PHP Expects you to send through that argument ($param4), because without it, the function won't be able to accomplish what you've set it out to do, at least so PHP assumes, because you would never create a function with 4 parameters where not one of them are used in the function body.

Make sense?

answered May 5, 2014 at 10:02

Zander Rootman

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1 Comment

bansi Over a year ago

in php you can have a function like function test(){} and call it as test('arg1',arg2)`. function signature is just the function name.

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