getline出错

Question

我在这里收到错误，我不知道为什么，我对c ++很陌生，如果你能查看我的其余代码，那么确保它是好的

我在这两行上收到错误。

getline(in, e.first, ',');
getline(in, e.last, ',');

它说班级员工没有会员第一，我知道它不在该功能中，我该如何解决？

以下是我的其余代码。

#include <iostream>
#include <string>
#include <fstream>

using namespace std;

struct Person {
    string first;
    string last;
};

struct Address {
    string street;
    string city;
    string state;
    string zipcode;
};

struct Employee {
    Person name;
    Address homeAddress;
    int eid;
};

void readEmployee(istream& in, Employee& e);
void displayEmployee(ostream& out, const Employee& e);

int main(int argc, const char* argv[]) {
    Employee e[50];

    ifstream fin;
    ofstream fout;

    fin.open("employeesIn.txt");

    if (!fin.is_open()) {
        cerr << "Error opening employeesIn.txt for reading." << endl;
        exit(1);
    }
    fout.open("employeesOut.txt");
    if (!fout.is_open()) {
        cerr << "Error opening employeesOut.txt for writing." << endl;
        exit(1);
    }
    int EmployeePopulation = 0;
    readEmployee(fin, e[EmployeePopulation]);
    while (!fin.eof()) {
        EmployeePopulation++;
        readEmployee(fin, e[EmployeePopulation]);
    }
    fin.close();
    for (int i = 0; i <= EmployeePopulation - 1; i++) {
        displayEmployee(fout, e[i]);
    }

    fout.close();

    cout << endl;

    return 0;
}

void readEmployee(istream& in, Employee& e) {
    string cidText;
    if (getline(in, cidText, ',')) {
        e.eid = stoi(cidText);

        getline(in, e.first, ',');
        getline(in, e.last, ',');

        getline(in, e.homeAddress.street, ',');
        getline(in, e.homeAddress.city, ',');
        getline(in, e.homeAddress.state, ',');

        string zipcodeText;
        getline(in, zipcodeText, ',');
        e.homeAddress.zipcode = stoi(zipcodeText);
    }
}

Answer 1

我们如何将 Person 结构重命名为名称结构呢？ （它毕竟只包含第一个和最后。）
这会给我们这个：

struct Name {
    string first;
    string last;
}; 

那么员工现在看起来像什么？ 它看起来像这样：

+-------------+
|  Employee   |
|             |
| +---------+ |
| |  Name   | |
| +---------+ |
|             |
| +---------+ |
| | Address | |
| +---------+ |
|             |
+-------------+

名称是员工的一部分，但第一个和最后一个在哪里？它们是名称的一部分。
这是相同的图片，除了它更深层次地显示第一个和最后：

+---------------+
|   Employee    |
|               |
| +-----------+ |
| |   Name    | |
| |           | |
| | +-------+ | |
| | | first | | |
| | +-------+ | |
| |           | |
| | +-------+ | |
| | | last  | | |
| | +-------+ | |
| |           | |
| +-----------+ |
|               |
|  +---------+  |
|  | Address |  |
|  +---------+  |
|               |
+---------------+

你需要使用两个点运算符（'。'）来访问第一个和最后一个，因为它们的深度是两倍。

Employee e;
e.name.first = "Joe";

您的代码经过重新考虑以反映这些变化：

#include <iostream>
#include <string>
#include <fstream>

using namespace std;

struct Name {
    string first;
    string last;
};

struct Address {
    string street;
    string city;
    string state;
    string zipcode;
};

struct Employee {
    Name name;
    Address homeAddress;
    int eid;
};

void readEmployee(istream& in, Employee& e);
void displayEmployee(ostream& out, const Employee& e);

int main(int argc, const char* argv[]) {
    Employee e[50];

    ifstream fin;
    ofstream fout;

    fin.open("employeesIn.txt");

    if (!fin.is_open()) {
        cerr << "Error opening employeesIn.txt for reading." << endl;
        exit(1);
    }
    fout.open("employeesOut.txt");
    if (!fout.is_open()) {
        cerr << "Error opening employeesOut.txt for writing." << endl;
        exit(1);
    }
    int EmployeePopulation = 0;
    readEmployee(fin, e[EmployeePopulation]);
    while (!fin.eof()) {
        EmployeePopulation++;
        readEmployee(fin, e[EmployeePopulation]);
    }
    fin.close();
    for (int i = 0; i <= EmployeePopulation - 1; i++) {
        displayEmployee(fout, e[i]);
    }

    fout.close();

    cout << endl;

    return 0;
}

void readEmployee(istream& in, Employee& e) {
    string cidText;
    if (getline(in, cidText, ',')) {
        e.eid = stoi(cidText);

        getline(in, e.name.first, ',');
        getline(in, e.name.last, ',');

        getline(in, e.homeAddress.street, ',');
        getline(in, e.homeAddress.city, ',');
        getline(in, e.homeAddress.state, ',');

        string zipcodeText;
        getline(in, zipcodeText, ',');
        e.homeAddress.zipcode = stoi(zipcodeText);
    }
}
void displayEmployee(ostream& out, const Employee& e){

  //you can access the specific name values this way:
  cout <<  e.name.first << " " << e.name.last << endl;

  return;
}