如何只为一个特定的函数和类声明friend函数？ [英] How can friend function be declared for only one particular function and class?

问题描述

我的代码有什么问题？

我试图在GNU G ++环境中编译下面的代码，我得到这些错误：

I tried to compile the code below in the GNU G++ environment and I get these errors:

friend2.cpp:30: error: invalid use of incomplete type ‘struct two’
friend2.cpp:5: error: forward declaration of ‘struct two’
friend2.cpp: In member function ‘int two::accessboth(one)’:
friend2.cpp:24: error: ‘int one::data1’ is private
friend2.cpp:55: error: within this context

#include <iostream>
using namespace std;

class two;

class one
{
    private:
        int data1;
    public:
        one()
        {
            data1 = 100;
        }

        friend int two::accessboth(one a);
};

class two
{
    private:
        int data2;

    public:
        two()
        {
            data2 = 200;
        }

        int accessboth(one a);
};

int two::accessboth(one a)
{
    return (a.data1 + (*this).data2);
}

int main()
{
    one a;
    two b;
    cout << b.accessboth(a);
    return 0;
}

推荐答案

在其类中声明（不在朋友声明中）。这必须意味着在朋友宣告之前，你应该定义它的类 - 只是向前的声明是不够的。

A member function must be first declared in its class (not in a friend declaration). That must mean that prior to the friend declaration, you should have the class of it defined - a mere forward declaration does not suffice.

class one;

class two
 {
    private:
  int data2;
    public:
  two()
  {
    data2 = 200;
  }
 // this goes fine, because the function is not yet defined. 
 int accessboth(one a);
 };

class one
 {
     private:
  int data1;
    public:
  one()
  {
    data1 = 100;
  }
    friend int two::accessboth(one a);
 };

 // don't forget "inline" if the definition is in a header. 
 inline int two::accessboth(one a) {
  return (a.data1 + (*this).data2);
 }

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