Thanks Curt.
This solution works indeed. BUT, there's one but. Isn't there a way to not
have to tell PHP that I want the return of $this-AddFoo() as a reference?
In the actual code I want to worry as little as possible about specific
things like this. Isn't it a little strange that you have to
On Thu, 9 Oct 2003 22:50:10 +0200, Esctoday.Com | Wouter Van Vliet
[EMAIL PROTECTED] wrote:
Thanks Curt.
This solution works indeed. BUT, there's one but. Isn't there a way to
not
have to tell PHP that I want the return of $this-AddFoo() as a
reference?
In the actual code I want to worry as
If I recall correctly, the reason that the function needs to indicate it
returns a reference AND you need to declare it when assigning variables
is because the function will return a reference when instructed;
however, you then need to inform the binary assignment operator that it
should assign a
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