#include<iostream.h>

#include<windows.h>
class Base1
{protected:

int number1;
public:
Base1()
{number1=0;
cout<<"number1:"<<number1<<endl;
}
};
class Base2
{protected:
int number2;
public:
Base2(int x)
{
number2=x;

cout<<"number2:"<<number2<<endl;
}
~Base2()
{
cout<<"xi gou:"<<endl;
}
};
class Base3
{protected:

int number3;
public:
Base3(int y)
{
number3=y;

cout<<"number3:"<<number3<<endl;
}
~Base3()
{
cout<<"xi gou:"<<endl;
}
};
class Derive:private Base1,private Base2,private Base3
{private:
Base1 number1base;
Base2 number2base;
Base3 number3base;
public:
Derive(int a,int b);

};
Derive::Derive(int a,int b):number2base(a),number3base(b)
{}

void main()
{
Derive st(4,3);
}
程序目的：三个基类，一个派生类，三个基类有析构，构造函数（一个基类的构造函数无参），派生类有三个私有成员
是三个基类的对象；
从头到尾查了一遍，剩一个错误不知道怎么改：
error C2512: 'Base2' : no appropriate default constructor available
error C2512: 'Base3' : no appropriate default constructor available
请问是什么意思，请高人指点一下．．

[此贴子已经被作者于2006-4-23 22:31:08编辑过]

你要一开始就说，我就不会搞错了
#include<iostream.h>

#include<windows.h>//去掉，没用
class Base1
{protected:

int number1;
public:
Base1()
{number1=0;
cout<<"number1:"<<number1<<endl;
}
};
class Base2
{protected:
int number2;
public:

Base2(int x = 0)
{
number2=x;

cout<<"number2:"<<number2<<endl;
}
~Base2()
{
cout<<"xi gou:"<<endl;
}
};
class Base3
{protected:

int number3;
public:

Base3(int y = 0)
{
number3=y;

cout<<"number3:"<<number3<<endl;
}
~Base3()
{
cout<<"xi gou:"<<endl;
}
};
class Derive:private Base1,private Base2,private Base3

{private:
Base1 number1base;
Base2 number2base; //
Base3 number3base;//用带默认参数的构造函数
public:
Derive(int a,int b);

};
Derive::Derive(int a,int b):number2base(a),number3base(b)

{

}

void main()
{
Derive st(4,3);
}

[此贴子已经被作者于2006-4-24 10:02:38编辑过]

谢谢gototheworld,调出来了,单我不明白:Base2(int x = 0)
Base3(int y = 0)为什么这么写,主函数中st(4,3)不是已经把x,y
赋值了吗?能在稍微解释一下嘛?

{private:
Base1 number1base;
Base2 number2base; //你原来只写了一个构造函数，即
// Base2(int x)
//{
// number2=x;

// cout<<"number2:"<<number2<<endl;
// }但是你这里要调用的是默认构造函数，可是你没写这样的默认构造函数即Base2() {},所以编译出错在这里。
//我是看到你题目的要求“构造函数（一个基类的构造函数无参）”，所以我就没用默认构造函数的写法，
//只是在构造数里加了默认参数，即

Base2(int x = 0)
{
number2=x;

cout<<"number2:"<<number2<<endl;
}这里是用0，去赋值

而主函数中st(4,3)，是重新调用构造函数用3，4赋值

Base3 number3base;//最主要的原因是你这里没有构造函数来初始化number3base这个对象，所以就错了，用上面的方法来解决
public:
Derive(int a,int b);

};