What's happening here is that each time foo is called, a new activation object, containing the local variables "i" and "rectArea", is created and returned. Since these activation objects are completely independent of each other, they seem to exhibit some of the behaviour of types that are passed by value.
But try this ( in the context of your sample code, of course ;) ) :
var increaseCount4:Function = increaseCount3;
increaseCount3(); //output 1
increaseCount4(); //output 2
cheers,
till
Bryan Choi wrote:
>
> Hi, all.
>
> I'll show you some test code about the closure.
> You can see that 3 variables that is type of Function were assigned with
> foo() method .
> Genually, as you know, the primitive types is passed by value that is
> copied.
> So I think that this is also.
>
> How about you think?
>
>
> private function foo () : Function
> {
> var i:int = 0;
> function rectArea() : int { // function closure defined
> return ++i;
> }
> return rectArea;
> }
> private function bar () : void
> {
> var increaseCount1:Function = foo();
> var increaseCount2:Function = foo();
> trace("increaseCount1 : " + increaseCount1() );
> trace("increaseCount1 : " + increaseCount1() );
> var increaseCount3:Function = foo();
> trace("increaseCount2 : " + increaseCount2() );
> trace("increaseCount3 : " + increaseCount3() );
> trace("increaseCount1 : " + increaseCount1() );
> trace("increaseCount1 : " + increaseCount1() );
> trace("increaseCount2 : " + increaseCount2() );
> trace("increaseCount3 : " + increaseCount3() );
> }
> bar();
> // Output
> increaseCount1 : 1
> increaseCount1 : 2
> increaseCount2 : 1
> increaseCount3 : 1
> increaseCount1 : 3
> increaseCount1 : 4
> increaseCount2 : 2
> increaseCount3 : 2
>
>
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