c++ - 如何从C++中的函数返回结构？

我曾在几个不同的论坛上尝试过，但似乎无法得到一个直接的答案，如何使此函数返回结构？如果我尝试“返回newStudent；”我收到错误消息“不存在从studentType到studentType的适当的用户定义转换。”

// Input function
studentType newStudent()
{
    struct studentType
    {
        string studentID;
        string firstName;
        string lastName;
        string subjectName;
        string courseGrade;

        int arrayMarks[4];

        double avgMarks;

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

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

}

最佳答案

您遇到范围问题。在函数之前而不是在函数内部定义结构。