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I have a list implementation in which [] operator has been ocerloaded .
Now when i have pointer to list say *list_ptr , and i write
list_ptr[0] .
Which [] operator will be called ?
Overloaded or default ?
I think default [] , but in may case i am not getting result
accordingly .

Regards ,
Mangesh Sawant .

mangesh wrote:
> I have a list implementation in which [] operator has been ocerloaded .
> Now when i have pointer to list say *list_ptr , and i write
> list_ptr[0] .
> Which [] operator will be called ?
> Overloaded or default ?
> I think default [] , but in may case i am not getting result
> accordingly .
>
> Regards ,
> Mangesh Sawant .
>
i think it uses the overloaded one

"mangesh" <> writes:

> I have a list implementation in which [] operator has been ocerloaded .
> Now when i have pointer to list say *list_ptr , and i write
> list_ptr[0] .
> Which [] operator will be called ?
> Overloaded or default ?

The default, because you have only overloaded [] for type Foo (where
Foo is you list class) and *not* for Foo* (which in fact is impossible
as Foo* is /built-in/ type). To use overloaded [] you have to do:
(*list_ptr)[0].

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