覆盖具有较低权限的方法 [英] Overriding methods with lower permission
问题描述
你好，
我有以下代码：
class Base {
public：
virtual void f（）{cout<< "碱:: f（）的" << endl;}
virtual void f（int）{cout<< "碱:: F（INT）" << endl;}
};
class派生：public Base {
void f（）{cout<< "衍生:: f（）的" << endl;}
};
int main（）{
Base * ptr = new Derived（）;
ptr-> f（）;
ptr-> f（1）;
返回0;
}
如您所见，我将Base :: f（）从公共覆盖到私有访问
权限。作为我的Java人，我期待编译器错误。
但令我惊讶的是，该程序运行良好，并调用私有
版本？这是一个编译器错误（我使用g ++），还是将函数
显式转换为public？谢谢。
Hello there,
I have the following code:
class Base {
public:
virtual void f() {cout << "Base::f()" << endl;}
virtual void f(int) {cout << "Base::f(int)" << endl;}
};
class Derived : public Base {
void f() {cout << "Derived::f()" << endl;}
};
int main() {
Base *ptr = new Derived();
ptr->f();
ptr->f(1);
return 0;
}
As you can see, I overridden Base::f() from public, to private access
permission. Being the Java guy that I am, I expected a compiler error.
But to my surprise, the program ran fine, and called the private
version? Is this a compiler bug (I use g++), or is the function
explicitly converted to public? Thank you.
推荐答案
7月28日，1：53 * pm，dani ... @ mail.ru写道：
On Jul 28, 1:53*pm, dani...@mail.ru wrote:
你好，
我有以下代码：
class Base {
public：
* virtual void f（）{cout<< "碱:: f（）的" << endl;}
* virtual void f（int）{cout<< "碱:: F（INT）" <<结束;}
};
类派生：public Base {
* void f（）{cout << "衍生:: f（）的" << endl;}
};
int main（）{
* Base * ptr = new Derived（ ）;
* ptr-> f（）;
* ptr-> f（1）;
*返回0;
}
正如您所看到的，我将Base :: f（）从公共覆盖到私有访问
权限。作为我的Java人，我期待编译器错误。
但令我惊讶的是，该程序运行良好，并调用私有
版本？这是一个编译器错误（我使用g ++），还是将函数
显式转换为public？谢谢。
Hello there,
I have the following code:
class Base {
public:
* virtual void f() {cout << "Base::f()" << endl;}
* virtual void f(int) {cout << "Base::f(int)" << endl;}
};
class Derived : public Base {
* void f() {cout << "Derived::f()" << endl;}
};
int main() {
* Base *ptr = new Derived();
* ptr->f();
* ptr->f(1);
* return 0;
}
As you can see, I overridden Base::f() from public, to private access
permission. Being the Java guy that I am, I expected a compiler error.
But to my surprise, the program ran fine, and called the private
version? Is this a compiler bug (I use g++), or is the function
explicitly converted to public? Thank you.
这在C ++中是允许的。但是，你不能提供比超类提供的更多的b $ b可访问性，换句话说，它是一个
私有成员，子类中的公共禁止当前
标准。
That''s allowed in C++. However, you''re not allowed to provide more
accessibility than provided by super class, in other words making it a
private member a public in subclass is prohibited under the current
standard.
2008-07-28 13:53:41 -0400， da ***** @ mail.ru 说：
On 2008-07-28 13:53:41 -0400, da*****@mail.ru said:
>
如您所见，我将Base :: f（）从公共覆盖到私有访问
权限。作为我的Java人，我期待编译器错误。
>
As you can see, I overridden Base::f() from public, to private access
permission. Being the Java guy that I am, I expected a compiler error.
Java造成许多误解。
Java creates many misapprehensions.
令我惊讶的是，程序运行正常，并调用了私人
版本？这是一个编译器错误（我使用g ++），还是将函数
显式转换为public？谢谢。
But to my surprise, the program ran fine, and called the private
version? Is this a compiler bug (I use g++), or is the function
explicitly converted to public? Thank you.
两者都没有。 Base :: f是公共的，Derived :: f是私有的，就像你一样
表示它们应该是。静态检查访问。也就是说，ptr->; f（）
可以调用f，因为f在Base中是公共的。尝试使用指针
派生。
-
Pete
Roundhouse Consulting，有限公司（ www.versatilecoding.com ）作者The
标准C ++库扩展：教程和参考
（ www。 petebecker.com/tr1book ）
Neither. Base::f is public, and Derived::f is private, just like you
said they should be. Access is checked statically. That is, ptr->f()
can call f because f is public in Base. Try it with a pointer to
Derived.
--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)
2008-07-28 14:35:25 -0400，puzzlecracker< ir *** ******：gmail.com说：
On 2008-07-28 14:35:25 -0400, puzzlecracker <ir*********@gmail.comsaid:
7月28日，1：53？* pm，dani ... @ mail.ru写道：
On Jul 28, 1:53?*pm, dani...@mail.ru wrote:
>你好，
我有以下代码：
class Base {
public：
？* virtual void f（）{cout<< "碱:: f（）的" << endl;}
？* virtual void f（int）{cout<< "碱:: F（INT）" << endl;}
};
类派生：public Base {
？* void f（）{cout<< "衍生:: f（）的" << endl;}
};
int main（）{
？* Base * ptr = new Derived（）;
？* ptr-> ; f（）;
？* ptr-> f（1）;
？*返回0;
}
如您所见，我将Base :: f（）从公共访问权限覆盖到私人访问权限。作为我的Java人，我期待编译器错误。
但令我惊讶的是，该程序运行良好，并称为私有
版本？这是一个编译器错误（我使用g ++），还是明确转换为public的函数？谢谢。
>Hello there,
I have the following code:
class Base {
public:
?* virtual void f() {cout << "Base::f()" << endl;}
?* virtual void f(int) {cout << "Base::f(int)" << endl;}
};
class Derived : public Base {
?* void f() {cout << "Derived::f()" << endl;}
};
int main() {
?* Base *ptr = new Derived();
?* ptr->f();
?* ptr->f(1);
?* return 0;
}
As you can see, I overridden Base::f() from public, to private access
permission. Being the Java guy that I am, I expected a compiler error.
But to my surprise, the program ran fine, and called the private
version? Is this a compiler bug (I use g++), or is the function
explicitly converted to public? Thank you.
这是C ++中允许的。但是，你不能提供比超类提供的更多的b $ b可访问性，换句话说，它是一个
私有成员，子类中的公共禁止目前
标准。
That''s allowed in C++. However, you''re not allowed to provide more
accessibility than provided by super class, in other words making it a
private member a public in subclass is prohibited under the current
standard.
不，它不是。我认为你对使用
指令的规则感到困惑，这些指令不允许增加访问权限。
-
Pete
Roundhouse Consulting，Ltd。（ www.versatilecoding.com ）作者
标准C ++库扩展：教程和参考
（ www.petebecker.com/tr1book ）
No, it''s not. I think you''re confusing this with the rule about using
directives, which are not allowed to increase access.
--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)
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