Forward declaration doesn't work with conversion operator

Consider the next code :

#include <iostream>
using namespace std;


class B;

class A
{
public:

  开发者_StackOverflow中文版  A() { p = 1;}
    int p;
    operator B() {B b; b.x = this->p; return b;}
};


class B
{
public:
    int x;
};

int main()
{

    A a;
    B b = a;
    return 0;
}

I'm trying to convert A to B , but I get the following compiler scream :

..\main.cpp:13: error: return type 'struct B' is incomplete

When I do this :

#include <iostream>
using namespace std;

class B
{
public:
    int x;
};

class A
{
public:

    A() { p = 1;}
    int p;
    operator B() {B b; b.x = this->p; return b;}
};


int main()
{
    A a;
    B b = a;
    return 0;
}

the code compiles , but the question is : is it possible to do that using the forward declaration I wrote above ?

Much thanks Ronen

---

Yes, it is possible, as long as the definition of A::operator B follows the definition of class B.

#include <iostream>
using namespace std;


class B;

class A
{
public:

    A() { p = 1;}
    int p;
    operator B();
};


class B
{
public:
    int x;
};

inline A::operator B() {B b; b.x = this->p; return b;}


int main()
{

    A a;
    B b = a;
    return 0;
}

---

No. Your A::operator B() creates an object of type B. The compiler needs to know the definition of B in order to be able to create an object of type B (for instance, it needs to know how big it is to perform stack-pointer calculations. It needs to know whether custom constructors need to be called.)

---

The line:

operator B() {B b; return b;}

creates an object of B. This cannot happen since B isn't defined.

Forward declarations let you declare pointers to objects which can be created later when the object definition is known, but you can't create the objects straight off the bat.

---

No, it's not possible. You cannot declare an object of type which is not completely defined.

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You can't do this with the forward declaration because the compiler needs to know the size of B (and its default ctor as well) for the operator return type. Without a complete type, it cannot know the size.

Yes, it probably could get this information from the definition you provide below, but for historical reasons, C and C++ compilers only consider definitions that come before in a translation unit.

---

It is not possible to create an object of a class without knowing its definition.

Separate your implementation into .h and .cpp files. So you can have a forward declaration class B in A.h and include its definition in A.cpp