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所以我有这个 do-while 循环，如果你按下 y 并通过 input() 函数输入信息，它应该重复询问你是否想为学生输入信息，然后将数据存储在一个向量中。如果按 q，程序应该打印向量中包含的信息并退出循环。

由于某种原因，循环会为您输入的第一个和第二个学生完全执行。然后，不是重复循环询问您是否要输入第三个学生，而是似乎只执行 input() 函数。它不会要求您“输入 y 继续或 q 退出”，并且您在此处输入的任何学生信息似乎都没有被存储。它会在执行完整循环和仅执行 input() 函数之间间歇地进行更改。我想知道是否有人知道为什么会发生这种情况以及我能做些什么来解决它。

干杯

#include <iostream>
#include <iomanip>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>

using namespace std;

const int NO_OF_TEST = 4;

struct studentType
{
    string studentID;
    string firstName;
    string lastName;
    string subjectName;
    string courseGrade;

    int arrayMarks[4];

    double avgMarks;
};

studentType input();
double calculate_avg(int marks[],int NO_OF_TEST); // returns the average mark
string calculate_grade (double avgMark); // returns the grade 

void main()
{
    unsigned int n=0; // no. of student
    vector<studentType> vec; // vector to store student info
    studentType s;

    char response;
    do
    {
        cout << "\nEnter y to continue or q to quit... ";
        cin >> response;
        if (response == 'y')
        {
            n++;
            for(size_t i=0; i<n; ++i)
            {
                s = input();
                vec.push_back(s);
            }
        }

        else if (response == 'q')
        {
            for (unsigned int y=0; y<n; y++)
            {
                cout << "\nFirst name: " << vec[y].firstName;
                cout << "\nLast name: " << vec[y].lastName;
                cout << "\nStudent ID: " << vec[y].studentID;
                cout << "\nSubject name: " << vec[y].subjectName;
                cout << "\nAverage mark: " << vec[y].avgMarks;
                cout << "\nCourse grade: " << vec[y].courseGrade << endl << endl;
            }
        }
    }
    while(response!='q');
}

studentType input()
{   
    studentType newStudent;

    cout << "\nPlease enter student information:\n";

    cout << "\nFirst Name: ";
    cin >> newStudent.firstName;

    cout << "\nLast Name: ";
    cin >> newStudent.lastName;

    cout << "\nStudent ID: ";
    cin >> newStudent.studentID;

    cout << "\nSubject Name: ";
    cin >> newStudent.subjectName;

    for (int x=0; x<NO_OF_TEST; x++)
    {   cout << "\nTest " << x+1 << " mark: ";
        cin >> newStudent.arrayMarks[x];
    }

    newStudent.avgMarks = calculate_avg(newStudent.arrayMarks,NO_OF_TEST );
    newStudent.courseGrade = calculate_grade (newStudent.avgMarks);

    return newStudent;
}

double calculate_avg(int marks[], int NO_OF_TEST)
{
     double sum=0;

     for( int i=0; i<NO_OF_TEST; i++)
     {
         sum = sum+ marks[i];
     }

     return sum/NO_OF_TEST;

}


string calculate_grade (double avgMark)
{
    string grade= "";

    if (avgMark<50)
    {
        grade = "Fail";
    }

    else if (avgMark<65)
    {
        grade = "Pass";
    }

    else if (avgMark<75)
    {
        grade = "Credit";
    }       

    else if (avgMark<85)
    {
        grade = "Distinction";
    }

    else
    {
        grade = "High Distinction";
    }

    return grade;
}
4

1 回答 1

0

我认为这段代码可以做到：

        n++;
        for(size_t i=0; i<n; ++i)
        {
            s = input();
            vec.push_back(s);
        }

它第二次要两个学生，第三次要三个学生，等等。

所以，做吧

        vec.push_back(input());
于 2013-10-06T14:31:07.433 回答