Question

嘿伙计们，我有问题试图让我的节目按上午和下午排序。这是我的任务要求的一部分，或者我将以24小时格式完成。到目前为止，这是我的代码

#include <stdio.h>


struct date
{
    int month;
    int day;
    int year;
    int hour;
    int min;
    int meridian;
};


typedef enum {undergraduate, graduate, post_doc } StudentLevel;


struct student
{
    char name[20];
    long int banner_id;
    struct date date_of_enrollment;
    struct date time_of_enrollment;
    StudentLevel level;
}data[5];


int lessThan(struct student S1 , struct student S2) {
    struct date A = S1.date_of_enrollment;
    struct date B = S2.date_of_enrollment;
    struct date C = S1.time_of_enrollment;
    struct date D = S2.time_of_enrollment;
    if (A.year < B.year) return 1;
    else if (A.year == B.year && A.month < B.month) return 1;
    else if (A.year == B.year && A.month == B.month && A.day < B.day) return 1;
    else if (A.year == B.year && A.month == B.month && A.day == B.day && C.hour < D.hour) return 1;
    else if (A.year == B.year && A.month == B.month && A.day == B.day && C.hour == D.hour && C.min < D.min) return 1;
     else if (A.year == B.year && A.month == B.month && A.day == B.day && C.hour == D.hour && C.min == D.min & C.meridian < D.meridian) return 1;
    else return 0;
}


int sort(struct student *data)
{
    int i, j;
    struct student x;
    for (i = 0; i < 5; i++){
        for (j = 0; j < 5; j++){
            if (lessThan(data[i], data[j])){
                x = data[j];
                data[j] = data[i];
                data[i] = x;
            }
        }
    }
    return 0;
}


int main ()
{

    int i;
    printf ("Enter Name, Banner ID, the date of Enrollment (Month Day Year), the time of enrollemtn (Hour Min Meridian(am or pm) and Student Level(0=Undergraduate, 1=graduate, or 2=Post Doc): \n");

    for (i=0;i<2;i++)

        scanf ("%s %ld %d %d %d %d %d %d %d",data[i].name,&data[i].banner_id,
               &data[i].date_of_enrollment.month,&data[i].date_of_enrollment.day, &data[i].date_of_enrollment.year, &data[i].time_of_enrollment.hour, &data[i].time_of_enrollment.min, &data[i].time_of_enrollment.meridian, &data[i].level);


    printf ("You entered : \n");
    sort(data);
    for (i=0;i<2;i++)
        printf("%s %ld %d %d %d %d %d %d %d \n",data[i].name, data[i].banner_id,data[i].date_of_enrollment.month,data[i].date_of_enrollment.day,data[i].date_of_enrollment.year,data[i].time_of_enrollment.hour, data[i].time_of_enrollment.min, data[i].time_of_enrollment.meridian,data[i].level);


    return 0;
}

该程序似乎不想按经络排序。我也不知道如何告诉它如何做到这一点，这完全是我的错。任何提示将非常感谢。

Answer 1

虽然很简单。您在上次比较中缺少一个&：

C.min == D.min & C.meridian < D.meridian

应该是

C.min == D.min && C.meridian < D.meridian

如何按C中的子午线（AM和PM）排序

1 个答案: