It's good to know though, that an object is a reference. This means that if $a is an object and you evaluate $b = $a; then both variables point to the same object and $a->Foo() is equivalent to $b->Foo();. Changing the value of $a doesn't change the value of $b though so if you later evaluate $a = new SomeObject('whatever'); then $a will refer to a different object than $b and $a->Foo() will no longer be equivalent to $b->Foo()!
Even if you do not use references and complex data structures yourself, it's good to understand this.
Jenda
Enoch was right!
Enjoy the last years of Rome.
