c++ - 不完整类型结构的使用无效，即使有前向声明

Question

我知道循环依赖，但即使有前向声明我也能得到这个区域。我做错了什么？

// facility.h
class Area;

class Facility {
public:
    Facility();
    Area* getAreaThisIn();
    void setAreaThisIsIn(Area* area);
private:
    Area* __area;
};

// facility.cpp
#include "facility.h"
#include "area.h"
{ ... }

// area.h
class Facility;
class Area {
public:
    Area(int ID);
    int getId();

private:
    std::list<Facility*> _facilities;
};

// area.cpp
#include "area.h"
#include "facility.h"

所以这编译得很好，但如果我这样做

// foo.h
#include "facility.h"
class Foo { .. };

// foo.cpp
#include "foo.h"
void Foo::function() {
    Facility* f = new Facility();
    int id = f->getAreaThisIsIn()->getId();

当我得到 invalid use of incomplete type struct Area

Answer 1

澄清一下:前向声明允许您以非常有限的方式对对象进行操作:

struct Foo; // forward declaration

int bar(Foo* f); // allowed, makes sense in a header file

Foo* baz(); // allowed

Foo* f = new Foo(); // not allowed, as the compiler doesn't
                    // know how big a Foo object is
                    // and therefore can't allocate that much
                    // memory and return a pointer to it

f->quux(); // also not allowed, as the compiler doesn't know
           // what members Foo has

前向声明在某些情况下会有所帮助。例如，如果头文件中的函数只采用指向对象的指针而不是对象的指针，那么您不需要为该头文件#include 整个类定义。这可以改善您的编译时间。但是几乎可以保证该 header 的实现需要 #include 相关定义，因为您可能想要分配这些对象、调用这些对象的方法等，而且您需要的不仅仅是前向声明(forward declaration)。