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
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#include <stdio.h>
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 %d\n" , aClass->a);
printf ( "a value of b %d\n" , bClass->a);
return 0;
}
|
运行结果:
2. 函数和Class都继承
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#include <stdio.h>
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 %d\n" , aClass->a);
printf ( "a value of b %d\n" , bClass->a);
return 0;
}
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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。
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class Base
{
public :
Base(){}
~Base(){}
int a;
virtual void setA()=0;
};
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3. 类A中不有函数,则继承自基类
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class A: public Base
{
public :
A(){}
~A(){}
};
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结果为:
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原文链接:https://www.cnblogs.com/anlia/p/9083877.html