378

假设我有课程FooBar设置如下:

class Foo
{
public:
    int x;

    virtual void printStuff()
    {
        std::cout << x << std::endl;
    }
};

class Bar : public Foo
{
public:
    int y;

    void printStuff()
    {
        // I would like to call Foo.printStuff() here...
        std::cout << y << std::endl;
    }
};

如代码中所述,我希望能够调用我正在覆盖的基类函数。在Java中有super.funcname()语法。这在 C++ 中可能吗?

4

8 回答 8

494

C++ 语法是这样的:

class Bar : public Foo {
  // ...

  void printStuff() {
    Foo::printStuff(); // calls base class' function
  }
};
于 2009-03-23T06:20:13.703 回答
130

是的,

class Bar : public Foo
{
    ...

    void printStuff()
    {
        Foo::printStuff();
    }
};

它与 Java 中的相同super,只是当您具有多重继承时,它允许从不同的基础调用实现。

class Foo {
public:
    virtual void foo() {
        ...
    }
};

class Baz {
public:
    virtual void foo() {
        ...
    }
};

class Bar : public Foo, public Baz {
public:
    virtual void foo() {
        // Choose one, or even call both if you need to.
        Foo::foo();
        Baz::foo();
    }
};
于 2009-03-23T06:20:03.627 回答
75

有时你需要调用基类的实现,当你不在派生函数中时......它仍然有效:

struct Base
{
    virtual int Foo()
    {
        return -1;
    }
};

struct Derived : public Base
{
    virtual int Foo()
    {
        return -2;
    }
};

int main(int argc, char* argv[])
{
    Base *x = new Derived;

    ASSERT(-2 == x->Foo());

    //syntax is trippy but it works
    ASSERT(-1 == x->Base::Foo());

    return 0;
}
于 2011-06-12T01:42:52.150 回答
29

以防万一您为班级中的许多功能执行此操作:

class Foo {
public:
  virtual void f1() {
    // ...
  }
  virtual void f2() {
    // ...
  }
  //...
};

class Bar : public Foo {
private:
  typedef Foo super;
public:
  void f1() {
    super::f1();
  }
};

如果你想重命名 Foo. 这可能会节省一些写作。

于 2009-03-23T15:11:39.077 回答
6

如果你想从它的派生类调用基类的函数,你可以简单地在重写的函数内部调用并提到基类名称(如Foo::printStuff())。

代码在这里

#include <iostream>
using namespace std;

class Foo
{
public:
    int x;

    virtual void printStuff()
    {
         cout<<"Base Foo printStuff called"<<endl;
    }
};

class Bar : public Foo
{
public:
    int y;

    void printStuff()
    {
        cout<<"derived Bar printStuff called"<<endl;
        Foo::printStuff();/////also called the base class method
    }
};

int main()
{
    Bar *b=new Bar;
    b->printStuff();
}

同样,您可以在运行时确定使用该类的对象(派生或基类)调用哪个函数。但这要求您在基类中的函数必须标记为虚拟。

下面的代码

#include <iostream>
using namespace std;

class Foo
{
public:
    int x;

    virtual void printStuff()
    {
         cout<<"Base Foo printStuff called"<<endl;
    }
};

class Bar : public Foo
{
public:
    int y;

    void printStuff()
    {
        cout<<"derived Bar printStuff called"<<endl;
    }
};

int main()
{

    Foo *foo=new Foo;
    foo->printStuff();/////this call the base function
    foo=new Bar;
    foo->printStuff();
}
于 2013-05-28T14:20:07.987 回答
2

如果有多个继承级别,您可以指定直接基类,即使实际实现处于较低级别。

class Foo
{
public:
  virtual void DoStuff ()
  {
  }
};

class Bar : public Foo
{
};

class Baz : public Bar
{
public:
  void DoStuff ()
  {
    Bar::DoStuff() ;
  }
};

在此示例中,类Baz指定Bar::DoStuff()尽管类Bar不包含DoStuff的实现。这是一个细节,巴兹不需要知道。

显然,调用Bar::DoStuff比调用Foo::DoStuff更好,以防更高版本的Bar也覆盖此方法。

于 2020-12-23T12:14:52.510 回答
0

检查这个...

#include <stdio.h>

class Base {
public:
   virtual void gogo(int a) { printf(" Base :: gogo (int) \n"); };    
   virtual void gogo1(int a) { printf(" Base :: gogo1 (int) \n"); };
   void gogo2(int a) { printf(" Base :: gogo2 (int) \n"); };    
   void gogo3(int a) { printf(" Base :: gogo3 (int) \n"); };
};

class Derived : protected Base {
public:
   virtual void gogo(int a) { printf(" Derived :: gogo (int) \n"); };
   void gogo1(int a) { printf(" Derived :: gogo1 (int) \n"); };
   virtual void gogo2(int a) { printf(" Derived :: gogo2 (int) \n"); };
   void gogo3(int a) { printf(" Derived :: gogo3 (int) \n"); };       
};

int main() {
   std::cout << "Derived" << std::endl;
   auto obj = new Derived ;
   obj->gogo(7);
   obj->gogo1(7);
   obj->gogo2(7);
   obj->gogo3(7);
   std::cout << "Base" << std::endl;
   auto base = (Base*)obj;
   base->gogo(7);
   base->gogo1(7);
   base->gogo2(7);
   base->gogo3(7);

   std::string s;
   std::cout << "press any key to exit" << std::endl;
   std::cin >> s;
   return 0;
}

输出

Derived
 Derived :: gogo (int)
 Derived :: gogo1 (int)
 Derived :: gogo2 (int)
 Derived :: gogo3 (int)
Base
 Derived :: gogo (int)
 Derived :: gogo1 (int)
 Base :: gogo2 (int)
 Base :: gogo3 (int)
press any key to exit

最好的方法是使用base::function作为@sth

于 2016-11-16T10:29:39.997 回答
0

是的,您可以调用它。在子类中调用父类函数的 C++ 语法是

class child: public parent {
  // ...

  void methodName() {
    parent::methodName(); // calls Parent class' function
  }
};

阅读有关函数覆盖的更多信息。

于 2018-02-17T19:48:06.257 回答