Why can't a forward declaration be used for a std::vector?

Question

If I create a class like so:

// B.h #ifndef _B_H_ #define _B_H_  class B { private:     int x;     int y; };  #endif // _B_H_ 

and use it like this:

// main.cpp #include <iostream> #include <vector>  class B; // Forward declaration.  class A { public:     A() {         std::cout << v.size() << std::endl;     }  private:     std::vector<B> v; };  int main() {     A a; } 

The compiler fails when compiling main.cpp. Now the solution I know is to #include "B.h", but I'm curious as to why it fails. Neither g++ or cl's error messages were very enlightening in this matter.

Answer 1

In fact your example would build if A's constructor were implemented in a compile unit that knows the type of B.

An std::vector instance has a fixed size, no matter what T is, since it contains, as others said before, only a pointer to T. But the vector's constructor depends on the concrete type. Your example doesn't compile because A() tries to call the vector's ctor, which can't be generated without knowing B. Here's what would work:

A's declaration:

// A.h #include <vector>  class B; // Forward declaration.  class A { public:     A(); // only declare, don't implement here  private:     std::vector<B> v; }; 

A's implementation:

// A.cpp #include "A.h" #include "B.h"  A::A() // this implicitly calls vector<B>'s constructor {     std::cout << v.size() << std::endl; } 

Now a user of A needs to know only A, not B:

// main.cpp #include "A.h"  int main() {     A a; // compiles OK }