Question

我有两个互相引用的类。类B有一个指向A对象的指针列表，每个A对象都需要一个指向创建它的B对象的指针。为什么在单个文件中写入时会编译，而在拆分为多个文件时却不能编译？

namespace.h

#ifndef NAMESPACES_H
#define NAMESPACES_H

#include "A.h"
#include "B.h"
#include "C.h"

namespace NAMESPACE
{
    class A;
    class B;
    class C;
}
#endif

A.h

#ifndef A_H
#define A_H
#include "namespaces.h"
#include <iostream>

namespace NAMESPACE{
class A
{
private:
    int x;
    B* ptrB;
public:
    void print();
    A(int g_x, B* g_ptrB);
};
}

#endif

A.cpp

#include "A.h"

namespace NAMESPACE{


A::print()
    {
        std::cout<<"I am class A, my x = "<<x<<std::endl;
    }
    A::A(int g_x, B* g_ptrB)
    {
        x = g_x;
        ptrB = g_ptrB;
    }
}

B.h

#ifndef B_H
#define B_H

#include "namespaces.h"
#include <vector>
#include <iostream>
namespace NAMESPACE{

class B
{
private:
    std::vector<A*> vA;
public:
    B(int g_x);
    void print();
};
}

#endif

B.cpp

#include "B.h"

namespace NAMESPACE{

B::B(int g_x)
    {
        vA.push_back(new A(g_x, this));
    }
    void B::print()
    {
        std::cout<<"I am class B\n";
        vA[0]->print();
    }
}

main.cpp

#include <iostream>
#include "namespaces.h"

int main()
{
    B *b = new B(3);
    b->print();
    return 0;
}

/ *相比之下，这个单个.cpp文件可以很好地编译* /

#include <iostream>
#include <vector>
class B;
class A
{
private:
    int x;
    B* ptrB;
public:
    void print()
    {
        std::cout<<"I am class A, my x = "<<x<<std::endl;
    }
    A(int g_x, B* g_ptrB)
    {
        x = g_x;
        ptrB = g_ptrB;
    }
};

class B
{
private:
    std::vector<A*> vA;
public:
    B(int g_x)
    {
        vA.push_back(new A(g_x, this));
    }
    void print()
    {
        std::cout<<"I am class B\n";
        vA[0]->print();
    }
};

class C:public A
{
public:
    C(int g_x, B* g_ptrB):A(g_x, g_ptrB){}
};

int main()
{
    B *b = new B(3);
    b->print();
    return 0;
}

编译单独的文件时，出现很多错误，例如

In file included from namespaces.h:4:0,
                 from B.h:4,
                 from B.cpp:1:
A.h:11:2: error: ‘B’ does not name a type
  B* ptrB;
  ^
A.h:14:13: error: ‘B’ has not been declared
  A(int g_x, B* g_ptrB);
             ^
In file included from namespaces.h:6:0,
                 from B.h:4,
                 from B.cpp:1:
C.h:10:13: error: ‘B’ has not been declared
  C(int g_x, B* g_ptrB):A(g_x, g_ptrB){}

但是当编译一个文件时，所有文件都运行顺利，我得到了

I am class B
I am class A, my x = 3

怎么来？

当将这些交叉引用的类拆分为不同的文件时，为什么不进行编译？

0 个答案: