[PHP] 'undef' as an argument value

[Rodent of Unusual Size]

An odd request..  I want to have optional function arguments
that *aren't* defined unless values are explicitly passed.  Sorta
the equivalent of

function foo($arg1_required, $arg2_optional=undef)
if (isset($arg2_optional)) {
print "explicit";
}
else {
print "not passed";
}
}

Something like this is possible in Perl through the 'undef'
construct; is anything like this available in PHP?
-- 
#kenP-)}

Ken Coar, Sanagendamgagwedweinini  http://Golux.Com/coar/
Author, developer, opinionist  http://Apache-Server.Com/

"Millennium hand and shrimp!"

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Re: [PHP] 'undef' as an argument value

[Rodent of Unusual Size]

Rick Emery wrote:
> 
> function foo($arg1_required, $arg2_optional="")
> if ($arg2_optional=="") {
> print "not passed";
> }
> else {
> print "explicit";
> }
> }

No, I said 'undef' because I mean 'undefined'.  The argument
can have *any* value; I need to know when it has *none*.  That
is, there is no magic cookie to which I can set the default that
might not be a valid call value.
-- 
#kenP-)}

Ken Coar, Sanagendamgagwedweinini  http://Golux.Com/coar/
Author, developer, opinionist  http://Apache-Server.Com/

"Millennium hand and shrimp!"

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Re: [PHP] 'undef' as an argument value

[Rodent of Unusual Size]

"Johnson, Kirk" wrote:
> 
> http://www.php.net/manual/en/function.func-num-args.php will
> return the number of arguments passed to the function.

Bingo!  That will do the trick..  Thanks *very* much!
-- 
#kenP-)}

Ken Coar, Sanagendamgagwedweinini  http://Golux.Com/coar/
Author, developer, opinionist  http://Apache-Server.Com/

"Millennium hand and shrimp!"

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