Eric,
Isnt there really no need for the 'return' though?
$test ($var)
{
addslashes($var)
}
$foo = "He's dreaming";
$foo = test($foo);
print($foo);
//should also print He\'s dreaming
Am I incorrect in thinking this?
-Jordan K. Martin
http://www.newimagedesign.com
Eric Coleman <[EMAIL PROTECTED]> wrote in message
018e01c1d93d$cd404be0$0201a8c0@zaireweb">news:018e01c1d93d$cd404be0$0201a8c0@zaireweb...
> But to answer your question
>
> The purpose of return, is to "return" a value..
>
> function test($var)
> {
> return addslashes($var);
> }
>
> $foo = "Yes, I'am Very Awsome";
> $foo = test($foo);
> echo($foo);
> // echo's, Yes I\'am Very Awsome
>
> Understand?
>
> ----- Original Message -----
> From: "Rasmus Lerdorf" <[EMAIL PROTECTED]>
> To: "Gary" <[EMAIL PROTECTED]>
> Cc: <[EMAIL PROTECTED]>
> Sent: Sunday, March 31, 2002 11:43 PM
> Subject: Re: [PHP] return
>
>
> > Nope, that code makes no sense. $_POST is an array containing the POST
> > variables. You make a copy of that array an put it in $foo thereby
> > overwriting the passed in $foo. Then you return $$foo which actually
ends
> > up returning a variable named $Array. It does not look like you have a
> > $Array variable in scope, and it is surely not what the misguided coder
> > behind this code was trying to achieve. In short, this is completely
> > bogus.
> >
> > -Rasmus
> >
> > On Sun, 31 Mar 2002, Gary wrote:
> >
> > > Can someone explain to me the reason for using return.
> > >
> > > function _getValue($foo)
> > > {
> > > $foo = $_POST;
> > > return ${$foo};
> > > }
> > >
> > > TIA
> > > Gary
> > >
> > >
> > > --
> > > PHP General Mailing List (http://www.php.net/)
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> > >
> >
> >
> > --
> > PHP General Mailing List (http://www.php.net/)
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> >
> >
> >
>
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