How can I initialize an array's argument to its default value? [duplicate]

Question

Here is my code:

function myfunc( $arg1, $arg2 = 'sth', $arg3 = 'sth else' ){
    // do stuff
}

Now I want to set a new value for arg3 and keep the default defined value for arg2. How can I do that?

Something like this:

myfunc( true, <keep everything defined as default>, $arg3 = 'new value' );

The expected result is sth in this fiddle.

as per your fiddle you haven't defined $arg2 before calling that function — Maitray Suthar
– Maitray Suthar, Commented Jul 5, 2017 at 6:05
your question is not clear please add some more explanation. — Yaseen Ahmad
– Yaseen Ahmad, Commented Jul 5, 2017 at 6:05
@YaseenAhmed I want to set a value for arg3 without setting any value for arg2. As you see both arg3 and arg2 are optional. And I want to use the defined default value for arg2. — stack
– stack, Commented Jul 5, 2017 at 6:07
you can use array as argument and set the value inside function if not found. — rahul singh
– rahul singh, Commented Jul 5, 2017 at 6:07

Md. Abutaleb · Accepted Answer · 2017-07-05 06:05:43Z

3

A possible alternative would not have your function take 3 parameters, but only one, an array:

function my_function(array $value = array()) {
    // if set, use $value['key1']
    // if set, use $value['key2']
    // ...
}

And call that function like this:

my_function(array(
    'key1' => 'google',
    'key2' => 'yahoo'
));

This would allow you to:

Hope it will helpful.

answered Jul 5, 2017 at 6:05

Md. Abutaleb

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