C++ 继承和函数覆盖

问题描述

在C++中，基类的成员函数是否会被其同名的派生类函数覆盖，即使它的原型(参数的数量、类型和常量)不同?我想这是一个愚蠢的问题，因为许多网站都说函数原型应该是相同的；但是为什么下面的代码不能编译?我相信这是一个非常简单的继承案例.

In C++, will a member function of a base class be overridden by its derived class function of the same name, even if its prototype (parameters' count, type and constness) is different? I guess this a silly question, since many websites says that the function prototype should be the same for that to happen; but why doesn't the below code compile? It's a very simple case of inheritance, I believe.

#include <iostream>
using std::cout;
using std::endl;

class A {};
class B {};

class X
{
public:
    void spray(A&)
    {
        cout << "Class A" << endl;
    }
};

class Y : public X
{
public:
    void spray(B&)
    {
        cout << "Class B" << endl;
    }
};

int main()
{
    A a;
    B b;
    Y y;

    y.spray(a);
    y.spray(b);

    return 0;
}

GCC 抛出

error: no matching function for call to `Y::spray(A&)'
note: candidates are: void Y::spray(B&)

推荐答案

用来描述这个的术语是隐藏"，而不是覆盖".默认情况下，派生类的成员将使具有相同名称的基类的任何成员无法访问，无论它们是否具有相同的签名.如果要访问基类成员，可以使用 using 声明将它们拉入派生类.在这种情况下，将以下内容添加到 class Y:

The term used to describe this is "hiding", rather than "overriding". A member of a derived class will, by default, make any members of base classes with the same name inaccessible, whether or not they have the same signature. If you want to access the base class members, you can pull them into the derived class with a using declaration. In this case, add the following to class Y:

using X::spray;

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