Question about this smart pointer.

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Original post by ManaStone First of all what does the line: T& operator*() {return *ptr;} Mean? It looks like it is just returning the contents of ptr, but what is the & operator for?

It's overloading the dereference operator. It's returning a reference to the pointed-to type:

auto_ptr< int > ap2i( 3 );*ap2i = 2;int i = *ap2i;

It's also overloading the pointer to member operator:

auto_ptr< MyClass > ap2myclass;
ap2myclass->myclassmemberfunction();

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Also how do I control whether the operator is going to before or after the variable. For example, how would I do an overload for this:

You would overload the address-of operator:

T * smart_ptr::operator & () { return pObj; }

Although doing so negates having a smart pointer to begin with, since anything can use that address, therefore your reference counting scheme is worthless.

EDIT: to expand, any pointer that holds the address of the encapsulated object when the smart pointer gets destroyed will be dangling. A safer way would be to force an explicit conversion. Having the user specifically ask for the address of the object is better than implicitly giving it to her, because you can assume she knows what she's doing with it.

Here is a good article explaining this and most other non-trivial issues with smart pointers: click

[Edited by - stylin on September 7, 2005 9:09:17 PM]

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Original post by ManaStone First of all what does the line: T& operator*() {return *ptr;} Mean? It looks like it is just returning the contents of ptr, but what is the & operator for?

It is returning a reference to the contents, so that you can modify them. Without the & you would return a copy of the contents, which would be fairly pointless.

That's all I can help with, sorry.

CM

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x->->DoSomething();

No. The -> operator is applied repeatably until it encounters a raw pointer.

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You would overload the address-of operator:

Don't do this. The address of an object is too fundamental a property to go mucking about with it, especially if you don't even know what a reference is.

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Original post by Deyja No. The -> operator is applied repeatably until it encounters a raw pointer.

Is there something special about the -> operator when it overloads? What if I wanted to change its behavior so that it does a multiplication of something instead or its normal behavior? Would I be able to do that?

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Don't do this. The address of an object is too fundamental a property to go mucking about with it, especially if you don't even know what a reference is.

I know what it is; I just never seen it used like that before.

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Original post by ManaStone Quote: Original post by Deyja No. The -> operator is applied repeatably until it encounters a raw pointer. Is there something special about the -> operator when it overloads? What if I wanted to change its behavior so that it does a multiplication of something instead or its normal behavior? Would I be able to do that?

The -> operator is applied repeatably until it encounters a raw pointer.

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Quote: Don't do this. The address of an object is too fundamental a property to go mucking about with it, especially if you don't even know what a reference is. I know what it is; I just never seen it used like that before.

The character & is the address of operator and is also used to declare a type as a reference