Re: [PLUG] C++/*this

["Sergey Ostrovsky" <sergey01@comcast.net>]

----- Original Message ----- 
From: "Tim Peeler" <t_peeler@bellsouth.net>
To: <plug@lists.phillylinux.org>
Sent: Monday, October 20, 2003 2:54 PM
Subject: [PLUG] C++/*this


>
> Hi, long time no post.  I have a C++ question, hopefully someone can
> direct me to some good documentation.
>
> I have always understood that the "this" pointer accessible in non-static
> member functions of C++ objects (classes, structs or unions) is actually
> an implicit argument added by the compiler.
>
>    Example:
>
>          void Object::Function();
>       is actually:
>          void Object::Function(Object *this);
>
>       so when you do this:
>          Object ob;
>          ob.foo();
>
>       it can be thought of as:
>          Object ob;
>          Object::foo(ob);
>
> This has always made sense to me.  Since there is only one instantiation
> of all member functions of an object, and these functions are not stored
> as a part of the object, the function must have the *this pointer as an
> implicit argument so that the function has access to the object's other
> members.  It also makes sense that static member functions, since they
> do not need an object from which they are called, don't get the implicit
> *this pointer (this would defeat the purpose of making the function
> static).
>
>
> My instructor seems to think otherwise.  Here is the explination I
> was given:
>
>    All classes, structs and unions contain a *this pointer that
>    references itself.  The object also contains pointers to the object
>    member functions, therefore when calling a member function in an
>    object the member function has access to the *this pointer.
>
>    Example:
>
>       class Foo {
>          private:
>             int x;
>
>          public:
>             void setx(int);
>       };
>
>       Foo::setx(int) is instantiated only once and each object of type
>       Foo has a function pointer to setx:
>
>          Basically, breaking down what he said results in an object
>          that "looks" like this:
>
>          Foo {
>             private:
>                int x;
>                Foo *this;
>
>             public:
>                void (*setx)(int); // this is the pointer to
Foo::setx(int);
>          };
>
>
> This makes no sense to me.  First there is the question of how does the
> function (which is instantiated only once, which I know to be true) know
> which object is "calling" it?  Is there a way for a function (any
> function) to "know" what object it was called from other than providing
> it with a pointer to that object?
>
> Furthur, a simple sizeof() test shows that this (the instructor's idea
> of how an object is created and stored) can't be true unless sizeof()
> doesn't return the true size of an object (which would be VERY bad
> especially if you tried to combine a malloc() call to get the required
> memory instead of new).
>
> By using the simple Foo class above, with only one private data member
> (x), and one public member function (void setx(int)), I see that the
> sizeof(Foo) is 4 (which happens to be sizeof(int)).  Is my instructor
> full of it

Yes he is.
>From what your saying it's clear he doesn't have a clue.
Me wonders if he ever read Stroustrup.
Ask him what's differnce between plain member function and virtual one,
and how it's implemented. And please share the fun...


Sergey.

 or am I missing something here?
>
> If I have it wrong, please someone point me to some documentation that
> shows me exactly how a class' member functions (non-static) "know" which
> object has called them and where they get *this.
>
> Thanks,
> Tim
>
>
> --
> -----BEGIN GEEK CODE BLOCK-----
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> ------END GEEK CODE BLOCK------
>
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