how to get the time of last access frm stat

skie_knite007 · 08-08-2005, 12:01 PM

how can we obtain the time of last access frm stat.

if i declared time as integer..I'm getting an unrecognized format..

like if

int time
stat struct buff

stat("file",& buff)

time=buff.st_atime;

The stat is showing time_t datatype for time......

suppose I'm declaring time as

time_t time;
time=buff.st_atime

If I get this frm the structure,,,how can I print it like we use %d for int.

instead of %d wat shld be used.....?

itsme86 · 08-08-2005, 12:23 PM

st_atime is a time_t which is usually defined as an unsigned long. This seems to work:

Code:

itsme@itsme:~/C$ cat atime.c
#include <stdio.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <unistd.h>

int main(int argc, char **argv)
{
  struct stat stbuf;

  stat(argv[0], &stbuf);

  printf("%lu\n", stbuf.st_atime);
  return 0;
}

itsme@itsme:~/C$ gcc -Wall atime.c -o atime
itsme@itsme:~/C$ ./atime
1123521643
itsme@itsme:~/C$
[/code]
You could do this too to make sure:

Code:

  printf("%lu\n", (unsigned long)stbuf.st_atime);

Hko · 08-08-2005, 12:30 PM

Itsme86's code outputs the time as the number of seconds since 00:00:00 on January 1, 1970 ("the epoch").

If you're looking for something like "Mon Jul 25 18:25:25 2005", use ctime() to convert the large number to a string:

Code:

#include <stdio.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <unistd.h>
#include <time.h>

int main(int argc, char *argv[])
{
    struct stat stbuf;
    char *timestring;

    if (argc != 2) {
        fprintf(stderr, "Usage: %s <path>\n", argv[0]);
        return 1;
    }
    if (stat(argv[1], &stbuf) < 0) {
        perror("Error while reading file information");
        return 1;
    }
    timestring = ctime(&stbuf.st_atime);
    printf("Last access time of %s is: %s\n", argv[1], timestring);
    return 0;
}