Question

我需要一个可以计算两个日期时间（年，月，日，小时，分钟，秒）之间差异的函数。然后以相同的格式返回差异。

int main (){

    struct datetime dt_from;
    init_datetime(&dt_from, 1995, 9, 15, 10, 40, 15);

    struct datetime dt_to;
    init_datetime(&dt_to, 2004, 6, 15, 10, 40, 20);

    struct datetime dt_res;

    datetime_diff(&dt_from, &dt_to, &dt_res);

    return 0;

}

void datetime_diff(struct datetime *dt_from, struct datetime *dt_to
, struct datetime *dt_res) {

    //What can I do here to calculate the difference, and get it in the dt_res?

}

Answer 1

这是基本的想法：

将您的datetime转换为整数类型（最好是long long或unsigned long long），代表您的datetime值作为它的最小单位（在您的情况下为第二位））。怎么实现呢？轻松将单个值转换为秒，并将所有内容添加到一起。（seconds + minutes * 60 + hours * 3600 ...）

对两个值执行此操作，然后减去整数值。

现在将单个整数值（时差）转换回datetime。怎么样？从最大单位（年）开始，将差异除以一年内的秒数（60 * 60 * 24 * 365）。现在您知道两个datetime之间有多少年了。取其余部分并将其除以每月的秒数，依此类推......

（显然我忽略了一切相当复杂的事情，例如夏令时）

但我强烈建议您使用评论中提到的struct tm time.h。它是便携式的，您可以使用difftime。

Answer 2

请查看并尝试使用time.h的示例，并且应该是可移植的。它会计算您问题中日期之间的天数差异。您可以稍微更改程序，使其按您希望的方式工作。

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

int main() {
    time_t start_daylight, start_standard, end_daylight, end_standard;;
    struct tm start_date = {0};
    struct tm end_date = {0};
    double diff;

    printf("Start date: ");
    scanf("%d %d %d", &start_date.tm_mday, &start_date.tm_mon, &start_date.tm_year);
    printf("End date: ");
    scanf("%d %d %d", &end_date.tm_mday, &end_date.tm_mon, &end_date.tm_year);

    /* first with standard time */
    start_date.tm_isdst = 0;
    end_date.tm_isdst = 0;
    start_standard = mktime(&start_date);
    end_standard = mktime(&end_date);
    diff = difftime(end_standard, start_standard);

    printf("%.0f days difference\n", round(diff / (60.0 * 60 * 24)));

    /* now with daylight time */
    start_date.tm_isdst = 1;
    end_date.tm_isdst = 1;
    start_daylight = mktime(&start_date);
    end_daylight = mktime(&end_date);
    diff = difftime(end_daylight, start_daylight);

    printf("%.0f days difference\n", round(diff / (60.0 * 60 * 24)));

    return 0;
}

测试

Start date: 15 9 1995
End date: 15 6 2004
3195 days difference

甚至更简单的非交互式代码以及标准或夏令时：

#include <stdio.h>
#include <time.h>
#include <math.h>
int main()
{
    time_t start_daylight, start_standard, end_daylight, end_standard;;
    struct tm start_date = {0};
    struct tm end_date = {0};
    double diff;

    start_date.tm_year = 1995;
    start_date.tm_mon = 9;
    start_date.tm_mday = 15;
    start_date.tm_hour = 10;
    start_date.tm_min = 40;
    start_date.tm_sec = 15;

    end_date.tm_mday = 15;
    end_date.tm_mon = 6;
    end_date.tm_year = 2004;
    end_date.tm_hour = 10;
    end_date.tm_min = 40;
    end_date.tm_sec = 20;

    /* first with standard time */
    start_date.tm_isdst = 0;
    end_date.tm_isdst = 0;
    start_standard = mktime(&start_date);
    end_standard = mktime(&end_date);
    diff = difftime(end_standard, start_standard);

    printf("%.0f days difference\n", round(diff / (60.0 * 60 * 24)));

    /* now with daylight time */
    start_date.tm_isdst = 1;
    end_date.tm_isdst = 1;
    start_daylight = mktime(&start_date);
    end_daylight = mktime(&end_date);
    diff = difftime(end_daylight, start_daylight);

    printf("%.0f days difference\n", round(diff / (60.0 * 60 * 24)));

    return 0;
}

用C计算日期时间差

2 个答案: