c++ - 如何以正确的方式将结构数组参数传递给函数？

Question

源代码:

#include <iostream>

using namespace std;

struct Employee {

   string name;
   int age;
   float salary;

};

void displayData(Employee);

int main() {

   Employee employee[3];
   int sizeOfEmployee = sizeof(employee) / sizeof(employee[0]);

   for(int i = 0; i < sizeOfEmployee; i++) {
       cout << "Enter name for employee " << i + 1 << ": ";
       cin >> employee->name;
       cout << "Enter age for employee " << i + 1 << ": ";
       cin >> employee->age;
       cout << "Enter salary for employee " << i + 1 << ": ";
       cin >> employee->salary;       
}

   for(int i = 0; i < sizeOfEmployee; i++) {
       displayData(employee[i]);
}

}

void displayData(Employee employee) {

   cout << "DISPLAYING INFORMATION" << endl;
   cout << "Name: " << employee.name << endl;
   cout << "Age: " << employee.age << endl;
   cout << "Salary: " << employee.salary << endl;

}

我的代码不显示我存储到数组中的所有数据，它只读取并显示我最后输入的数据。

代码输出:

我可以用什么方式获取我存储到数组中的所有数据并显示它？

Answer 1

您正在将数据读取到指向数组第一个元素的指针中。要读入第 n 个元素，您的代码应该使用索引，例如 cin >> employee[i].name;。

然而...

“裸”数组是许多错误、细微错误和通常不安全代码的来源。考虑改用 std::array - 这将消除对像 int sizeOfEmployee = sizeof(... 这样的代码的需要，并将很好地将该数组封装为函数的参数:

#include <array>

int main() {

   std::array<Employee, 3> employee;

   for(int i = 0; i < employee.size(); i++) {
       cout << "Enter name for employee " << i + 1 << ": ";
       cin >> employee[i].name;
   // the rest of that for loop...
   }
// and the second loop becomes:
   for(const auto& e: employee) {
       displayData(e);
}