[RESOLVED] passing values to a function (and defaults)

[kansaschuck]

Do I have this right? My PHP book is saying that if I have a function with 6 variables, all coded with defaults. And I want to call the function and only override the 6th default.... I have to supply all six values.

function fake_example ( a=1, b=2, c=3, d=4, e=5, f=7 )

and I wish to call

fake_ example ( ,,,,, 6 )

That ain't going to work? (It isn't cause I know) So this post is 25% complaint, 25% lets fix the next release and introduce the concept of positional parameters and 50% request if any knows an easy way to get way to get around this?

My books says to code so that the one one that needs replaced is first in the list.
Life isn't always that predictable.

any wisdom appreciated.

Chuck

[Weedpacket]

I've typically ended up doing this:

function fake_example($a=null, $b=null, $c=null, $d=null, $e=null, $f=null)
{
    if(is_null($a)) $a=1;
    if(is_null($b)) $b=2;
//...
}

fake_example(null,null,null,null,null, 6);

In no small part because I might want the default to be something more sophisticated than can fit in a function signature. If that happens a lot I write additional specialised functions that call the generic one - save typing all those nulls.