类继承有限个数据成员，结果显示烫烫

Question

题目要求

定义一个Teacher和Student类，二者有num,name,sex三个数据成员的相同的。编写程序将一个Student对象转换为Teacher类，只将这3个相同的数据成员移植过去。

用代码块功能插入代码，请勿粘贴截图

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include<string>
using namespace std;

class Student{
public:
    Student(int n, const char* nam, const char* s, int a) {
        num = n;
        strcpy(name, nam);
        strcpy(sex, s);    
        age=a;
    }
    int get_num() {
        return num;
    }
    char* get_name() {
        return name;
    }
    char* get_sex() {
        return sex;
    }
    void display() {
        cout << "Student's info:" << endl << "Num:" << num << endl << "Name:" << name << endl << "Sex:" << sex << endl << "Age:" << age << endl;
    }
private:
    int num;
    char* name;
    char* sex;
    int age;
};

class Teacher {
public:
    Teacher(){}
    Teacher(Student&);
    Teacher(int n,const char* nam,const char* s, int sa) {
    num = n;
    strcpy(name, nam);
    strcpy(sex, s);
    salary = sa;
    }
    void display();
private:
    int num;
    char* name;
    char* sex;
    int salary = 1000;
};

Teacher::Teacher(Student& s) {
    num = s.get_num();
    strcpy(name, s.get_name());
    strcpy(sex, s.get_sex());
}

void Teacher::display() {
    cout << "Teacher's info:" << endl << "Num:" << num << endl << "Name:" << name << endl << "Sex:" << sex << endl << "Salary:" << salary << endl;
}

int main() {
    Student s(123456, "Zhang", "female", 18);
    s.display();
    cout << endl;
    Teacher t = Teacher(s);
    t.display();
    return 0;
}

运行结果及报错内容

我的解答思路和尝试过的方法

name和sex的类型改为char*，运行结果为空

我想要达到的结果

正确显示name和sex

Answer 1

Teacher::Teacher(Student& s) {
num = s.get_num();
name[10] = s.get_name();
sex[7] = s.get_sex();
}
name和sex复制都是错的
get_name和get_sex函数返回值改为char *
然后改为 strcpy(name,s.get_name());和strcpy(sex,s.get_sex());