作业如下,要求您编写一个程序,对有关 COP 3223 的 TA 何时可用的信息进行排序。您的程序将读入所有 TA 办公时间轮类的数据,对数据进行排序,然后以预先指定的格式打印出信息。您可以假设最大类次数为 100。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct Shift
{
char name[100];
char day_of_week[100];
int start_hour;
int end_hour;
};
struct Shift shift_data[100];
struct Shift temp;
int read_data(struct Shift shift_data[], int *num_shifts);
void sort_data(struct Shift shift_data[], int *num_shifts);
void print_data(struct Shift shift[], int num_shifts);
int main(void)
{
int num_shifts;
read_data(shift_data, &num_shifts);
sort_data(shift_data, &num_shifts);
print_data(shift_data, num_shifts);
return 0;
}
/*** Preconditions: array of structure "Shift" to store data
Postconditions: number of shifts read in from data file
Actions: Ask user for name of input file. Read the number of shifts, then read in the data for all
of the shifts. Return the number of shifts. ***/
int read_data(struct Shift shift_data[], int *num_shifts)
{
char input_schedule[100];
int i;
char shift_name[100], shift_day[100];
printf("Enter the name of the input file.\n");
scanf("%s", &input_schedule[100]);
FILE *ifp;
ifp = fopen("input_schedule.txt", "r");
fscanf(ifp, "%d", &num_shifts);
for(i=0; i<(*num_shifts); i++)
{
fscanf(ifp, "%s %s %d %d\n", &shift_data[i].name, &shift_data[i].day_of_week, &shift_data[i].start_hour, &shift_data[i].end_hour);
}
fclose(ifp);
return num_shifts;
}
/*** Preconditions: array of structure "Shift" integer value indicating number of shifts
Postconditions: none - this function does not return anything.
Actions: Sort the shifts by the TA's first name. ***/
void sort_data(struct Shift shift_data[], int *num_shifts)
{
int i,j;
for(i=0; i<(*num_shifts); i++)
{
for(j=i+1; j<(*num_shifts); j++)
{
if(strcmp(shift_data[i].name,shift_data[i].name) < 0)
{
temp.name = shift_data[i].name;
shift_data[i].name = shift_data[j].name;
shift_data[j].name = temp.name;
}
if(strcmp(shift_data[i].day_of_week,shift_data[i].day_of_week) < 0)
{
temp.day_of_week = shift_data[i].day_of_week;
shift_data[i].day_of_week = shift_data[j].day_of_week;
shift_data[j].day_of_week = temp.day_of_week;
}
if(strcmp(shift_data[i].start_hour,shift_data[i].start_hour) < 0)
{
temp.start_hour = shift_data[i].start_hour;
shift_data[i].start_hour = shift_data[j].start_hour;
shift_data[j].start_hour = temp.start_hour;
}
if(strcmp(shift_data[i].end_hour,shift_data[i].end_hour) < 0)
{
temp.end_hour = shift_data[i].end_hour;
shift_data[i].end_hour = shift_data[j].end_hour;
shift_data[j].end_hour = temp.end_hour;
}
}
}
}
/*** Preconditions: array of structure "Shift" integer value indicating number of shifts
Postconditions: none - this function does not return anything.
Actions: Print the sorted data in the format described below. **/
void print_data(struct Shift shift[], int num_shifts)
{
int i;
printf("\n\n\n");
for(i=0; i<num_shifts; i++)
{
printf("%s\t\t", shift_data[i].name);
printf("%s\t", shift_data[i].day_of_week);
printf("%d to ", shift_data[i].start_hour);
printf("%d\n", shift_data[i].end_hour);
if(shift_data[i].start_hour > 12)
{
shift_data[i].start_hour = shift_data[i].start_hour - 12;
printf("%2d:00 pm to ", shift_data[i].start_hour);
}
else if(shift_data[i].start_hour < 12)
{
printf("%2d:00 am to ", shift_data[i].start_hour);
}
else if(shift_data[i].start_hour == 12)
{
printf("%2d:00 pm to ", shift_data[i].start_hour);
}
if(shift_data[i].end_hour > 12)
{
shift_data[i].end_hour = shift_data[i].end_hour - 12;
printf("%2d:00 pm\n", shift_data[i].end_hour);
}
else if(shift_data[i].end_hour < 12)
{
printf("%2d:00 am\n", shift_data[i].end_hour);
}
else if(shift_data[i].end_hour == 12)
{
printf("%2d:00 pm\n", shift_data[i].end_hour);
}
}
}
但是,我的临时变量在 sort_data 函数中遇到问题。我不断收到错误消息,错误:从类型“char *”分配给类型“char[100]”时不兼容的类型
最佳答案
知道轮类次数后,接下来的事情就很简单了。您只需循环 num_shifts
次并扫描相应的值即可。
像这样的事情:
for(i = 0; i < num_shifts; i++)
{
int ret = fscanf(ifp, "%99s %99s %d %d\n",
shift_data[i].name,
shift_data[i].day_of_week,
&shift_data[i].start_hour,
&shift_data[i].end_hour);
}
关于c - 使用结构体和数组编写函数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34012949/