我不知道为什么这个 void function 给我一个错误

Question

案例 2 中的 function 由于某种原因无法正常工作。 预期的 output 说参数是未定义的，但是当我定义参数时，它说它只能是 integer 但是当我用变量替换字符串时，调试器仍然发出错误说“C++ 确实不支持默认 Int。” 我不断更改变量以适应调试的需要，但程序仍然无法运行。

#include <iostream>
#include <string>
#include <sstream>
#include <vector>
#include <fstream>
#include <time.h>
#include <stdio.h>
#include <Windows.h>//Secret to play music: Go to project file properties, go to linker and find input, in output write this Winmm.lib and press okay and done.
#include <mmsystem.h>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <string>
#include <vector>
#include <iostream>
#include <iterator>
#include <algorithm>
#include <deque>


void email(string emailbackup);

using std::cout;
using std::cin;
using namespace std;
int main()
{


    int x;

    std::cout << "What would you like to do?" << std::endl;
    std::cout << "1)Would you like to see the archive?" << std::endl;
    std::cout << "2)Would you like to register student information?" << std::endl;
    std::cout << "3)Would you like to find a student within the registry?" << std::endl;
    std::cout << "4)Delete all student information?" << std::endl;
    std::cout << "5)Exit Program?" << std::endl;
    std::cin >> x;

    switch (x)
    {
    case 1:

        break;
    case 2:
        email(emailbackup);//This is where the error is located within int main.//
//emailbackup is the parameter I am describing within int main()//

            break;

    case 3:

        break;
    case 4:

        break;
    case 5:

        break;
    }
}

void email(string emailbackup)
{
    string emailbackup;
    int studentID = 0;
    std::string let("abcdefghijklmnopqrstuvwxyz");
    int numbers = (1, 3, 0);
    std::string str;
    std::string name;
    std::string studID;
    std::string studID2;
    string email;

    string studFinal;
    std::stringstream concatenate;
    std::stringstream concatenatetwo;


    int amountofStudents;
    std::cout << "How many student archives would you like to enter?" << std::endl;
    std::cin >> amountofStudents;
    amountofStudents + 1;
    for (int i = 0; i < amountofStudents; i++)
    {
        cout << "Please enter your name.." << std::endl;
        std::cin >> name;
        while (std::string::npos != name.find_first_of("0123456789"))
        {
            cout << "You must have letter within user input." << std::endl;
            cout << "Please enter your name." << std::endl;
            std::cin >> name;
        }
        //I need to check to see if the first 3 numbers are correct?
        //The student ID must be at least 6 digits, and the first 3 numbers must be 130. 
        cout << "Please enter Your student I.D." << std::endl;
        std::cin >> studID;

        stringstream(studID) >> studentID;

        while (/*std::string::npos != studID.find_first_of("130") */ studentID != 130 /*&& studID.length() <= 6*/)
        {
            stringstream(studentID) >> studID;
            cout << "You must enter 130 as the first 3 digits" << std::endl;
            std::cin >> studID;
            stringstream(studID) >> studentID;

        }
        //==============
        cout << "Please enter the second part of the student I.D. " << studentID << "-" << std::endl;
        std::cin >> studID2;

        while (/*std::string::npos != studID.find_first_of("130") */ studID2.length() < 3 || studID2.length() > 3)
        {
            //stringstream(studentID) >> studID;
            cout << "Add 3 more digits." << std::endl;
            std::cin >> studID2;


        }
        concatenate << studentID << "-" << studID2 << endl;
        studFinal = concatenate.str();
        /*while (studID.length() != 6)
        {

            cout << "You must enter 130 as the first 3 digits and you must have 6 digits." << std::endl;
            std::cin >> studID;

        }*/

        cout << "Please enter your email.." << endl;
        cin >> email;
        while (/*string::npos != email.find_first_not_of("@atlanticu.edu")*/ email == emailbackup || email.empty())
        {
            cout << "Please enter your email..." << endl;
            cin >> email;
        }
        concatenatetwo << email << "@atlanticu.edu" << endl;
        email = concatenatetwo.str();
        emailbackup = email;
        cout << "Your email is.. " << email << endl;
        system("pause");
    }

}

我只需要知道如何修复 function 参数，这样我就可以正确执行程序。

Answer 1

想想你的 function 签名。

void email(string emailbackup);

您是否真的希望调用者将字符串传递给 function email() ？

因为您随后在 function 中再次定义 emailbackup。

你可能想要这样的东西......

...
    case 2:
        const string backup = "backup@email";
        email(backup);
        break;
...
void email(string emailbackup)
{
    // string emailbackup; // delete this line
...
}

另外，删除using namespace std; 并与您的命名空间使用保持一致。 为什么要添加using namespace cout, cin等，如果您使用命名空间显式调用它们，即 std::cin、std::cout。

添加using namespace std::cout并使用cout而不是std::cout ，反之亦然。 不要混淆它。 与std::string和string相同。 让它看起来像你定义了自己的字符串......

最后一件事，为了便于阅读，我建议将这个 email function 分解为更小的函数。 例如：

void email(string emailbackup)
{
...
    int studentID = getStudentID();
    string studentEmail = getStudentEmail();
...
}

int getStudentID()
{
    int studentID;
    cout << "Please enter Your student I.D." << std::endl;
    std::cin >> studID;

    stringstream(studID) >> studentID;

    while (/*std::string::npos != studID.find_first_of("130") */ studentID != 130 /*&& studID.length() <= 6*/)
   {
       stringstream(studentID) >> studID;
       cout << "You must enter 130 as the first 3 digits" << std::endl;
       std::cin >> studID;
       stringstream(studID) >> studentID;

    }
    return studentID;
}