使用<vector>使用 cin 和 for() 循環 C++

Question

我從我的練習中得到了這段代碼，我需要：

• 添加學生的姓名、身份證、教育程度和 3 個年級
• 修改學生成績
• 刪除學生
• 顯示批准的學生名單
• 顯示推薦學生名單
• 顯示完整列表

所以我使用向量完成了這個，這是我的代碼：

int resp = 0;

vector<int> ids;
vector<string> names;
vector<string> levels;
vector<double> grades1;
vector<double> grades2;
vector<double> grades3;
vector<double> average;
vector<double> approved;
vector<double> repproved;

string name, level;
int id, elim, id_delete, i=0;
double grade=0, sum=0;

do {

    cout<<"######### OPTIONS MENU #########"<<"\n\n";

    cout<<"[1] Add a student"<<"\n";
    cout<<"[2] Add grades of a student"<<"\n";
    cout<<"[3] Modify grades of a student"<<"\n";   
    cout<<"[4] Delete a student"<<"\n";
    cout<<"[5] Show approved list"<<"\n";
    cout<<"[6] Show repproved list"<<"\n";
    cout<<"[7] Show the full list"<<"\n";
    cout<<"[8] Exit"<<"\n\n";

    cin>>resp;

    switch(resp) {
        case 1:
            cout<<"Enter the student's name: \n";
            cin.sync(); cin.clear();
            getline(cin, name);
            names.push_back(name);
            cout<<"Enter the student's id: \n";
            cin>>id;
            ids.push_back(id);
            cout<<"Enter the student's school level: \n";
            cin.sync(); cin.clear();
            getline(cin, level);
            levels.push_back(level);

            cout<<"Student added";

        break;
        case 2:
            cout<<"Enter the grade 1 of the student: \n";
            cin>>grade;
            grades1.push_back(grade);
            cout<<"Enter the grade 2 of the student: \n";
            cin>>grade;
            grades2.push_back(grade);
            cout<<"Enter the grade 3 of the student: \n";
            cin>>grade;
            grades3.push_back(grade);

            cout<<"Grades added";
        break;
        case 3:
            cout<<"Enter the student's id you wish to modify grades on: \n";
            cin>>id_delete;

            for(int i=0; i<ids.size(); i++) {
                if(id_delete==ids[i]) {
                    id_delete = i;
                    break;
                }
            }

            i=id_delete;
            cout<<"Enter the new grade 1 of the student: \n";
            cin>>grade;
            grades1[i] = grade;

            cout<<"Enter the new grade 2 of the student: \n";
            cin>>grade;
            grades2[i] = grade;

            cout<<"Enter the new grade 3 of the student: \n";
            cin>>grade;
            grades3[i] = grade;

            cout<<"Grades modified\n";

        break;
        case 4:
            cout<<"Enter the student's id you wish delete: \n";
            cin>>id_delete;

            for(int i=0; i<ids.size(); i++) {
                if(id_delete==ids[i]) {
                    id_delete = i;
                    break;
                }
            }

            ids.erase (ids.begin()+(id_delete-1));
            names.erase (names.begin()+(id_delete-1));
            levels.erase (levels.begin()+(id_delete-1));
            grades1.erase (grades1.begin()+(id_delete-1));
            grades2.erase (grades2.begin()+(id_delete-1));
            grades3.erase (grades3.begin()+(id_delete-1));

            cout<<"Student deleted\n";

        break;
        case 5:
            for(i = 0; i<ids.size(); i++) {
                average[i] = (grades1[i] + grades2[i] + grades3[i])/3;
            }

            for(i = 0; i<ids.size(); i++) {
                if(average[i] >= 7 && average[i] <= 10) {
                    approved[i] = average[i];
                } else {
                        repproved[i] = average[i];
                    }

            }

            for(i = 0; i<ids.size(); i++) {
                cout<<approved[i]<<endl;
            }
        break;
        case 6:
            for(i = 0; i<ids.size(); i++) {
                cout<<repproved[i]<<endl;
            }
        break;
        case 7:
            for(i = 0; i<ids.size(); i++) {
                cout<<approved[i]<<endl;
            }

            for(i = 0; i<ids.size(); i++) {
                cout<<repproved[i]<<endl;
            }

但是在選擇選項 5、6 和 7 時它會拋出“exercise.exe 已停止工作”。

附注。 我是 C++ 的初學者，遺憾的是，它是我的第一門編程語言......

Answer 1

發生崩潰是因為您訪問了超出其末端的多個向量的成員。

例如：

for(i = 0; i<ids.size(); i++) {
    average[i] = (grades1[i] + grades2[i] + grades3[i])/3;
}

在訪問它之前，您永遠不會擴展平均數組。 您確實會擴展成績數組，但僅限於響應特定的用戶輸入。

每次添加學生時，您可能希望在所有數組中添加條目。 並且您應該在添加成績之前詢問學生ID，並將這種情況視為修改。

更好的設計是使用從學生 ID 到結構的映射。

Answer 2

作為向量

vector<double> average;
vector<double> approved;
vector<double> repproved;

創建后，它們的大小為 0，即它們不包含任何元素，因此當您第一次將上述向量的元素作為平均值 [i] 訪問時，該元素不在內存中，因此您必須先插入平均值作為，

average.push_back((grades1[i] + grades2[i] + grades3[i])/3);

不像

average[i] = (grades1[i] + grades2[i] + grades3[i])/3;

與其他兩個向量相同

批准和重新批准

我希望你很清楚。