Forward declaration with unique_ptr? [duplicate]

问题:

This question already has answers here:

Is std::unique_ptr<T> required to know the full definition of T? (8 answers)

Closed 4 years ago.

I have found it useful to use forward declaration of classes in combination with std::unique_ptr as in the code below. It compiles and works with GCC, but the whole thing seem kind of strange, and I wonder if this is standard behaviour (i.e. required by the standard)? Since B isn't a complete type when I declare the unique_ptr.

A.hpp

#include <memory>

class B;

class A {
    std::unique_ptr<B> myptr;
    // B::~B() can't be seen from here
public:
    ~A();
};

A.cpp

#include "B.hpp"
//B.hpp has to be included, otherwise it doesn't work.

A::~A() = default; // without this line, it won't compile 
// however, any destructor definiton will do.

I suspect this has to do with the destructor (and therefore the need to call the destructor of unique_ptr<B>) is defined in a specific compilation unit (A.cpp).

回答1:

It's explicitly legal. The rule is that the types used to instantiate a template in the standard library must be complete, unless otherwise specified. In the case of unique_ptr, §20.7.1/5 says “[...] The template parameter T of unique_ptr may be an incomplete type.”

There are certain operations on the pointer which require a complete type; in particular, when the object will actually be destructed (at least with the default deleter). In your example, for example, if A::~A() were inline, this might cause problems. (Note that if you don't declare the destructor yourself, it will be inline. Which partially defeats the purpose of using std::unique_ptr.)