C语言 日期比较以查找 C 中哪个更大
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Date comparison to find which is bigger in C
提问by Vivekanandan
I want to know how to find which is bigger date using a c program
我想知道如何使用 ac 程序找到哪个日期更大
kindly help me out plz....
请帮我解决lz....
Thanks
谢谢
回答by Jo?o Silva
You can use the difftimefunction:
您可以使用该difftime功能：
#include <time.h>
#include <stdio.h>
int main(void) {
time_t date1, date2;
// initialize date1 and date2...
double seconds = difftime(date1, date2);
if (seconds > 0) {
printf("Date1 > Date2\n");
}
return 0;
}
If your dates are not of type time_t, you can use the function mktimeto convert them.
如果您的日期不是 type time_t，您可以使用该函数mktime来转换它们。
回答by SharpUrBrain
#include <stdio.h>
struct date
{
int month;
int date;
int year;
};
int main(void)
{
int i=compare_dates (struct date d1, struct date d2);
switch(i)
{
case -1:
printf("%d/%d/%d is earlear date than %d/%d %d", D1.day, D1.month, D1.year, D2.day
case 1:
printf("%d/%d/%d is later date than %d/%d/%d",D1.day,D1.month,D1.year,D2.day…
case 0:
printf("%d/%d/%d is the same date than %d/%d/%d", D1.day, D1.month, D1.year, D2.day
}
return 0;
}
int compare_dates (struct date d1, struct date d2)
{
if (d1.year < d2.year)
return -1;
else if (d1.year > d2.year)
return 1;
if (d1.year == d2.year)
{
if (d1.month<d2.month)
return -1;
else if (d1.month>d2.month)
return 1;
else if (d1.day<d2.day)
return -1;
else if(d1.day>d2.day)
return 1;
else
return 0;
}
}
回答by Sylvain Defresne
Can you give more information about what you want to achieve ? Because comparing date is really easy. After all, they are just number of seconds (or milli, micro, nano, ...) since a given past date, or a structure containing year, month, day, ... Whatever the format, the comparison should be pretty easy to perform.
你能提供更多关于你想要实现的目标的信息吗？因为比较日期真的很容易。毕竟，它们只是自给定的过去日期以来的秒数（或毫、微、纳……），或者包含年、月、日等的结构……无论格式如何，比较都应该很容易去表演。
Maybe you want to compare two date given by the user as strings (something like "2011-03-12 18:38") ? Then, you can use strptimeto convert the string to a struct tm, and then do the comparison.
也许您想比较用户作为字符串给出的两个日期（类似于“2011-03-12 18:38”）？然后，您可以使用strptime将字符串转换为 a struct tm，然后进行比较。
#include <time.h>
#include <stdio.h>
#include <stdlib.h>
int parse_date(char* date, struct tm* tm)
{
char* format;
char* formats[] = {
"%F %I", /* "2011-03-12 06:38:05 AM" */
"%F %T", /* "2011-03-12 18:38:05" */
"%F %R", /* "2011-03-12 18:38" */
NULL,
};
for (format = formats[0]; format; ++ format) {
if (strptime(date, format, &tm)) {
return 1;
}
}
return 0;
}
int main(int argc, char** argv)
{
float diff;
char* date1;
char* date2;
struct tm tm1;
struct tm tm2;
time_t time1;
time_t time2;
if (argc != 3) {
fprintf(stderr, "usage: compare-date date1 date2\n");
exit(1);
}
date1 = argv[1];
date2 = argv[2];
if (!parse_date(date1, &tm1)) {
fprintf(stderr, "unsupported date: %s\n", date1);
exit(1);
}
if (!parse_date(date2, &tm1)) {
fprintf(stderr, "unsupported date: %s\n", date2);
exit(1);
}
time1 = mktime(&tm1);
time2 = mktime(&tm2);
diff = difftime(time1, time2);
printf("%s %c %s\n",
date1,
(diff < 0 ? '<' : (diff > 0 ? '>' : '==')),
date2);
return 0;
}
回答by Bernd Elkemann
You didn't say in which format you have the date, so I will name two common examples:
你没有说日期的格式，所以我举两个常见的例子：
If you are using GNU lib-c (or MinGW) and query the time with:
time_t time (time_t *result)
如果您使用 GNU lib-c（或 MinGW）并使用以下命令查询时间：
time_t time (time_t *result)
Then time_tis just a long int, numbers of seconds since epochand you can subtract one date from the other to find out the number of seconds difference.
然后time_t只是long int, 秒数epoch，您可以从另一个日期中减去一个日期以找出秒数差异。
- If you are using the Windows API and have a filetime-structure:
- 如果您使用的是 Windows API 并具有文件时间结构：
typedef struct _FILETIME {
DWORD dwLowDateTime;
DWORD dwHighDateTime;
} FILETIME, *PFILETIME;
you can cast the pointer to a pointer to ULARGE_INTEGER, and subtract that one giving you the number of 100-nanosecond intervals difference.
您可以将指针转换为指向 的指针ULARGE_INTEGER，然后减去该指针，得到 100 纳秒间隔差异的数量。
