所以我现在正在尝试学习C,完成任务并且我不断遇到相同的错误
上下文:
我期望编写一个包含4个函数的程序 -
read_employee (读取名称,ID和工资并将其存储在 然后返回员工变量),
print_employee (读取员工结构,打印员工详情),
employee_total_salary (在一个参数中接受一组员工值 和另一个参数中的数组大小,并返回所有工资的总和。)
employee_index_search (从员工结构中搜索ID)
这就是我告诉过的伪代码:
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[];
有人能告诉我哪里出错了吗?该程序也不会调用print_employee并打印数组 (我评论了一些,因为我想先尝试至少1-2个工作函数)
答案 0 :(得分:0)
scanf("%d", &read.emp_id);
你应该改为
scanf("%d", &read[i].emp_id);
当您将read
作为结构数组
您还缺少()
和main
的参数括号employee_total_salary