我无法获得预期的结果我希望我的程序显示任何人都可以帮助我。该程序用于打印结构中列出的 worker 姓名,如果您不输入任何这些姓名,我应该打印不存在的 worker 姓名。有人可以告诉我要使用的代码/语法吗
这是我的
#include <stdio.h>
#include <conio.h>
#include <string.h>
//Program Purpose: To accept a specific set of worker names and worker id number and accept the time they came to work and determine if they were early or late for the day.`
struct workers {
char worker_name[10];
int worker_id;
} workers;
int main ()
{
struct workers worker1;
struct workers worker2;
strcpy (worker1.worker_name, "sean");
worker1.worker_Id = 1234;
strcpy (worker2.worker_name,"tajae");
worker2.worker_Id = 7890;
char worker_name [30];
int Worker_Id;
float Time_Arrived;
float Minutes_Late;
float Extra_Minutes;
float Minutes_Early;
float lunch_time;
float Departure;
printf("******************Produce Pro Time Management System********************\n\n");
printf("Good morning. Welcome to Produce Pro, Hope you had a good nights rest and ready to have a successful day at work today.\n\n");
printf("Please follow the instruction and answer with the required details\n");
printf("Note brief: All time are in army hours\n\n");
printf("Enter your Worker Name\n");
scanf("%S",&worker_name[30]);
if (worker_name= worker1,worker2) // this is the error in the program//
{
printf(&worker_name[30]);
}
else
{
printf ("Worker Name doesn't exist");
}
}
当我更改if语句并放置
if (worker_name == worker1.worker_name || worker_name == worker2.worker_name)
{
printf("Welcome %s\n",worker_name);
}
else printf ("Worker Name doesn't exist\n");
我得到的是 worker 不来
最佳答案
如果您只检查名称,是否不需要在 If 条件中指定 Struct 成员?请看下面。
if (Worker_name == worker1.Worker_name || Worker_name == worker2.Worker_name)
{
printf("Welcome %s\n",Worker_name);
}
或
if (strcmp(Worker_name,worker1.Worker_name) != 0 || strcmp(Worker_name,worker2.Worker_name) != 0)
{
printf("Welcome %s\n",Worker_name);
}
关于c++ - If语句打印 worker 姓名,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29398550/