为什么仅在使用 public 继承时派生 类 的友元函数 "available"？
Why are friend functions "available" for derived classes only when using public inheritance?
当派生 class 通过 public 访问从基础 class 继承时，问题与 中的问题相同。但是，如果它通过受保护或私有访问继承，则会出现可见性错误。
当它通过 public 访问继承时，A 的私有成员的可访问性与通过私有访问继承相同。它们有什么区别？
class A {
private:
int a;
friend void f();
};
class B : private A {
};
void f() {
B obj;
int x = obj.a;
}
int main() {
f();
return 0;
}
正如上面链接的答案中已经指出的那样，友谊不会被继承。因此，A 的朋友不是 B 的朋友。 B 通过 private 访问继承自 A 意味着 A 的所有成员都可以作为 B [class.access.base]/1. Since f is not a friend of B, it cannot access private members of B [class.access]/1.1. Since f is not a friend of B, the base class A of B is also not accessible from f [class.access.base]/4. Since the base class A of B is not accessible from f, there's also no way you could get to the A subobject of a B (of which you could access the members) in f [class.access.base]/5…[=32 的私有成员访问=]
当派生 class 通过 public 访问从基础 class 继承时，问题与
当它通过 public 访问继承时，A 的私有成员的可访问性与通过私有访问继承相同。它们有什么区别？
class A {
private:
int a;
friend void f();
};
class B : private A {
};
void f() {
B obj;
int x = obj.a;
}
int main() {
f();
return 0;
}
正如上面链接的答案中已经指出的那样，友谊不会被继承。因此，A 的朋友不是 B 的朋友。 B 通过 private 访问继承自 A 意味着 A 的所有成员都可以作为 B [class.access.base]/1. Since f is not a friend of B, it cannot access private members of B [class.access]/1.1. Since f is not a friend of B, the base class A of B is also not accessible from f [class.access.base]/4. Since the base class A of B is not accessible from f, there's also no way you could get to the A subobject of a B (of which you could access the members) in f [class.access.base]/5…[=32 的私有成员访问=]