php function arguments with a variable

Question

Is there a way to set up default function arguments with variables in them such as

function example($test = 'abc', $image = $scripturl . 'abc.png')
{
        // ...
}

The reason I want to do this is because I have sources that I have global variable settings with the path set up so that it is easy to grab when I need to include a css file or image.

The code above gives an unexpected T_VARIABLE...

zerkms · Accepted Answer · 2012-04-04 22:09:16Z

4

Default values should be a constant. They should have the value, that is already available in compile time, not in a runtime.

answered Apr 4, 2012 at 22:09

zerkms

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1 Comment

MLM Over a year ago

What I am proposing would be constant at compile since those settings would only change if the sources moved. I get your point but is there any way to do something the way I am proposing? I can always add in the variable in the function though... I think I will go with @ThiefMaster solution

ThiefMaster · Accepted Answer · 2012-04-05 23:01:11Z

3

That's not possible; default arguments must be constant.

You can easily emulate it like this though:

function example($test = 'abc', $image = null) {
    if($image === null) {
        global $scripturl;
        $image = $scripturl . 'abc.png';
    }
}

edited Apr 5, 2012 at 23:01

answered Apr 4, 2012 at 22:10

ThiefMaster

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3 Comments

Eric Cope Over a year ago

some OOP may be useful. Make $scripturl an object variable set upon construction, then it could be $this->scripturl.

MLM Over a year ago

Ye, I guess this would be the only solution and to use it in the function :/ - Thanks for the reply!

craniumonempty Over a year ago

don't forget global $scripturl; for external variables that aren't superglobals

Marc B · Accepted Answer · 2012-04-04 22:10:14Z

1

No. Default values in function arguments must be constant values, not the results of expressions: http://php.net/manual/en/functions.arguments.php#functions.arguments.default

At the point the arguments are parsed, $scripturl will not exist anyways, and would always come out as NULL, since you couldn't make it global before it's used.

answered Apr 4, 2012 at 22:10

Marc B

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Comments

poncha · Accepted Answer · 2012-04-04 22:10:57Z

1

well, as the error already stated, you cannot use a variable as (part of) default value in function signature.

what you can do however, is pass some known illegal value (null for instance) and then check inside the function and assign if needed:

function example($test = 'abc', $image = null)
{
   global $scripturl;
   if($image === null) $image = $scripturl . 'abc.png';
   ...
}

answered Apr 4, 2012 at 22:10

poncha

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