员工结构

Question

所以我目前正在尝试学习 C，完成一项任务，但我一直遇到同样的错误

对于上下文：

我希望写一个有 4 个函数的程序 -

read_employee（读取姓名、ID 和薪水，并将它们存储在 然后返回的员工变量），

print_employee（读取员工结构，打印员工详细信息），

employee_total_salary（在一个参数中接受一组员工值 和数组的大小在另一个参数中，并返回他们所有薪水的总和。）

employee_index_search（从员工结构中搜索 ID）

这是我告诉我要通过的伪代码：

1: Print the message "-- Enter Array 1 Employee Data –"
2: Loop i for each index of test_array1
3: Assign test_array1 at index i, the result of calling read_employee
4: Print the message '-- Enter Array 2 Employee Data --'
5: Loop i for each index of test_array2
6: Assign test_array2 at index i, the result of calling read_employee
7: Print the message "-- Test Array 1 --"
8: Loop i for each index of test_array1
9: Call print_employee, passing in test_array1 at index i
10: Print 'Total : ', employee_total_salary ( test_array1, 5 )
11: Print the message '-- Test Array 2 --'
12: Loop i for each index of test_array2
13: Call print_employee, passing in test_array2 at index i
14: Print "Total : ", employee_total_salary ( test_array2, 10)
15: Store in i, the employee_index_search passing in test_array1 , and the id 123 and
size 5
16: if something was found
17: Call print_employee passing in the located computer from testArray1
18: else
19: Print the message " array 1 does not contain a employee with id 123"

这是我写的代码：

     #include <stdio.h>

struct employee{
    char name[20];
    int emp_id;
    float salary;
}; struct employee e[];

void read_employee ()
{
    int i = 0;
        printf("%dEnter Employee Name: ", i);
        scanf("%s", &e[i].name);
        printf("Enter in ID: ");
        scanf("%d", &e[i].emp_id);
        printf("Enter in Salary: ");
        scanf("%f", &e[i].salary);
       i++;
}

void print_employee(){
    for (int i = 0; i > 10; ++i)
    {
        if (e[i].salary < 4000)
        {
            printf("%s(%d):%f - Level A", e[i].name, e[i].emp_id, e[i].salary);
        }
        else if (e[i].salary > 5000)
        {
            printf("%s(%d):%f - Level B", e[i].name, e[i].emp_id, e[i].salary);
        }
        else
        {
            printf("%s(%d):%f", e[i].name, e[i].emp_id, e[i].salary);
        }
    }
}

//int employee_total_salary(){
    //int total;
    //sizeof(e.salary);

    //total = sizeof(e.salary) + e.salary;
    //return total;
//}

int employee_index_search(){
    int *id;
    int size = 5;

    for (int i = 0; i < size; i++)
    {
        if (e[i].emp_id == id)
        {
            printf("%s(%d):%f", e[i].name, e[i].emp_id, e[i].salary);
        }
        else
        {
            printf("Array 1 does not contain an employee with ID 123");
            return -1;
        }
    }
return 0;
}

int main(){
    int test_array1[5];
    int test_array2[10];
    int i;

    printf("\n-- Enter Array 1 Employee Data --\n");
    for (int i = 0; i < 5; i++)
    {
        read_employee(&test_array1[i]);
    }
    printf("\n-- Enter Array 2 Employee Data --\n");
    for (int i = 0; i < 10; i++)
    {
        read_employee(&test_array2[i]);
    }
    printf("-- Test Array 1 --\n");
    for (int i = 0; i < 5; ++i)
    {
        print_employee(&test_array1[i]);
    }
    //printf("Total:");
    //gets(employee_total_salary(test_array1, 5));
    printf("-- Test Array 2 --\n");
    for (int i = 0; i < 10; ++i)
    {
        print_employee(&test_array2[i]);
    }
    return 0;
    //printf("Total:");
    //gets(employee_total_salary(test_array2, 10));
    //employeee_index_search();
}

我收到以下错误：

employee.c:13:21: warning: format specifies type 'char *' but the argument has
      type 'char (*)[20]' [-Wformat]
        scanf("%s", &e[i].name);
               ~~   ^~~~~~~~~~
employee.c:53:19: warning: comparison between pointer and integer
      ('int' and 'int *')
                if (e[i].emp_id == id)
                    ~~~~~~~~~~~ ^  ~~
employee.c:74:32: warning: too many arguments in call to 'read_employee'
                read_employee(&test_array1[i]);
                ~~~~~~~~~~~~~                ^
employee.c:79:32: warning: too many arguments in call to 'read_employee'
                read_employee(&test_array2[i]);
                ~~~~~~~~~~~~~                ^
employee.c:84:33: warning: too many arguments in call to 'print_employee'
                print_employee(&test_array1[i]);
                ~~~~~~~~~~~~~~                ^
employee.c:91:33: warning: too many arguments in call to 'print_employee'
                print_employee(&test_array2[i]);
                ~~~~~~~~~~~~~~                ^
employee.c:7:20: warning: tentative array definition assumed to have one element
}; struct employee e[];

谁能告诉我哪里出错了？该程序也不会调用 print_employee 并打印数组 （我评论了其中一些，因为我想先尝试至少 1-2 个工作功能）

Answer 1

scanf("%d", &read.emp_id);

你应该改成

scanf("%d", &read[i].emp_id);

当您将 read 作为结构数组传递时

对于main 和employee_total_salary，您还缺少参数括号()