PHP Default Parameters Overwriting Passed Values

Question

I have a function like this:

protected function example($param = null)
{
}

and when I call it like this:

$this->example("string");

or like this: $string = "string"; $this->example("string");

the parameter value is still null. What am I doing wrong?

I'd suggest putting in a var_dump and tracing through your code to make sure it is doing what you expect it to and for sanity reasons. Make sure you're calling the right file as well. — Newbi3
– Newbi3, Commented Sep 22, 2015 at 21:22
Can you show a full example that shows this issue? Are you sure you are calling that function? This looks to be in a class (because of the protected keyword), so you'd need $this->example("string");. — gen_Eric
– gen_Eric, Commented Sep 22, 2015 at 21:23
example() is not the same as $this->example() or $class->example() in OOP — scrowler
– scrowler, Commented Sep 22, 2015 at 21:24
Newbi3, I am actually calling the function from inside the same class. I have had it write the $param value to a file, and the value is always whatever I set the default to, never the value I pass in. — amorrise
– amorrise, Commented Sep 22, 2015 at 21:40

ngen · Accepted Answer · 2015-09-22 21:25:05Z

1

Try returning the variable.

protected function example($param = null)
{
    return $param;
}

answered Sep 22, 2015 at 21:25

ngen

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Comments

Cameron Spanos · Accepted Answer · 2015-09-22 21:27:02Z

0

Having the parameter set to NULL only assigns the variable $param to NULL when no parameter is passed.

like when you call it like this $this->example();

Otherwise $param will be set to whatever you pass it.

answered Sep 22, 2015 at 21:27

Cameron Spanos

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