36

我在几个不同的论坛上尝试过，似乎无法得到一个直接的答案，我怎样才能让这个函数返回结构？如果我尝试'return newStudent;' 我收到错误“不存在从 studentType 到 studentType 的合适的用户定义转换。”

// 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);

}
4

5 回答 5

42

这是您的代码的编辑版本，它基于ISO C++并且适用于 G++：

#include <string.h>
#include <iostream>
using namespace std;

#define NO_OF_TEST 1

struct studentType {
    string studentID;
    string firstName;
    string lastName;
    string subjectName;
    string courseGrade;
    int arrayMarks[4];
    double avgMarks;
};

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

    return newStudent;
}

int main() {
    studentType s;
    s = input();

    cout <<"\n========"<< endl << "Collected the details of "
        << s.firstName << endl;

    return 0;
}
于 2013-10-06T05:21:18.953 回答
19

你有一个范围问题。在函数之前定义结构，而不是在函数内部。

于 2013-10-06T03:13:17.753 回答
7 
studentType newStudent() // studentType doesn't exist here
{   
    struct studentType // it only exists within the function
    {
        string studentID;
        string firstName;
        string lastName;
        string subjectName;
        string courseGrade;

        int arrayMarks[4];

        double avgMarks;

    } newStudent;
...

将其移到函数之外：

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

    int arrayMarks[4];

    double avgMarks;

};

studentType newStudent()
{
    studentType newStudent
    ...
    return newStudent;
}
于 2013-10-06T03:15:13.713 回答
7

您现在可以 (C++14) 返回一个本地定义的（即在函数内部定义的），如下所示：

auto f()
{
    struct S
    {
      int a;
      double b;
    } s;
    s.a = 42;
    s.b = 42.0;
    return s;
}

auto x = f();
a = x.a;
b = x.b;
于 2020-02-18T21:19:42.253 回答
0

正如其他人所指出的，在函数之外定义 studentType 。还有一件事，即使您这样做，也不要在函数内创建本地 studentType 实例。该实例位于函数堆栈上，当您尝试返回它时将不可用。但是，您可以做的一件事是动态创建 studentType 并在函数外部返回指向它的指针。

于 2013-10-06T03:39:29.390 回答