CODEWITHSUNDEEP
phpparameter-passingoptional-parameters
xDaizu
xDaizu

Reputation: 1061

Behaviour when declaring non-optional parameter after optional parameter

Let's define this function:

function fooFunction($a, $b='foo', $c){}

If I call it like: 

$foo = fooFunction("bar", "buzz");

...will 'buzz' be asigned to $c or to $b?

Upvotes: 0

Views: 523

Answers (1)

deceze
deceze

Reputation: 522636

The presence or absence of default values for a parameter does not in any way influence what arguments will be assigned to what parameters; the first passed argument always goes to the first parameter, the second to the second and so on. If no value was passed for a parameter and it has a default value, its default value is used instead.

In PHP <7.1, not supplying an argument for parameters without default value merely produced a warning. The parameter would then be undefined inside the function:

function fooFunction($a, $b='foo', $c) {
    var_dump($c);
}

$foo = fooFunction("bar", "buzz");

Warning: Missing argument 3 for fooFunction()

Notice: Undefined variable: c
NULL

Since PHP 7.1 it finally acts sane and throws an exception:

Fatal error: Uncaught ArgumentCountError:
  Too few arguments to function fooFunction(),
  2 passed and exactly 3 expected
Stack trace:
#0 fooFunction('bar', 'buzz')

It's still insane that you can define parameters without default value after parameters with, since in practice that makes little sense; perhaps PHP is counting on you not to do that precisely because there's no practical purpose.

Upvotes: 1

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