c语言中取日期为dd/mm/yy

Question

    # include <stdio.h>
int main(){
    int d1, d2, m1, m2, year;
    printf("Enter date (dd/mm/yy): ");
    scanf("%d,%d/%d,&,d/%d", &d1,&d2,&m1,&m2,&year);

**大家好，我是 c 编程的新手，所以我试图将日期设为 dd/mm/yy 格式，但是当我这样写时，它给我的输出为 191/033/0 19/07/14 输入我该如何解决这个问题？ **

Answer 1

# include <stdio.h>
int main(){
    int d, m, year;
    printf("Enter date (dd/mm/yy): ");
    scanf("%d/%d/%d", &d,&m,&year);
    if (d%10==1 && d!=11) printf("%d st",d);
    else if (d%10==2 && d!=12) printf("%d nd",d);
    else if (d%10==3 && d!=13) printf("%d rd",d);
    else printf("%d th",d);
    return 0;
}

顺便说一下，dd/mm/yy 不是的意思 'two days, two months and two years'。这意味着 'two digits for day, month and year'。这就是为什么你不需要那么多变量。

Answer 2

您需要 strftime() 函数，并正确处理输入，以免发生意外情况

这是如何操作的示例

#include <stdio.h>
#include <time.h>
#include <string.h>

int main()
{
    char      buffer[100];
    struct tm date;

    memset(&date, 0, sizeof(date));

    printf("Enter date (dd/mm/yy): ");
    if (fgets(buffer, sizeof(buffer), stdin) == NULL)
        return -1;
    if (sscanf(buffer, "%d/%d/%d", &date.tm_mday, &date.tm_mon, &date.tm_year) == 3)
    {
        const char *format;

        format = "Dated %A %dth of %B, %Y";
        if (strftime(buffer, sizeof(buffer), format, &date) > sizeof(buffer))
            fprintf(stderr, "there was a problem converting the string\n");
        else
            fprintf(stdout, "%s\n", buffer);
    }
    return 0;
}

c语言中取日期为dd/mm/yy

taking date as dd/mm/yy in c language

c

date