使用派生类的对象访问基类的方法，该对象具有相同的方法 [英] Accessing base class's method with derived class's object which has a method of same name

问题描述

使用派生类的对象访问base的foo（）时。

when accessing foo() of "base" using derived class's object.

#include <iostream>

class  base
{
      public:
      void foo()
     {
        std::cout<<"\nHello from foo\n";
     }
};

class derived : public base
{
     public:
     void foo(int k)
     {
        std::cout<<"\nHello from foo with value = "<<k<<"\n";
     }
};
int main()
{
      derived d;
      d.foo();//error:no matching for derived::foo()
      d.foo(10);

}

如何访问具有同名方法的基类方法在派生类中。
生成的错误已显示。
如果我不清楚我会道歉，但我觉得我已经把自己弄清楚了。
提前感谢。

how to access base class method having a method of same name in derived class. the error generated has been shown. i apologize if i am not clear but i feel i have made myself clear as water. thanks in advance.

推荐答案

您可以使用base :: foo c $ c>到派生类：

You could add using base::foo to your derived class:

class derived : public base
{
public:
     using base::foo;
     void foo(int k)
     {
        std::cout<<"\nHello from foo with value = "<<k<<"\n";
     }
};

编辑 这个问题解释了为什么你的 base :: foo（）不能直接使用派生的而不使用使用声明。

The answer for this question explains why your base::foo() isn't directly usable from derived without the using declaration.

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