Re: [PHP4BETA] return references branch merged into the main repository

On Thu, 16 Dec 1999, Zeev Suraski wrote: > It's partially double work, true, but if you think about it, there's no > real other consistent way to do it. > First, you don't want to let users take references from functions that > don't intend to return references; That's a basic encapsulation rule. So, > you must note that you're returning a reference in the function > declaration, as the function author. Right, & in function declaration should stay there. > Second, an expression that is a reference, when assigned to a variable, > does not create another reference, but a copy of its value: > $a = &$foo; /* $a is a reference to $foo */ > $b = $a; /* $b is NOT a reference to $foo, it just contains $foo's > original value */ > $c = &$a; /* $c IS a reference to $foo */ > For that reason, if you do > $a = foo() > even if foo() evaluates to a reference, $a = ref_expr causes $a to be a > copy of the value of ref_expr, not a reference to it. So what you're saying is there is no way to tell whether foo() returns reference when it's being executed? -Andrei Nobody tried to design Windows - it just grew in random directions without any kind of thought behind it.