C++ 继承详解及实例代码

　C++继承可以是单一继承或多重继承，每一个继承连接可以是public,protected,private也可以是virtual或non-virtual。然后是各个成员函数选项可以是virtual或non-virtual或pure virtual。本文仅仅作出一些关键点的验证。

　　public继承，例如下：

1 class base
2 ｛...｝
3 class derived:public base
4 ｛...｝

　　如果这样写，编译器会理解成类型为derived的对象同时也是类型为base的对象，但类型为base的对象不是类型为derived的对象。这点很重要。那么函数形参为base类型适用于derived，形参为derived不适用于base。下面是验证代码，一个参数为base的函数，传入derived应该成功执行，相反，一个参数为derived的函数

#include <iostream>
#include <stdio.h>

class base
{
  public:
  base()
  :baseName(""),baseData(0)
  {}
  
  base(std::string bn,int bd)
  :baseName(bn),baseData(bd)
  {}
  
  std::string getBaseName() const
  {
    return baseName;
  }
  
  int getBaseData()const
  {
    return baseData;
  }
  
  private:
    std::string baseName;
    int baseData;
};

class derived:public base
{
  public:
    derived():base(),derivedName("")
    {}
    derived(std::string bn,int bd,std::string dn)
    :base(bn,bd),derivedName(dn)
    {}
    std::string getDerivedName() const
    {
      return derivedName;
    }
  private:
    std::string derivedName;
};

void show(std::string& info,const base& b)
{
  info.append("Name is ");
  info.append(b.getBaseName());
  info.append(", baseData is ");
  char buffer[10];
  sprintf(buffer,"%d",b.getBaseData());
    info.append(buffer);
}

int main(int argc,char* argv[])
{
  base b("test",10);
  std::string s;
  show(s,b);
  std::cout<<s<<std::endl;
  derived d("btest",5,"dtest");
  std::string ss;
  show(ss,d);
  std::cout<<ss<<std::endl;
  return 0;
}

void show2(std::string& info,const derived& d)
{
  info.append("Name is ");
  info.append(d.getBaseName());
  info.append(", baseData is ");
  char buffer[10];
  sprintf(buffer,"%d",d.getBaseData());
  info.append(buffer);
}

1 derived_class.cpp: In function `int main(int, char**)':
2 derived_class.cpp:84: error: invalid initialization of reference of type 'const derived&' from expression of type 'base'
3 derived_class.cpp:70: error: in passing argument 2 of `void show2(std::string&, const derived&)'