在 C++ 中使用变量声明结构数组

Question

尝试通过键入请求的限制来更改结构数组。当我尝试通过使数组声明成为变量来做到这一点时，它只是说“分段错误（核心转储）”我该如何解决这个问题并允许数组工作？ 这是我正在尝试制作的代码示例：

#include <stdio.h>
struct Employee {
    char firstName[20];
    char lastName[20];
    unsigned int age;
    char gender;
    double hourlySalary;
};
int main(){
    int limit;
    printf("How much employees do you wish to input\n");
    scanf("%d", &limit);
    int number = limit;
    struct Employee Employ[number];
    
    for(unsigned int i = 0; i < limit; ++i){
        ("\n\t Enter employee first name\n");
        scanf("%s", &Employ[i].firstName);
        ("\n\t Enter employee last name\n");
        scanf("%s", &Employ[i].lastName);
        ("\n\t Enter employee age\n");
        scanf("%d", &Employ[i].age);
        ("\n\t Enter employee hourly salary\n");
        scanf("%lf", &Employ[i].hourlySalary);
    }
    for(unsigned int i = 0; i < limit; i++){
        printf("\n\tfirstname\tlastname\tage\thourlysalary\n%s\t%s\t%d\t%lf",Employ[i].firstName,Employ[i].lastName,Employ[i].age,Employ[i].gender,Employ[i].hourlySalary);
    }
}

预期结果和输入：

Answer 1

我使用在线 CPP 编译器来测试代码 - 即使使用古老的 cpp98 标准进行编译，我也无法使其运行。当使用其他人指出的 C 构造时。

但是我使用 c++11 标准并使用了流和向量 - 我做了一个工作得很好的例子。

#include <iostream>
#include <string>
#include <vector>

struct Employee 
{
    char firstName[20];
    char lastName[20];
    unsigned int age;
    char gender;
    double hourlySalary;
};

int main()
{
    int limit = 0;

    std::cout << "How many employees?\n";
    std::cin >> limit;

    //Employee Employ[size];
    std::vector<Employee> employees;

    for(int i = 0; i < limit; ++i)
    {
        employees.push_back(Employee());
        auto& employee = employees[i];
    
        std::cout << "first name?\n";
        std::cin >> employee.firstName;
    
        std::cout << "last name?\n";
        std::cin >> employee.lastName;
    
        int temp = 0;
        std::cout << "age?\n";
        std::cin >> temp; 
        employee.age = temp;
    
        std::cout << "gender?\n";
        std::cin >> employee.gender;
    
        std::cout << "salary?\n";
        std::cin >> employee.hourlySalary;
    }

    std::cout << "\n======================================================= \n";
    for(auto it = employees.begin(); it != employees.end(); ++it)
    {
        std::cout << (*it).firstName << "\t";
        std::cout << (*it).lastName << "\t";
        std::cout << (*it).age << "\t";
        std::cout << (*it).gender << "\t";
        std::cout << (*it).hourlySalary << "\n";
    }
}

这确实有效.. 但是有一些代码闻起来像是将原始字符串读取到大小受限的变量中。 （就像评论建议的大小限制变量是 20 个字符的 char 数组）

Answer 2

你应该扫描一个不带&的char数组，数组的名字已经是一个指针了。

scanf("%s", Employ[i].firstName)

Answer 3

设法自己最终让它工作，可能是 & 就像上面所说的帖子

#include<stdio.h>
#include<string.h>

struct employee
{
    char firstName[20];
    char lastName[20];
    unsigned int age;

    double hourlySalary;
};

int main()
{
    int i;
    int limit;
    printf("insert number of employees to enter\n");
    scanf("%d", &limit);
    struct employee arr_employ[limit];
    

    for(i = 0; i < limit;i++)
    {
        

        printf("Enter name: ");
        scanf("%s", arr_employ[i].firstName);

        printf("Enter last name: ");
        scanf("%s", arr_employ[i].lastName);

        printf("Enter age:");
        scanf("%u", &arr_employ[i].age);
        
        printf("Enter hourly Salary:");
        scanf("%lf",&arr_employ[i].hourlySalary);
    }

    printf("\n");

    for(i = 0; i < limit; i++)
    {
        printf("%s\t%s\t%u\t%.2f\n",arr_employ[i].firstName,arr_employ[i].lastName,arr_employ[i].age,arr_employ[i].hourlySalary);
    }

    // signal to operating system program ran fine
    return 0;
}