指向继承中的对象的指针

Pointer to an object in inheritance

本文关键字:对象 指针 继承      更新时间:2023-10-16

在这里我没有得到正确的输出,这里我使用了继承概念,但我不知道如何使用指针类型的对象调用方法。

请有人给我一些解决方案。

#include <iostream>
#include <cstring>
#include <iomanip>
using namespace std;    
class person
{
private:
    char *name,*blood,*dob;
    int height,weight;
public:
    static int count;
    person()
    {
        strcpy(name,"Name");
        strcpy(blood,"Blood");
        height=0;
        weight=0;
    }
    ~person()
    {
    }
    friend class person_info;
};

class person_info
{
private:
    char *add,*tel,*lic,*insu;
public:
    void getdata(person *obj,int n);
    void display(person *obj);
};
void person_info::getdata(person *obj,int n)
{
    for(int i=0;i<n;i++)
    {
        cin.ignore();
        cout<<"Enter Name Of the Person : ";
        cin.getline(obj->name,30);
        cout<<"Enter Blood group Of the Person : ";
        cin.getline(obj->blood,30);
        cout<<"Enter date of birth of the Person : ";
        cin.getline(obj->dob,30);
        cout<<"Enter Height of the Person : ";
        cin>>obj->height;
        cout<<"Enter Weight of the Person : ";
        cin>>obj->weight;
        cout<<"Enter Address Of the Person : ";
        cin.getline(this->add,30);
        cout<<"Enter Insurance no. Of the Person : ";
        cin.getline(this->insu,30);
        cout<<"Enter Telephone no Of the Person : ";
        cin.getline(this->tel,30);
        cout<<"Enter License no the Person : ";
        cin.getline(obj->blood,30);
    }
}
void person_info::display(person *obj)
{
    cout<<setw[10]<<"Name"<<setw[10]<<"Address"<<setw[10]<<"D. O. B."<<setw[3]<<"Blood G."<<setw[4]<<"Height"<<setw[10]<<"Weight"<<setw[10]<<"Insrn No."<<setw[10]<<"Tele No."<<setw[11]<<"Licence No.";
    cout<<setw[10]<<obj->name<<setw[10]<<this->add<<setw[10]<<obj->dob<<setw[3]<<obj->blood<<setw[4]<<obj->height<<setw[10]<<obj->weight<<setw[10]<<this->insu<<setw[10]<<this->tel<<setw[11]<<this->lic;
}
int main()
{
int ch=0,i=0,n;
    do
    {
        cout<<"1.getdata"<<endl;
        cout<<"2.display data"<<endl;
        cout<<"Enter choice";
        cin>>ch;
        person_info *p[2];        \I think this part is not correct
        p[2]=new person_info();   \I think this part is not correct
        person *p1[2];            \I think this part is not correct
        p1[2]=new person();       \I think this part is not correct
        switch(ch) 
        { 
        case 1: 
            cout<<"Enter No. Entries to be Entered :"; 
            cin>>n; 
            p[2]->getdata(p1[2],n);
            break;
        case 2: 
            for(int j=0;j<i;j++)
            {
                p[i]->display(p1[i]); 
            }
            break;
        }
    }while(ch!=3); 

您既没有初始化成员char *name,*blood,*dob;,也没有为它们分配内存。我建议使用std::string而不是char*

#include <string>
class person
{
   private:
   std::string name;
   std::string blood;
public:
    person()
        : name( "Name" )
        , blood( "Blood" )
    {}
    person( const char *n, const char *b )
        : name( n )
        , blood( b )
    {}
};

注意:strcpy(name,"Name");将字符串"Name"复制到 char *name 引用的内存中,但您永远不会为 name 分配任何动态内存。 name未初始化和未定义。