function argument variables

Anon

Hello all..
I tried searching around for this answer, but couldn't find it, so i ask you..
is it possible to do

$default_arg = "can't figure it out";
function yak($arg=$default_arg) {
echo $arg;
}

echo yak() should print "can't figure this out" but instead gives me a parse error. I know how to kludge my way around it, but is the above possible?
Thanks,
-Si

spacemax

function yak($arg) {
echo $arg;
}

$default_arg = "can't figure it out";

yak($default_arg);

should work..

ame12

or an alternative might be:

$defaultarg = "can't figure it...";

function yak($arg = "") {
global $defaultarg;
if ($arg == "") {
$arg = $defaultarg;
}
echo($arg);
}

yak(); ==> "can't figure it out...";
yak("blah"); ==> "blah";

Dave

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Anon

thanks for the reply..
but that defeats the purpse.. I want to set a configuration variable (i.e. global variable), and i'm trying to get a function w/out arguments to call that var..
so, still using my original code, yak() should use $default_arg or something else if given an argument.
still stumped,
-Si

Anon

yeah, your reply is a variant of the kludge i came up with, and it's lookin like the only answer..
thanks Dave
-Si