hiding inner class implementation using namespace

Question

I am developing a library and a would like to provide my users a public interface separate from the real implementation that is hidden in a namespace. This way, I could change only the class HiddenQueue without changing myQueue that will be exposed to users only.

If I put the C++ code of HiddenQueue in the myQueue.cpp file the compiler complains saying _innerQueue has incomplete type. I thought that the linker was able to resolve this. What I am doing wrong here?

// myQueue.h
namespace inner{
    class HiddenQueue;
};

class myQueue{

public:
    myQueue();
);

private:

    inner::HiddenQueue _innerQueue;

};

///////////////////////////

// myQueue.cpp
namespace inner{
    class HiddenQueue{};
};

Answer 1

The compiler needs to know the exact memory layout of an object by looking at the header file it's defined in.

Your code says that class MyQueue has a member of type InnerQueue, which will be part of the memory layout of MyQueue objects. Therefore, to deduce the memory layout of MyQueue it needs to know the memory layout of InnerQueue. Which it does not, because you say "oh well, it's defined elsewhere".

What you are trying to do is closely related to "the PIMPL idiom"/"compiler firewall" technique.

To solve the problem, you must either include HiddenQueue.h in your header or declare _innerqueue as a pointer:

class myQueue {

public:
    myQueue();

private:

    inner::HiddenQueue* _pinnerQueue;
};

Using a pointer is possible because a pointer has a known memory size (dependent on your target architecture), therefore the compiler doesn't need to see the full declaration of HiddenQueue.

Answer 2

To be able to make a member of a class, you need to have a definition for it and not only a declaration. (A declaration is enough for a pointer or a reference to the class).