Forward declaration of nested types/classes in C++

问题:

I recently got stuck in a situation like this:

class A
{
public:
    typedef struct/class {...} B;
...
    C::D *someField;
}

class C
{
public:
    typedef struct/class {...} D;
...
    A::B *someField;
}

Usually you can declare a class name:

class A;

But you can\'t forward declare a nested type, the following causes compilation error.

class C::D;

Any ideas?

回答1:

You can\'t do it, it\'s a hole in the C++ language. You\'ll have to un-nest at least one of the nested classes.

回答2:

class IDontControl
{
    class Nested
    {
        Nested(int i);
    };
};

I needed a forward reference like:

class IDontControl::Nested; // But this doesn\'t work.

My workaround was:

class IDontControl_Nested; // Forward reference to distinct name.

Later when I could use the full definition:

#include <idontcontrol.h>

// I defined the forward ref like this:
class IDontControl_Nested : public IDontControl::Nested
{
    // Needed to make a forwarding constructor here
    IDontControl_Nested(int i) : Nested(i) { }
};

This technique would probably be more trouble than it\'s worth if there were complicated constructors or other special member functions that weren\'t inherited smoothly. I could imagine certain template magic reacting badly.

But in my very simple case, it seems to work.

回答3:

If you really want to avoid #including the nasty header file in your header file, you could do this:

hpp file:

class MyClass
{
public:
    template<typename ThrowAway>
    void doesStuff();
};

cpp file

#include \"MyClass.hpp\"
#include \"Annoying-3rd-party.hpp\"

template<> void MyClass::doesStuff<This::Is::An::Embedded::Type>()
{
    // ...
}

But then:

1. you will have to specify the embedded type at call time (especially if your function does not take any parameters of the embedded type)
2. your function can not be virtual (because it is a template)

So, yeah, tradeoffs...

回答4:

This can be done by forward declare the outer class as a namespace.

Sample: We have to use a nested class others::A::Nested in others_a.h, which is out of our control.

others_a.h

namespace others {
struct A {
    struct Nested {
        Nested(int i) :i(i) {}
        int i{};
        void print() const { std::cout << i << std::endl; }
    };
};
}

my_class.h

#ifndef MY_CLASS_CPP
// A is actually a class
namespace others { namespace A { class Nested; } }
#endif

class MyClass {
public:
    MyClass(int i);
    ~MyClass();
    void print() const;
private:
    std::unique_ptr<others::A::Nested> _aNested;
};

my_class.cpp

#include \"others_a.h\"
#define MY_CLASS_CPP // Must before include my_class.h
#include \"my_class.h\"

MyClass::MyClass(int i) :
    _aNested(std::make_unique<others::A::Nested>(i)) {}
MyClass::~MyClass() {}
void MyClass::print() const {
    _aNested->print();
}

回答5:

I would not call this an answer, but nonetheless an interesting find: If you repeat the declaration of your struct in a namespace called C, everything is fine (in gcc at least). When the class definition of C is found, it seems to silently overwrite the namspace C.

namespace C {
    typedef struct {} D;
}

class A
{
public:
 typedef struct/class {...} B;
...
C::D *someField;
}

class C
{
public:
   typedef struct/class {...} D;
...
   A::B *someField;
}

回答6:

This would be a workaround (at least for the problem described in the question -- not for the actual problem, i.e., when not having control over the definition of C):

class C_base {
public:
    class D { }; // definition of C::D
    // can also just be forward declared, if it needs members of A or A::B
};
class A {
public:
    class B { };
    C_base::D *someField; // need to call it C_base::D here
};
class C : public C_base { // inherits C_base::D
public:
    // Danger: Do not redeclare class D here!!
    // Depending on your compiler flags, you may not even get a warning
    // class D { };
    A::B *someField;
};

int main() {
    A a;
    C::D * test = a.someField; // here it can be called C::D
}