作者：陈炘宇_573 | 来源：互联网 | 2022-01-18 02:30
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：
代码如下:
class base
｛...｝
class derived:public base
｛...｝
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
代码如下:
#include
#include
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< derived d("btest",5,"dtest");
std::string ss;
show(ss,d);
std::cout< return 0;
}
运行结果为：
base:baseName is test, baseData is 10
base:baseName is btest, baseData is 5
下面改改代码，将函数参数变为derived
代码如下:
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);
}
调用show(ss,d);编译器报错
代码如下:
derived_class.cpp: In function `int main(int, char**)':
derived_class.cpp:84: error: invalid initialization of reference of type 'const derived&' from expression of type 'base'
derived_class.cpp:70: error: in passing argument 2 of `void show2(std::string&, const derived&)'第二点对各种形式的继承作出验证，首先给出表格
| 继承方式\成员类型 |
public |
protected |
private |
| public |
public |
protected |
无法继承 |
| protected |
protected |
protected |
无法继承 |
| private |
private |
private |
无法继承 |
这里解释一下，这里仅仅表达基类的成员，被public,protected,private三种方式继承后，在原基类为public,protectedc,private的成员在继承类里类型为表格里内容
代码如下:
class base
{
public:
std::string testPublic()
{
return std::string("this is public base");
}
protected:
std::string testProtected()
{
return std::string("this is protected base");
}
private:
std::string testPrivate()
{
return std::string("this is private base");
}
};
class derivedPublic:public base
{
public:
std::string testPubPublic()
{
return testPublic()+= "in derived";
}
std::string testProPublic()
{
return testProtected()+= "in derived";
}
std::string testPriPublic()
{
return testPrivate()+= "in derived";
}
};
int main(int argc,char* argv[])
｛
derivedPublic dpub;
std::cout <｝
报下面错误，说明testPrivate()不是derived私有函数而是base的私有函数
derived11.cpp:16: error: `std::string base::testPrivate()' is private
derived11.cpp:36: error: within this context这样验证private类型成员无法被继承（public,private,protected）注：private，protected略去不做证明
下面只要验证 testProtected 能被第三层继承类继承，但是无法被第三层类直接调用就说明是public继承后继承类型为protected，而基类为Public类型成员则即可被继承又可以直接调用。
代码如下:
#include
#include
class base
{
public:
std::string testPublic()
{
return std::string("this is public base");
}
protected:
std::string testProtected()
{
return std::string("this is protected base");
}
private:
std::string testPrivate()
{
return std::string("this is private base");
}
};
class derivedPublic:public base
{
public:
std::string testPubPublic()
{
return testPublic()+= "in derived";
}
std::string testProPublic()
{
return testProtected()+= "in derived";
}
// std::string testPriPublic()
// {
// return testPrivate()+= "in derived";
// }
};
class deepDerived:public derivedPublic
{
public:
std::string deepProtected()
{
return testProtected() +="in deep";
}
std::string deepPublic()
{
return testPublic() +="indeep";
}
};
int main(int argc,char* argv[])
{
derivedPublic dpub;
std::cout < deepDerived deepdpub;
std::cout< std::cout< std::cout< std::cout<}
这里服务器报错
derived12.cpp:13: error: `std::string base::testProtected()' is protected
derived12.cpp:62: error: within this context这样就验证了一个是public，一个是protected，protected是不能直接调用的，但是被继承后是可以被public成员调用的。
下面的已经证明，详细步骤就略去如果对该部分验证感兴趣，可以看下面代码。
代码如下:
#include
2 #include
3 class base
4 {
5 public:
6 std::string testPublic()
7 {
8 return std::string("this is public base");
9 }
protected:
std::string testProtected()
{
return std::string("this is protected base");
}
private:
std::string testPrivate()
{
return std::string("this is private base");
}
};
class derivedPublic:public base
{
public:
std::string testPubPublic()
{
return testPublic()+= "in derived";
}
std::string testProPublic()
{
return testProtected()+= "in derived";
}
// std::string testPriPublic() //私有成员并没有被继承下来
// {
// return testPrivate()+= "in derived";
// }
};
class deepDerived:public derivedPublic
{
public:
std::string test()
{
return testPublic() +="in 3";
}
};
class derivedProtected:protected base
{
public:
std::string testPubProtected()
{
return testPublic()+= "in derived";
}
std::string testProProtected()
{
return testProtected()+= "in derived";
}
};
class deepDerived2:public derivedProtected
{
public:
std::string test()
{
return testPublic() +="in 3";
}
};
class derivedPrivate:private base
{
public:
std::string testPubPirvate()
{
return testPublic()+= "in derived";
}
std::string testProPrivate()
{
return testProtected()+= "in derived";
}
};
//class deepDerived3:public derivedPrivate
//{
// public:
// std::string test()
// {
// return testPublic() +="in 3";
// }
//};
int main(int argc,char* argv[])
{
derivedPublic dpub;
//derivedProtected dpro;
//derivedPrivate dpri;
std::cout< //std::cout< //cout< std::cout< std::cout< //std::cout<
deepDerived dd;
std::cout<
derivedProtected dpro;
//std::cout< std::cout< std::cout<
deepDerived2 dd2;
std::cout<
derivedPrivate dpri;
std::cout< std::cout<
// deepDerived3 dd3;
// std::cout<｝
京公网安备 11010802041100号 | 京ICP备19059560号-4 | PHP1.CN 第一PHP社区 版权所有 
