Static Member Function Working Like A Pointer To Data Members

[Nosophorus]

Hi!

I was reading about static member functions when I met the following
interesting piece of code:

#include <iostream>
using namespace std;

class Egg {
private:
static Egg e;
int i;
Egg(int ii) : i(ii) {}
Egg(const Egg&); // Prevent copy-construction
public:
static Egg* instance() { return &e; } //<--- THIS IS WHAT INTERESTS
ME
int val() const { return i; }
};

Egg Egg::e(47);

int main() {
cout << Egg::instance()->val() << endl; //<--- THIS IS TOO
}

It's interesting, because the function "instance()", as it returns an
Egg pointer/address, is pointing to (in main) the member function "val
()" -- which returns an int. It seems to me the "instance()" member
function works like a pointer of type Egg. Am I right?

I would appreciate any further clarifications/explanations.

Thank You!

Marcelo de Brito

[aceh...@gmail.com]

On Dec 30, 11:36 am, Nosophorus <Nosopho...@gmail.com> wrote:

> class Egg {
> private:
> static Egg e;

[...]

> static Egg* instance() { return &e; } //<--- THIS IS WHAT INTERESTS
> ME

The instance member function returns the address of the static member
e.

> cout << Egg::instance()->val() << endl; //<--- THIS IS TOO

> It's interesting, because the function "instance()", as it returns an
> Egg pointer/address,

Correct.

> is pointing to (in main) the member function "val
> ()" -- which returns an int.

That's not correct. The Egg* that instance() returns "is pointing to
e". It is not pointing to a member function.

> It seems to me the "instance()" member
> function works like a pointer of type Egg. Am I right?

You are right on that one.

> I would appreciate any further clarifications/explanations.

Would this make it clearer:

Egg * p = Egg::instance();
p->val();

That's the equivalent of what you have above:

Egg::instance()->val()

Maybe you confuse -> operator as "pointing to"? Possibly because it
looks like an arrow? If so, no, -> operator "dereferences" the pointer
on it's left hand side.

So, when you say

p->val();

you mean "call the val() member of what p points to." Or more simply:

SomeType p = /* ... */
p->i = 42;

The last statement says "assign 42 to the 'i' member of the object
that 'p' points to."

Ali

[Paavo Helde]

Nosophorus <Nosop...@gmail.com> kirjutas:

> Hi!
>
> I was reading about static member functions when I met the following
> interesting piece of code:
>
> #include <iostream>
> using namespace std;
>
> class Egg {
> private:
> static Egg e;

Here you declare there is a static instance of Egg class, called e.

> int i;
> Egg(int ii) : i(ii) {}
> Egg(const Egg&); // Prevent copy-construction
> public:
> static Egg* instance() { return &e; } //<--- THIS IS WHAT INTERESTS

Here you return a pointer to this instance.

> ME
> int val() const { return i; }
> };
>
> Egg Egg::e(47);

Here the static instance e is defined (incl. memory allocation and
initialization).

>
> int main() {
> cout << Egg::instance()->val() << endl; //<--- THIS IS TOO

Here you use this pointer for calling val() on the e object. val()
returns an int which is passed to the cout via the relevant operator<<.

> }
>
> It's interesting, because the function "instance()", as it returns an
> Egg pointer/address, is pointing to (in main) the member function "val
> ()" -- which returns an int. It seems to me the "instance()" member
> function works like a pointer of type Egg. Am I right?

It returns a pointer to a single static object of that type, namely e.

>
> I would appreciate any further clarifications/explanations.

It seems you have troubles with the static data members, look them up in
your textbook. The static data objects have a couple of problems, namely
regarding their multi-threading behavior and also the order of
initialization, so they should be used with caution. In this example it
is also not clear why a pointer is returned from instance(); a reference
would serve better here IMO.

Paavo

[Nosophorus]

Hi!

Thank you very much, Ali and Paavo! :)

Ali, you are right. I should have expressed me more clearly concerning
C++ dereferencing. :)

Let me ask you an extra question.

From what Ali wrote above:

Here "p" is pointing to the address returned by "instance()" (you
assign the address returned by "instance()" to "p"):

Egg * p = Egg::instance();

And now you dereference/access it:

p->val();

Correct? :)

[Paavo Helde]

Nosophorus <Nosop...@gmail.com> kirjutas:

> Hi!
>
> Thank you very much, Ali and Paavo! :)
>
> Ali, you are right. I should have expressed me more clearly concerning
> C++ dereferencing. :)
>
> Let me ask you an extra question.
>
> From what Ali wrote above:
>
> Here "p" is pointing to the address returned by "instance()"

Not exactly, p *is* the address, not pointing to one.

(you
> assign the address returned by "instance()" to "p"):

Right.

>
> Egg * p = Egg::instance();
>

> And now you dereference/access it:
>
> p->val();
>
> Correct? :)

Yes, the -> operator dereferences the pointer, and during this operation
the value of p is certainly accessed (what is "access" is not so clear in
the C++ standard, but using the pointer value for dereferencing it should
certainly qualify).

Paavo

[Nosophorus]

Hi!

Thank you, Paavo! :)

Now I have a more clarifying conception of dereferencing in C++!

Best Wishes!

Marcelo de Brito