C++类继承 继承后函数的值实现详解

类的继承会首先寻找基类，若基类未实现，则会寻找派生类的函数

1. class继承，函数不继承

#include 
 
class base
{
public:
  base(){}
  ~base(){}
 
  int a;
  void setA()
  {
    a = 1;
  }
};
 
class A:public base
{
public:
  A(){}
  ~A(){}
   
  void setA()
  {
    a = 2;
  }
   
};
 
class B:public base
{
public:
  B(){}
  ~B(){}
   
  void setA()
  {
    a = 3;
  }
};
 
int main()
{
  A *ax = new A();
  B *bx = new B();
  base *aClass = ax;
  base *bClass = bx;
  aClass->setA();
  bClass->setA();
 
  printf("a value of a %dn", aClass->a);
  printf("a value of b %dn", bClass->a);
  return 0;
}

运行结果：

2. 函数和Class都继承

#include 
 
class base
{
public:
  base(){}
  ~base(){}
 
  int a;
  virtual void setA()
  {
    a = 1;
  }
};
 
class A:public base
{
public:
  A(){}
  ~A(){}
   
  virtual void setA()
  {
    a = 2;
  }
   
};
 
class B:public base
{
public:
  B(){}
  ~B(){}
   
  virtual void setA()
  {
    a = 3;
  }
};
 
int main()
{
  A *ax = new A();
  B *bx = new B();
  base *aClass = ax;
  base *bClass = bx;
  aClass->setA();
  bClass->setA();
 
  printf("a value of a %dn", aClass->a);
  printf("a value of b %dn", bClass->a);
  return 0;
}

运行结果：注意派生类中可以不写virtual，最好写上，以辨别是函数继承

若将2中基类的函数写为纯虚函数，运行的结果一样，但是如果基类是纯虚函数，派生类必须实现相应的函数。

class base
{
public:
  base(){}
  ~base(){}
 
  int a;
  virtual void setA()=0;
   
};

3. 类A中不有函数，则继承自基类

class A:public base
{
public:
  A(){}
  ~A(){}
   
};

结果为：

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