A simple qustion about C++ syntax

[Yong W]

if A and B are Class type, what's mean about the next sentence?

" int A::*B::*pMember; "

And how to use this pointer ?

we assume A and B's definition is:
class A
{
int a;
};
class
{
B obj;
};

[Gianni Mariani]

Yong W wrote:

if A and B are Class type, what's mean about the next sentence?

" int A::*B::*pMember; "

And how to use this pointer ?

we assume A and B's definition is:
class A
{
int a;
};
class
{
B obj;
};

What you wrote above makes no sense - is this below what you meant to
write ?

struct A
{

struct B
{
int a;
};

B obj;
};
int A::B::* pMember;

[Yong W]

Sorry, I have a mistake. the class A and B's definition is:

class B
{
int b;
};

class A
{
B obj;
};

[Jaspreet]

Yong W wrote:

Sorry, I have a mistake. the class A and B's definition is:

class B
{
int b;
};

class A
{
B obj;
};

Thanks for posting the correct code. You are giving the full scope for
member 'b' when accessing it.
Assuming you have the pubic access specifier. If you need to give full
scope for member B you would give - A::B.
Now if you need to access 'b' you would further use the scope
resolution to get :

A::B::b

(pay special care to upper and lower case Bs)

[Neelesh Bodas]

Jaspreet wrote:

Yong W wrote:

Sorry, I have a mistake. the class A and B's definition is:

class B
{
int b;
};

class A
{
B obj;
};

Thanks for posting the correct code. You are giving the full scope for
member 'b' when accessing it.
Assuming you have the pubic access specifier. If you need to give full
scope for member B you would give - A::B.
Now if you need to access 'b' you would further use the scope
resolution to get :

A::B::b

That doesnot work because it is completely incorrect. When you say A::B
, you mean that B is a subclass inside A, and not that an instance of B
is a member of A. Saying A::B::b means that you are talking about a
static member b from a class B which is nested inside A. Something like
this :

class A
{
public:
class B
{
public:
static int b;
};
};

The Code that OP posted was much different than this. ::B::b doesnot
make any sense in the context of OP's post.

[Neelesh Bodas]

Jaspreet wrote:

Yong W wrote:

Sorry, I have a mistake. the class A and B's definition is:

class B
{
int b;
};

class A
{
B obj;
};

Thanks for posting the correct code. You are giving the full scope for
member 'b' when accessing it.
Assuming you have the pubic access specifier. If you need to give full
scope for member B you would give - A::B.
Now if you need to access 'b' you would further use the scope
resolution to get :

A::B::b

That doesnot work because it is completely incorrect. When you say A::B
, you mean that B is a subclass inside A, and not that an instance of B
is a member of A. Saying A::B::b means that you are talking about a
static member b from a class B which is nested inside A. Something like
this :

class A
{
public:
class B
{
public:
static int b;
};
};

The Code that OP posted was much different than this. A::B::b doesnot
make any sense in the context of OP's post.

[Victor Bazarov]

Jaspreet wrote:

Yong W wrote:

Sorry, I have a mistake. the class A and B's definition is:

class B
{
int b;
};

class A
{
B obj;
};

Thanks for posting the correct code. You are giving the full scope for
member 'b' when accessing it.
Assuming you have the pubic access specifier. If you need to give full
scope for member B you would give - A::B.
Now if you need to access 'b' you would further use the scope
resolution to get :

A::B::b

(pay special care to upper and lower case Bs)

This again makes no sense. 'B' is not declared in 'A's scope, so you
cannot write 'A::B'. You can write 'A::obj', possibly. You can write
B A::*. From the latter you can try writing

int A::*B::*p;

which declares a pointer to an int member of 'B' that is a member of 'A'.
Problem is, I have no idea how to initialise it. I don't see any syntax
for that.

V
--
Please remove capital As from my address when replying by mail

[Jaspreet]

Neelesh Bodas wrote:

Jaspreet wrote:

Yong W wrote:

Sorry, I have a mistake. the class A and B's definition is:

class B
{
int b;
};

class A
{
B obj;
};

Thanks for posting the correct code. You are giving the full scope for
member 'b' when accessing it.
Assuming you have the pubic access specifier. If you need to give full
scope for member B you would give - A::B.
Now if you need to access 'b' you would further use the scope
resolution to get :

A::B::b

That doesnot work because it is completely incorrect. When you say A::B
, you mean that B is a subclass inside A, and not that an instance of B
is a member of A. Saying A::B::b means that you are talking about a
static member b from a class B which is nested inside A. Something like
this :

class A
{
public:
class B
{
public:
static int b;
};
};

The Code that OP posted was much different than this. A::B::b doesnot
make any sense in the context of OP's post.

Not neccesary. You could have a containership like below.

A and B are declared like below :

class B
{
public:
static int b;
};

class A
{
public:
B bObj;
};

If you need to access static member b from B, you would probably write
B::b.
Now if you need to access that static member b from A, you would
probably write A::B::b.

[Greg]

Victor Bazarov wrote:

Jaspreet wrote:

Yong W wrote:

Sorry, I have a mistake. the class A and B's definition is:

class B
{
int b;
};

class A
{
B obj;
};

Thanks for posting the correct code. You are giving the full scope for
member 'b' when accessing it.
Assuming you have the pubic access specifier. If you need to give full
scope for member B you would give - A::B.
Now if you need to access 'b' you would further use the scope
resolution to get :

A::B::b

(pay special care to upper and lower case Bs)

This again makes no sense. 'B' is not declared in 'A's scope, so you
cannot write 'A::B'. You can write 'A::obj', possibly. You can write
B A::*. From the latter you can try writing

int A::*B::*p;

which declares a pointer to an int member of 'B' that is a member of 'A'.
Problem is, I have no idea how to initialise it. I don't see any syntax
for that.

V

I believe the expression is a member pointer in B that points to a
member pointer in A. Like so:

struct A
{
int i;
};

struct B
{
int A::*p1;
};

int main()
{
// this is slightly nuts:

int A::*B::*p = &B::p1;
}

Greg