指针麻烦

Jay*_*sen 4 c++ pointers class

#include "stdafx.h"
#include <iostream>
#include <vector>
#include <string>

using namespace std;

class Person
{
    public:
    Person();
    Person(string input_name, int input_age);
    void read(string input_name, int input_age);
    void print() const;

    private:
    string name;
    int age;
};

Person::Person()
{
name = " ";
age = 0;
}

Person::Person(string input_name, int input_age)
{
name = input_name;
age = input_age;
}
void Person::read(string input_name, int input_age)
{
name = input_name;
age = input_age;
}

void Person::print() const
{
cout << "Name: " << name << " Age: " << age << endl;
}




class Car
{
public:
Car();
Car(string input_model, Person* p_input_owner, Person* p_input_driver);
void read(string input_model, Person* p_input_owner, Person* p_input_driver);
void print() const;

private:
string model;
Person* p_owner;
Person* p_driver;
};

Car::Car()
{
model = " ";
p_owner = NULL;
p_driver = NULL;
}

Car::Car(string input_model, Person* p_input_owner, Person* p_input_driver)
{
model = input_model;
p_owner = p_input_owner;
p_driver = p_input_driver;
}

void Car::read(string input_model, Person* p_input_owner, Person* p_input_driver)
{
model = input_model;
p_owner = p_input_owner;
p_driver = p_input_driver;
}

void Car::print() const
{
cout << "Model: " << model << " Owner: " << p_owner << " Driver: " << p_driver << endl;
}



int main()
{
vector<Person*> people(4);
vector<Car*> cars;
string input_name;
int input_age = 0;
string remainder;

for(int i = 0; i < people.size(); i++)          
{
    cout << "Enter the person's name: ";
    getline(cin, input_name);
    cout << "Enter the person's age: ";
    cin >> input_age;

    people[i]->read(input_name, input_age);
}

for(int i = 0; i < people.size(); i++)
{
    people[i]->print();
}

return 0;
}
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我正在尝试为学校编写一个程序,使用Person*和Car*类型的向量.这就是我到目前为止,代码编译得很好但是当我填充人员矢量时程序崩溃并且我得到异常错误.

我不希望任何人为我做作业,只是指出我正确的方向,所以我可以继续前进.

感谢你给与我的帮助.

显然代码是不完整的,我目前只是试图填充人物矢量.一旦有效,我将转向汽车矢量.

Mic*_*ker 5

您调用从未实例化的对象的成员函数,即指针从未分配.也就是说,在打电话之前people[i]->read你应该分配一些东西people[i],比如new Person()new Person(input_name,input_age).

一般来说,如果你有一行崩溃你的问题并且你无法通过阅读你的代码弄明白,你可能只打印出所涉及的变量.喜欢people[i].这可能NULL会给你一些想法.