Rasmus,
Makes sense. I didn't think much when writing that piece...but...what is
the & for? wouldn't it work the same without it?
function test ($var)
{
$var = addslashes($var)
}
$foo = "He's dreaming";
test($foo);
print($foo);
Jordan K. Martin
http://www.newimagedesign.com
"Rasmus Lerdorf" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Well, you could do it without a return, but first, addslashes() returns
> the modified string, it does not do in-place replacement. And second, you
> would need to pass the string in by reference, like this:
>
> function test (& $var)
> {
> $var = addslashes($var)
> }
>
> $foo = "He's dreaming";
> test($foo);
> print($foo);
>
> -Rasmus
>
> On Mon, 1 Apr 2002, Jordan wrote:
>
> > 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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> > > > >
> > > >
> > > >
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> > > >
> > > >
> > > >
> > >
> >
> >
> >
> > --
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> >
>
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