我在一个帖子中遇到了以下响应:
可以从派生类访问受保护的成员。私人不能。
class Base {
private:
int MyPrivateInt;
protected:
int MyProtectedInt;
public:
int MyPublicInt;
};
class Derived : Base
{
public:
int foo1() { return MyPrivateInt;} // Won't compile!
int foo2() { return MyProtectedInt;} // OK
int foo3() { return MyPublicInt;} // OK
};
class Unrelated
{
private:
Base B;
public:
int foo1() { return B.MyPrivateInt;} // Won't compile!
int foo2() { return B.MyProtectedInt;} // Won't compile
int foo3() { return B.MyPublicInt;} // OK
};
...
1)我的问题是: 我读过:“类派生列表命名了一个或多个基类,其格式为:
class derived-class:access-specifier base-class
其中access-specifier是public,protected或private之一,而base-class是先前定义的类的名称。如果未使用访问说明符,则默认情况下它是私有的。 “和”私有继承:从私有基类派生时,基类的公共成员和受保护成员将成为派生类的私有成员。
“
SO ...在我们的示例类中Derived:Base相当于类Derived:private Base,因为没有定义访问说明符,但代码工作正如编写者所说,所以我缺少什么? - 我认为类Derived访问说明符的基类是私有因此Base的公共成员和受保护成员对于Derived类应该是私有的,并且无法访问...谢谢!
答案 0 :(得分:0)
它是一种类似的想法。它不是应用于您可以访问的类的成员,而是适用于您可以访问的基类。
class A
{
public:
void f();
};
class B : public A
{
public:
void g()
{
f(); // OK
A *a = this; // OK
}
};
class B2 : public B
{
public:
void h()
{
f(); //OK
A *a = this; // OK
};
};
B b;
A& ba = b;
class C : protected A
{
public:
void g()
{
f(); // OK, can access protected base
A *a = this; // OK, can access protected base
}
};
class C2 : public C
{
public:
void h()
{
f(); // OK, can access protected base
A *a = this; // OK, can access protected base
};
};
C c;
c.f(); // Not OK, allthough A::f() is public, the inheritance is protected.
A& ca = c; // Not OK, inheritence is protected.
class D : private A
{
public:
void g()
{
f(); // OK because A is a base of D
A *a = this;
}
};
class D2 : public D
{
public:
void h()
{
f(); //Not OK, A is inherited with private in D
A *a = this; //Not OK
};
};
D d;
d.f(); // Not OK, allthough A::f() is public, the inheritance is private.
D& da = d; // Not OK, inheritence is private.