Question

我一直在使用struct一段时间，但我总是有问题。看这个例子： - `

#include <iostream>
struct Employee
{
    short id;
    int age;
    double wage;
};

void printInformation(Employee employee)
{
    std::cout << "ID:   " << employee.id << "\n";
    std::cout << "Age:  " << employee.age << "\n";
    std::cout << "Wage: " << employee.wage << "\n";
}

int main()
{
    Employee joe = { 14, 32, 24.15 };
    Employee frank = { 15, 28, 18.27 };
    // Print Joe's information
    printInformation(joe);

    std::cout << "\n";

    // Print Frank's information
    printInformation(frank);

    return 0;
}`

此代码完全正常，但何时我将使用字符串而不是“Joe”和“Frank”。我试过但失败了。这是我工作的代码。

 '#include <bits/stdc++.h>
 using namespace std;
 struct People{
   string name;
   int id;
   int age;
   int wage;
   };                
int main(){

string iname;
int iid = 0;
int iage; 
int iwage;     
while(1){
 iid++;
 cout << "ID No." << iid << endl <<"Enter Name";
 std::getline(std::cin,iname);
 cout << "Enter your Age:-";
 cin >> iage;       
  cout << "Enter your Wage :-";
  cin >> iwage; 
  cout << "See your details."<<endl <<"Name"<<iname<< endl<< "ID."<< iid << 
   endl << "Age" << iage << endl<< "Wage" << iwage << endl;
  People a = static_cast<People>(iid);
    People a={iname,iid,iage,iwage};  
    std::cout << "Name:" << a.name << "\n";
 std::cout << "ID:   " << a.id << "\n";
 std::cout << "Age:  " << a.age << "\n";
std::cout << "Wage: " <<  a.wage << "\n";
    }
    return 0;
}

此处用户输入其数据。我想以struct的形式保存太多数据，所以我用'a'。根据我，它必须计算1.age =（bla ..），3.age =（bla ..）请帮帮我。

Answer 1

我认为你正在寻找类似std :: map

的东西

http://en.cppreference.com/w/cpp/container/map

示例：

std::map<std::string, Employee> employees;
Employee joe = { 14, 32, 24.15 };
employees["joe"] = joe;

printInformation(employees["joe"]);

Answer 2

您还可以为每个员工<div id="navbarContentHamburger"> <svg width="25" height="25"> <path d="M0,5 50,5" stroke="#fff" stroke-width="3"/> <path d="M0,10 50,10" stroke="#fff" stroke-width="3"/> <path d="M0,15 50,15" stroke="#fff" stroke-width="3"/> </svg> </div>提供另一个变量，然后将员工存储在一个向量中，然后在函数循环中通过员工，直到找到具有该名称的员工：

string name;

我认为它比使用地图更好，因为你只需要写#include <iostream> #include <string> #include <vector> using namespace std; struct Employee{ string name; short id; int age; double wage; }; vector<Employee> Employees; void printInformation(string employee){ for(int i = 0; i < Employees.size(); i++){ if(Employees[i].name == employee){ cout << "Name: " << Employees[i].name << "\n"; cout << "ID: " << Employees[i].id << "\n"; cout << "Age: " << Employees[i].age << "\n"; cout << "Wage: " << Employees[i].wage << "\n"; } } } int main(){ Employee joe = {"joe", 14, 32, 24.15 }; Employee frank = {"frank", 15, 28, 18.27 }; Employees.push_back(joe); Employees.push_back(frank); printInformation("joe"); printInformation("frank"); return 0; }，当你使用地图时你写printInformation("joe")。你决定了。

如何将struct name更改为整数？

2 个答案: