Question

我尝试编写一个程序，要求用户提供电影信息。将影片的信息作为结构存储在向量中，然后将结果输出到具有返回类型为void的2个函数的屏幕。

#include <iostream> 
#include <iomanip> 
#include <vector>
#include <string>
using namespace std;

void make_movie(struct movie *film);
void show_movie(vector <movie> data, int cnt);

struct movie {
    string name;
    string director;
    int year;
    int duration;
};

int main() {

    int count = 0;
    char input;
    vector <movie> record;
    movie *entry = nullptr;

    do {

        make_movie(entry);
        record.push_back(*entry);
        count++;

        cout << endl;
        cout << "Do you have more movie info to enter?\n";
        cout << "Enter y / Y for yes or n / N for no: ";
        cin.ignore();
        cin >> input;
        cout << endl;


    } while (input == 'y' || input == 'Y');

    show_movie(record, record.size());

    return 0;
}

void make_movie(struct movie *film) {

    cout << "Enter the title of the movie: ";
    cin.ignore();
    getline(cin, film -> name);

    cout << "Enter the director's name: ";
    cin.ignore();
    getline(cin, film -> director);

    cout << "Enter the year the movie was created: ";
    cin >> film -> year;

    cout << "Enter the movie length (in minutes): ";
    cin >> film -> duration;

}

void show_movie(vector <movie> data, int cnt) {

    cout << "Here is the info that you entered: " << endl;

    for (int i = 0; i < cnt; i++) {

        cout << "Movie Title: " << data[i].name << endl;
        cout << "Movie Director: " << data[i].director << endl;
        cout << "Movie Year: " << data[i].year << endl;
        cout << "Movie Length: " << data[i].duration << endl;
        cout << endl;
    }
 }

我收到一条错误消息，表示我正在尝试访问禁止的内存地址。

Answer 1

您需要做的最少量更改是：

movie *entry = nullptr;

do {
    make_movie(entry);
    record.push_back(*entry);

为：

movie entry;

do {
    make_movie(&entry);
    record.push_back(entry);

进一步改善：

将make_movie更改为通过引用接受参数，然后您的程序不使用任何指针，因此不容易受到与指针相关的任何问题的影响。
将make_movie更改为按值返回，而不是使用参考参数。
cin.ignore();使用不当。您的程序将丢失几个输入字符串的第一个字符。相反，删除所有这些调用，并在make_movie函数的末尾忽略当前行的其余部分。另外，请将cin >> input;更改为使用getline。

Answer 2

你的错误
movie *entry = nullptr;

和
你有额外的cin.ignore();

cout << "Enter the title of the movie: "; // cin.ignore(); getline(cin, film -> name); cout << "Enter the director's name: "; // cin.ignore(); getline(cin, film -> director);

如何修复

movie main_info; movie* entry = &main_info;

<强>测试

intput：

Enter the title of the movie: any_thing Enter the director's name: yourself Enter the year the movie was created: 2016 Enter the movie length (in minutes): 120 Do you have more movie info to enter? Enter y / Y for yes or n / N for no: n

输出

Here is the info that you entered: Movie Title: any_thing Movie Director: yourself Movie Year: 2016 Movie Length: 120

我该如何解决这个指针/内存问题？

2 个答案: