UC知道为什么定义了类的构造函数，却提示“没有合适的默认的构造函数可用”
来源：百度知道 编辑：UC知道 时间：2024/05/22 13:53:25
为什么定义了类的构造函数，却提示“没有合适的默认的构造函数可用”，我想在构造子类的构造函数的时候也调用基类的带参构造函数，该怎么写呀，急呀，那位高手知道呀，在下不胜感激呀！
#include <iostream>
using namespace std;
class animal
{
public:
int _age;
char* _name;
animal(int age,char* name)
{
_age=age;
_name=name;
}
};
class Fish:public animal
{
public:
int _rate;
char* _hobic;
Fish(int rate,char* hobic)
{
_rate=rate;
_hobic=hobic;
}
};
void main()
{
Fish fish(18,"打篮球")
cout<<fish._age<<endl;
cout<<fish._name<<endl;
cout<<fish._rate<<endl;
cout<<fish._hobic<<endl;
}
#include <iostream>
using namespace std;
class animal
{
public:
int _age;
char* _name;
animal(int age,char* name)
{
_age=age;
_name=name;
}
};
class Fish:public animal
{
public:
int _rate;
char* _hobic;
Fish(int rate,char* hobic)
{
_rate=rate;
_hobic=hobic;
}
};
void main()
{
Fish fish(18,"打篮球")
cout<<fish._age<<endl;
cout<<fish._name<<endl;
cout<<fish._rate<<endl;
cout<<fish._hobic<<endl;
}
#include <iostream>
using namespace std;
class animal
{
public:
int _age;
char* _name;
animal(int age,char* name)
{
_age=age;
_name=name;
}
animal( void )
{
}
};
class Fish : public animal
{
public:
int _rate;
char* _hobic;
Fish(int rate,char* hobic, int age, char* name )
{
_rate=rate;
_hobic=hobic;
_age = age;
_name = name;
}
};
void main()
{
Fish fish( 1,"hobic", 18, "打篮球");
cout<<fish._age<<endl;
cout<<fish._name<<endl;
cout<<fish._rate<<endl;
cout<<fish._hobic<<endl;
}
或者————————————
#include <iostream>
using namespace std;
class animal
{