C / C++ / MFC

George_George wrote:

1.

I think it means an operator &, which returns const Inner* type and takes no arguments, right?

const Inner* pInner = &objOuter\\ object of Outer or the class within which the operator is defined
error C2440

const Outer::Inner * operator &();

Outer OuterObj;
const   Outer::Inner *   innerObj = (const   Outer::Inner * )(OuterObj);
or
const   Outer::Inner *   &innerObj = (const   Outer::Inner * &)(OuterObj);

Rajkumar_R wrote:

const Outer::Inner * innerObj = (const Outer::Inner * )(OuterObj);

Rajkumar_R wrote:

const Outer::Inner * &innerObj = (const Outer::Inner * &)(OuterObj);

class O
{
public:
  class I
  { } m;
  const I* operator &() {
    printf("const I* operator & \n");
    return &m;
  }
};
void main()
{
  O  v;
  const O::I* p = &v;
}

const I* operator &

Maxwell Chen wrote:

This does not compile.

Maxwell Chen wrote:

class I { } m;
const I* operator &()
{
printf("const I* operator & \n");
return &m;
}

public:  
class I  { } m;  
operator const I* &() 
{    
printf("const I* operator & \n");    
return &m;  
}

class I  { } 
	const *m;  

	operator const I*  &()  
	{    printf("const I* typecast operator & \n");    
	return m;  
	}

Maxwell Chen wrote:

This compiles, but the member operator function is not invoked.

Rajkumar_R wrote:

class I { } const *m;
operator const I* &() {
printf("const I* typecast operator & \n");
return m;
}

class O
{
public:
  class I
  {} const* m;
  O() : m(new I()) {}
  operator const I* &() {
    printf("operator const I* & \n");
    return m;
  }
};
void main()
{
  O v;
  const O::I* &p = v; // got it!
  const O::I* &q = (const O::I* & )v; // the same.
}

Rajkumar_R wrote:

So now you can compile without error

pther wrote:

I do not have any idea of shell menu.

AbhijitAkude wrote:

Win32 Application doesn't provide that message handler

Mark Salsbery
Microsoft MVP - Visual C++

azusakt wrote:

I have no much idea on the calculations or algorithms.