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;}
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
voID show2(std::string& info,const derived& d){ info.append("name is "); info.append(d.getBasename()); info.append(",d.getBaseData()); info.append(buffer);}
调用show(ss,d);编译器报错
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&)'
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,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
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 << dpub.testPublic() << std::endl; }
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 privatederived11.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。
#include <iostream>#include <string>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 << dpub.testProtected() << std::endl; deepDerived deepdpub; std::cout<<deepdpub.testPublic() <<std::endl; std::cout<<deepdpub.testProtected() <<std::endl; std::cout<<deepdpub.deepProtected() <<std::endl; std::cout<<deepdpub.deepPublic() <<std::endl;}
这里服务器报错
derived12.cpp:13: error: `std::string base::testProtected()' is protectedderived12.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。
#include <iostream>#include <string>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() //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<<dpub.testPublic()<<std::endl; // //std::cout<<dpub.testProtected()<<std::endl; //用户被继承也是无法使用 //cout<<dpub.testPrivate()<<std::endl; //基类都是私有函数 std::cout<<dpub.testPubPublic()<<std::endl; std::cout<<dpub.testProPublic()<<std::endl; //std::cout<<dpub.testPriPrivate()<<std::endl; //没有被继承 deepDerived dd; std::cout<<dd.test()<<std::endl; derivedProtected dpro; //std::cout<<dpro.testPublic()<<std::endl; //变成protected类型 std::cout<<dpro.testPubProtected()<<std::endl; std::cout<<dpro.testProProtected()<<std::endl; deepDerived2 dd2; std::cout<<dd2.test()<<std::endl; derivedPrivate dpri; std::cout<<dpri.testPubPirvate()<<std::endl; std::cout<<dpri.testProPrivate()<<std::endl; // deepDerived3 dd3;// std::cout<<dd3.test()<<std::endl;}
以上就是对C++ j继承的资料整理,后续继续补充相关资料,谢谢大家对本站的支持!
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