Question

If I do:

fd2 = open ("file", O_RDONLY);

and then

fd1 = open ("file", O_RDONLY);

in the SAME PROCESS.
Do I get two different file pointers? I mean, can I move the “cursor” 100 bytes with fd2 and
fd1’s cursor will remain zero?

In addition, even if I open both for READONLY .. Does the filesystem creates TWO entries in the File table? or only one ? (Not the Inode table)

thanks!

Answer 1

Note: the initial version had a copy and paste bug which affects the result. Fixed now.

On a try it and see basis, I wrote

#include <stdio.h>
#include <fcntl.h>
#include <unistd.h>

int main(int argc, char *argv[]){
  int fd1 = open("/etc/passwd",O_RDONLY);
  int fd2 = open("/etc/passwd",O_RDONLY);
  printf("%d %d\n",fd1,fd2);
  printf("FD1 position = %d\n", lseek(fd1,0,SEEK_CUR));
  printf("FD2 position = %d\n", lseek(fd2,0,SEEK_END));
  printf("FD1 position = %d\n", lseek(fd1,0,SEEK_CUR));
}

Which returns

$ ./a.out 
3 4
FD1 position = 0
FD2 position = 2888
FD1 position = 0

on my Mac OS 10.5 box and something functionally identical on a Scientific Linux box (differs only in the size of /etc/passwd).

You’ll notice that you get back numerically different fds, and they each their own position cursor.