open() doesn't set O_CLOEXEC flag

Question

I try to set O_CLOEXEC flag using open() and have no sucess.

Consider the following microtest:

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

int main() {
  int fd = open("test.c", O_RDONLY | O_CLOEXEC);
  int ret = fcntl(fd, F_GETFL);
  if(ret & O_CLOEXEC) {
    printf("OK!\n");
  } else {
    printf("FAIL!\n");
  }
  printf("fd = %d\n", fd);
  printf("ret = %x, O_CLOEXEC = %x\n", ret, O_CLOEXEC);
  return 0;
} 

When running on linux with kernel version 2.6 the test succeeds and prints "OK!", but fails with 3.8 or 3.9 kernels.

What's wrong? Thanks!

Answer 1

It was decided that exposing O_CLOEXEC flag to fcntl(fd, F_GETFL) is security leak. Change was made by this commit in kernel 3.6-rc7:

commit c6f3d81115989e274c42a852222b80d2e14ced6f
Author: Al Viro <viro@zeniv.linux.org.uk>
Date:   Sun Aug 26 11:01:04 2012 -0400

don't leak O_CLOEXEC into ->f_flags

Signed-off-by: Al Viro <viro@zeniv.linux.org.uk>

In other words, you should not have relied on O_CLOEXEC being visible in first place.

Answer 2

The fcntl call parameter F_GETFD, the flag is FD_CLOEXEC and the O_CLOEXEC support appeared in 2.6.23. Read the manuals:

 File descriptor flags
   The following commands manipulate the flags associated with a file descriptor.
   Currently,  only one such flag is defined: FD_CLOEXEC, the close-on-exec flag.
   If the FD_CLOEXEC bit is 0, the file descriptor will  remain  open  across  an
   execve(2), otherwise it will be closed.

   F_GETFD (void)
          Read the file descriptor flags; arg is ignored.

   F_SETFD (int)
          Set the file descriptor flags to the value specified by arg.

Answer 3

You are doing it wrong. You should do by this way:

int ret = fcntl(fd, F_GETFD);
if (ret & FD_CLOEXEC) {
...
}