C++语法基础--恢复继承成员的访问级别,默认继承保护级别,友元关系与继承,继承与静态成员

淡淡的烟草味﹌ 2021-12-20 20:37 226阅读 0赞

**1.恢复继承成员的访问级别（using 声明）  
\*派生类可以恢复继承成员的访问级别，但是不能使访问级别比基类中原来指定的更严格或更宽松  
  
Example：  
 class base  
\{  
public:  
 int x;  
  
\};  
  
class private\_derived:private base  
\{  
public:  
 using base::x;  
  
 \};  
  
  
 int main()  
 \{   
private\_derived pd;  
pd.x=1;// 如果去除using base::x,则出错： 'int base::x' is inaccessible   
cout<<pd.x<<endl;//1  
 \}  
  
  
  
  
2.默认继承保护级别  
 \*class保留字定义的派生类默认具有private继承  
\*struct保留字定义的派生类默认具有public继承  
  
Example:  
  
 class base  
\{  
  
\};  
  
class private\_derived: base  
\{  
  
\};  
  
等效于  
 class private\_derived: private base  
\{  
  
\};  
  
  
3.友元关系与继承  
 \*友元关系不能继承，基类的友元对派生类的成员没有特殊访问权限  
Example：  
 class base  
 \{   
 friend class frd;  
 public:  
base(int i):x(i)  
\{   
\}  
 private:  
int x;  
  
 \};  
  
 class derived: public base   
 \{   
public:  
 derived(int j):base(j)  
 \{   
  
 \}   
 private:  
int y;  
 \};  
  
  
 class frd   
 \{   
public:  
void fun1(base b)  
\{   
cout<<"base:: "<<b.x<<endl;   
\}  
void fun2(derived d)  
\{   
//cout<<"derived:: "<<d.y<<endl;//error, 'int derived::y' is private  
\}  
 \};   
 int main()  
 \{   
base b(1);  
derived d(2);  
frd f;  
f.fun1(b);//base:: 1  
  
 return 0;  
 \}  
  
  
4.继承与静态成员  
 \*如果基类定义了static成员，则整个继承层次中只有一个这样的成员  
\*static成员遵循常规访问控制  
  
  
Example：  
 class base  
\{  
  
public:  
static void fun()  
\{  
cout<<"static base::fun()"<<endl;  
\}  
  
\};  
  
class derived: public base  
\{  
  
\};  
  
  
  
int main()  
\{  
  
derived d;  
derived \*pd=new derived;  
derived::fun();//static base::fun()  
d.fun();//static base::fun()  
pd->fun();//static base::fun()  
  
 return 0;  
 \}**