"Sascha Mantscheff" <[EMAIL PROTECTED]> schrieb im Newsbeitrag
02051311162204.02523@pico">news:02051311162204.02523@pico...
> When I pass an object as a parameter to a function as "$this", this is an
> object reference (according to the docs).
it depends on the function. if you call it by value, not by reference,
then the function will contain a copy of the original not a pointer.
Doesn't matter if you call the function with $this or some other reference.
$this is a self-reference to the object. you can only use it inside the
object itself.
but as soon as you send it to a function as "by value"-parameter, a copy of
the
object will be used inside the function. if you call the function "by
reference"
you'll hold a pointer inside the function
> I store this reference in a local variable called $someObject. $someObject
> now contains an object pointer.
Ok, let's say the parameter is send to the functions "by reference". Then
you are
right, it'll point to the original object, but:
function (&$objRef) {
$blah = $objRef;
}
-> $blah will hold a copy!
function (&$objRef) {
$blah = &$objRef;
}
-> $blah will hold a reference!
> I pass $someObject to another function. Is this a reference to the
original
> $this, or is it a copy of the object?
Read what i wrote above and answer it yourself.
Regards Michael.
> All function calls are by value, not by reference (without the "&" name
> modifier).
>
>
> s.m.
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