C - time_t string representation

Question

I am new to C programming and I would like some help with a simple task.

So I have this function:

char *time2str(time_t time) {
     static_char *str_fmt = "%02d/%02d/%4d %02d:%02d";
     char *time_s = "";

     return time_s;
}

What I would like to do is get in DD/MM/YYYY HH:MM format the time and store it in time_s variable so that I can return it. My problem is that I cannot figure out how to format the string and store it WITHOUT printing it. All I have found so far is the sprint method that actually prints the formatted string when storing it too, which is not what I want.

In a few words, I would just like the time from time_t time in the format I mentioned stored in time_s.

Sorry if I didn't explain everything properly but I'm new in C programming.

score 1 · Answer 1 · answered Nov 25 '16 at 14:32

The function ctime takes a time_t and outputs a string in a fixed format. If that is not sufficient for your task you should use strftime. This has a whole bunch of format specifiers that should suit your need.

It works on struct tm, so you'd first have to convert your time_t to that using localtime or gmtime.

Jabberwocky · Accepted Answer · 2016-11-25T14:28:24.817

You need sprintf which is similar to printf but "prints" the result into a buffer rather than on the screen.

Returning the pointer to a static buffer:

char *time2str(time_t time) {
     static_char *str_fmt = "%02d/%02d/%4d %02d:%02d";

     static char time_s[20];                             // ugly, 20 is hopefully enough
  // ^ static is important here, because we return a pointer to time_s
  //   and without static, the time_s buffer will no longer exist once the
  //   time2str function is finished

     sprintf(time_s, str_fmt, time.....);
     return time_s;
}

or (better), we provide a buffer (long enough) where the converted string is to be placed:

void time2str(time_t time, char *time_s) {
     static_char *str_fmt = "%02d/%02d/%4d %02d:%02d";
     sprintf(time_s, str_fmt, time.....);
     return time_s;
}
...
char mytime[20];                                         // ugly, 20 is hopefully enough
time2str(time, mytime);
printf("mytime: %s\n, mytime);

or the time2str function returns a newly allocated buffer that will contain the converted string. That buffer must be freed later with free.

char *time2str(time_t time) {
     static_char *str_fmt = "%02d/%02d/%4d %02d:%02d";
     char *time_s = malloc(20);                          // ugly, 20 is hopefully enough
     sprintf(time_s, str_fmt, time.....);
     return time_s;
}
...
char *mytime = time2str(time);
printf("mytime: %s\n, mytime);
free(mytime);

Completing the arguments of sprintf is left as an exercise to the reader.

Disclaimer: non tested, non error checking code meant only for demonstration purposes.

Generally speaking, always use at least `snprintf`. It is *so* easy to create buffer overflow with plain `sprintf`! — hyde, Nov 25 '16 at 14:30

score 0 · Answer 3 · answered Nov 25 '16 at 14:29

0

You can use strftime(), after converting your time_t value to struct tm type.

answered Nov 25 '16 at 14:29

pmg

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score 0 · Answer 4 · answered Nov 25 '16 at 14:40

in general,

call time() to get the current time into a time_t variable
call localtime() to get a pointer to a struct tm structure with the time information in separate fields
call strftime() to get the desired time/date format.

Suggest reading the man pages for localtime() and for strftime() to get all the details

here is a sample code, where you would need to modify the strftime() format string for what you want to have in the resulting char array.

time_t     currentTimeSec = time(NULL);
struct tm *tmStructptr;
char       timeStamp[128];

tmStructptr = localtime( &currentTimeSec );
strftime( timeStamp, sizeof(timeStamp), "YOUR FORMAT STRING" tmStructptr );

Note: to only extract the day of the week, the format string would be: "%w"

C - time_t string representation

4 Answers4