想问一下，为什么多态继承必须要求有默认构造函数啊？

#include <iostream>
#include <string>
#include <conio.h>
using namespace std;

class person
{
public:
    person()
    {
        cout<<endl<< "person construct!" <<endl;
    }
    virtual void display() = 0;
protected:
    int id;
    string name;
};

class student: virtual public person
{
public:
    student(){}    //为什么一定要有个默认构造函数啊？否这就报错...？
    student(int pid, string pname, int pnum, float pscore)
    {
        cout<<endl<< "student constructor";
        id = pid;
        name = pname;
        num = pnum;
        score = pscore;
    }
    void display()
    {
        cout<<endl<<"ID: "<<id;
        cout<<endl<<"Name: "<<name;
        cout<<endl<<"Num: "<<num;
        cout<<endl<<"score: "<<score;
    }
protected:
    int num;
    float score;
};

class teacher: virtual public person
{
public:
    teacher(){}   //还有这里？
    teacher(int pid, string pname, string pjob, string pdep)
    {
        cout<<endl<< "teach constructor";
        id = pid;
        name = pname;
        job = pjob;
        dep = pdep;
    }
    void display()
    {
        cout<<endl<<"ID: "<<id;
        cout<<endl<<"Name: "<<name;
        cout<<endl<<"Job: "<<job;
        cout<<endl<<"Department: "<<dep;
    }
protected:
    string job;
    string dep;

};


class graduate_student: virtual public student,virtual public teacher
{
public:
    graduate_student(int pid, string pname, int pnum, string pjob)
    {
        cout<<endl<<"graduate construct";
        id = pid;
        name = pname;
        num = pnum;
        job = pjob;
    }
    void display()
    {
        cout<<endl<<"ID: "<<id;
        cout<<endl<<"Name: "<<name;
        cout<<endl<<"Num: "<<num;
        cout<<endl<<"Job: "<<job;
    }
};

int main()
{
    student s1(2011201,"jack",1,100);   
    s1.display();
    teacher t1(1022,"john","teacher","school");
    t1.display();
    graduate_student gs1(13561,"xue",23,"techer student");
    gs1.display();
    getch();
    return 0;
}

[ 本帖最后由 飞扬冲天 于 2012-4-25 21:36 编辑 ]

为什么多态继承必须要求有默认构造函数啊？
------ 不需要呀

student(){}    //为什么一定要有个默认构造函数啊？否这就报错...？
------ 因为graduate_student的构造函数调用了student的默认构造函数
你可以不让它调用默认构造函数，比如
graduate_student(int pid, string pname, int pnum, string pjob) : student(pid,pname,pnum,0) // 逻辑对不对我不管，我说的只是语法
{

嗯，明白了...第一次看C++，我只是过了一次..现在回过头来发现好多东西理解的不透彻
谢谢了