c - 具有数组和指针的结构

Question

我必须编写一个程序来执行此操作，该程序将对有关 COP 3223 的 TA 何时可用的信息进行排序。该程序预计将读取所有 TA 办公时间轮类的数据，然后对数据进行排序，然后以预先指定的格式打印出信息。假设最大类次为100。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
/* Given structure format that is needed to be used in the program. */
struct Shift
{
char name[100];
char day_of_week[100];
int start_hour;
int end_hour;
};

/* Functions that will be used in the program. */
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)
{
struct Shift shift_data[100];
struct Shift temp;

/*Declare and initialize variables, and statements to call each function. */
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)
{
/*Declare and initialize variables. */
char input_schedule[100];
int i;
char shift_name[100], shift_day[100];
/*Prompt user to enter the input file. */
printf("Enter the name of the file name.\n");
scanf("%s", &input_schedule[100]);
/*Declare file and read in the needed values. */
FILE *ifp;
ifp = fopen("input_schedule.txt", "r");
fscanf(ifp, "%d", &num_shifts);
/*For loop to read in each value from the file and store them in the shift_data array. */
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);
}
/*Close file.*/
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)
{
/*Declare and initialize variables. */
int i,j;
/*For loop to run through each array.*/
for(i=0; i<(*num_shifts); i++)
{
/*For loop to use the next element in the array to compare to the element before it.*/
    for(j=i+1; j<(*num_shifts); j++)
    {
/*If statements using strcmp to compare the two strings and sort the data in order.*/
        if(strcmp(shift_data[i].name,shift_data[i].name) < 0)
        {
            memcpy(temp.name, shift_data[i].name, strlen(shift_data[i].name) + 1);
            memcpy(shift_data[i].name, shift_data[j].name, strlen(shift_data[j].name) + 1);
            memcpy(shift_data[j].name, temp.name, strlen(temp.name) + 1);
        }
        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_data[], int num_shifts)
{
/*Declare and initialize variables.*/
int i;

printf("\n\n\n");
/*For loop to print out the sorted data to the user.*/
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 statements to change the shift times from military time to standard time.*/
    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);
    }
}
}

我遇到的问题是 sort_data 函数。当从类型“char *”分配到类型“char[100]”时，我不断收到一个错误，提示类型不兼容。它与我的临时变量有关，但我似乎无法弄清楚如何修复它。

Answer 1

当你在 C 中对 char 数组进行赋值时，它不会复制数组的内容。

而不是使用:

temp.name = shift_data[i].name;

做:

strncpy(temp.name, shift_data[i].name, 99);
temp.name[99] = '\0';

对不起，我建议您使用 memcpy()，但您必须检查字符串的长度才能知道要复制多少字节，这不是正确使用的函数。

使用 strncpy 更好，因为复制在到达空字节或可以存储的最大字符数。请注意，您的字符串的长度是 100。 99 将使 strncpy 从字符串中的索引 0 复制到 98。 temp.name[99] = '\0'; 行将用 \0 填充字符串的最后位置。

在此页面上:http://linux.die.net/man/3/strncpy , 有关于 strncpy 的更好描述和此函数的简单代码。