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Can you just clear this up for me real quick

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This code :

class AutoPtr{
//...
//...
T& operator*(){ return *ptr;}
T* operator->(){ return ptr;} //[1]
~AutoPtr(){ delete ptr;}
};

//....

struct F{
int a;
};
int main(){
AutoPtr<F> p(new F);
p.operator->().a; //[2]
}


In [1] why are we returning a T*. In [2], p.operator->() could be substituted
with F*, right? So we are basically saying, F*.a ? Is that correct?
I would think operator-> should return T& or something. I'm sure
i'm missing something.

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In [1], we're returning a T* because the we want AutoPtr to behave like a a pointer. In other words we want to write: ptr->a = 5 instead of ptr.__internal_pointer->a = 5.

AFAIK, [2] shouldn't compile. It should be: p.operator->()->a, since operator->() returns T*.


auto_ptr<Foobar> ptr( new Foobar );

// Compare:
(*ptr).a = 5; // == ptr.operator*().a;
ptr->a = 5; // == ptr.operator->()->a;

// With:
(*ptr.__internal_pointer).a = 5;
ptr.__internal_pointer->a = 5;



[Edited by - _fastcall on July 19, 2010 10:24:19 PM]

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