Pointer to class???

[Sina]

Hi, can any body explain to me what the function of the line 'class class_name_2 *class_pointer;' is in the following code? Could you please introduce me to a webpage in this regard?

namespace namespace_name {

class class_name_1 {
 public:                       
  class class_name_2 *class_pointer;
};

}

Any help would be pretty much appreciated.

Sina

[filipe]

http://www.cplusplus.com/forum/beginner/39243/

[Sina]

Can you elaborate a little bit more?

[hanst99]

You do know what a pointer is I suppose?

[Sina]

Yes I do. The thing is I did not find pointer to class in such a way.

[filipe]

That class is being forward declared. This:

class Foo {
    class Bar *ptr;
};

is the same as:

class Bar;
class Foo {
    Bar *ptr;
};

You're telling the compiler that the name Bar refers to some class that hasn't been defined yet. Because all (non-function) pointers have the same size, the compiler can deal with a pointer to Bar (it knows its size and it knows what a pointer is). It can't dereference it, though, because it doesn't know anything about class Bar besides its name.

[webJose]

All the following are the same thing:

class class_name_2 *class_pointer;
class_name_2 *class_pointer;
class class_name_2* class_pointer;
class_name_2* class_pointer;

Does this help? The keyword class is optional. I believe this has something to do with backwards compatibility with C or at least with C coders (as classes did not exist in C). As far as I know, in C you would need to declare struct and enum variables using the keywords struct and enum:

//C-Style
struct MyStructType
{
    int myInt;
    unsigned myOtherInt;
};

//The struct keyword here is necessary
struct MyStructType myVariable;

I believe this is why the Windows SDK will always typedef the structs.

[filipe]

All the following are the same thing

They're not quite the same thing because of forward declarations.

[hanst99]

I believe this is why the Windows SDK will always typedef the structs.

Rather, because the win32 api is pure C, which enforces using structs that way.

[webJose]

Hello filipe. Kindly explain how they would be different if class_name_2 were forward-declared. I don't see it.

[hanst99]

If you do not forward declare class_name_2, and it's actual declaration comes after class_name_1, you may get compiler errors.

[webJose]

And how does that affect the pointers? Why is this going to affect the declarations? I surely can write stuff that doesn't compile, but stuff that doesn't compile shouldn't be presented here. What would be the point?

In my examples, I assume that class_name_2 has either been declared or forward-declared so that it can be compiled. My question is: Why was it implied that my pointers were not the same if the class had been forward-declared? I don't see the difference.

[filipe]

You didn't say you were assuming anything, just that they were the same. This:

class Foo *ptr;

will always compile. If the class hasn't been declared yet (as in the OP), it will forward declare it and declare ptr.

Now this:

Foo *ptr;

will never compile if Foo hasn't been declared. So they are not the same thing.

[webJose]

Ah, I see. I thought we were all under the assumption that class_name_2 had been declared already. Apparently I was the only one. Interesting.

In the example that you show there, filipe, if class Foo hadn't been declared or forward declared, what will ptr be?

[filipe]

The first example forward declares Foo and declares ptr to be a pointer to Foo. The second one is not legal if Foo hasn't been declared.

[Sina]

So what I understood from your replies is that the statement:

class Foo *ptr;

is equivalent to:

Foo *ptr;

if the class 'Foo' is declared before.
Is that correct?

[filipe]

Yes.

[webJose]

@filipe: Understood.

@Sina: Yes, they are the same thing so long the class Foo has been declared or forward-declared. It actually doesn't make a difference. The only difference arises if Foo hasn't been declared in any way at all, as explained by filipe.

[Sina]

Thank you guys

[Sina]

one last question guys: can you tell me why the following code is not working?

#include <iostream>
using namespace std;
namespace Sina_NS {
class Sina {

public:
    int width, height;
	class Dim *dim;
	int area () {return (width * height);}
};
}
using namespace Sina_NS;
class Dim: public Sina{
	
public:
    int width, height;
	int pp () {return (width * height);}
};

int main () {
	dim->width=2;
	return 0;
}

it gives the error 'dim was not declared in this scope'. I thought by this way forward declared dim as a pointer to Dim