私人班的朋友

nam*_*ame 5 c++ friend

如何为私人课程定义朋友?

#include <iostream>

class Base_t{
    private:
        struct Priv_t{
            friend std::ostream & operator<<(std::ostream &os, const Priv_t& obj);
        } p;
    friend std::ostream & operator<<(std::ostream &os, const Base_t& obj);
};

std::ostream & operator<<(std::ostream &os, const Base_t& obj) {
    return os << "base: " << obj.p;
}

std::ostream & operator<<(std::ostream &os, const Base_t::Priv_t& obj) {
    return os << "priv";
}

int main() {
    Base_t b;
    std::cout << b << std::endl;
}
Run Code Online (Sandbox Code Playgroud)

错误:

:!make t17 |& tee /tmp/vB5G5ID/54
g++     t17.cpp   -o t17
t17.cpp: In function 'std::ostream& operator<<(std::ostream&, const Base_t::Priv_t&)':
t17.cpp:5:16: error: 'struct Base_t::Priv_t' is private
         struct Priv_t{
                ^
t17.cpp:15:59: error: within this context
 std::ostream & operator<<(std::ostream &os, const Base_t::Priv_t& obj) {
                                                           ^
make: *** [t17] Error 1

shell returned 2
Run Code Online (Sandbox Code Playgroud)

它在我直接在Priv_t中定义朋友时有效

 friend std::ostream & operator<<(std::ostream &os, const Priv_t& obj) { return os << "priv"; }
Run Code Online (Sandbox Code Playgroud)

如何在类/结构定义之外执行此操作?

mas*_*oud 3

虽然Priv_t是私人声明,但您应该移动

friend std::ostream & operator<<(std::ostream &os, const Base_t::Priv_t& obj);
Run Code Online (Sandbox Code Playgroud)

进入Base_t

class Base_t
{
private:
  struct Priv_t
  {
  } p;
  friend std::ostream & operator<<(std::ostream &os, const Base_t& obj);
  friend std::ostream & operator<<(std::ostream &os, const Base_t::Priv_t& obj);
};
Run Code Online (Sandbox Code Playgroud)

实时代码。