如何在子类的无参构造函数中,引用父类的有参构造函数
怎么在子类的无参构造函数中,引用父类的有参构造函数?
典型的是这样的:
#include <iostream.h>
class animal
{
public:
animal(int height, int weight)
{
cout < < "animal construct " < <endl;
}
…
};
class fish:public animal
{
public:
fish():animal(400,300)
{
cout < < "fish construct " < <endl;
}
…
};
void main()
{
fish fh;
}
-----------------------------
现在要求写成如下形式。
class fish:public animal
{
public:
fish()
{
..... //部分代码
:animal(400,300)//这句话怎么写啊?
cout < < "fish construct " < <endl;
}
…
};
------解决方案--------------------
fish():animal(400,300)
{
cout < < "fish construct " < <endl;
}
这个就是对的嘛。
为什么要修改?
典型的是这样的:
#include <iostream.h>
class animal
{
public:
animal(int height, int weight)
{
cout < < "animal construct " < <endl;
}
…
};
class fish:public animal
{
public:
fish():animal(400,300)
{
cout < < "fish construct " < <endl;
}
…
};
void main()
{
fish fh;
}
-----------------------------
现在要求写成如下形式。
class fish:public animal
{
public:
fish()
{
..... //部分代码
:animal(400,300)//这句话怎么写啊?
cout < < "fish construct " < <endl;
}
…
};
------解决方案--------------------
fish():animal(400,300)
{
cout < < "fish construct " < <endl;
}
这个就是对的嘛。
为什么要修改?