Question

2>&1 redirect in Bourne shell takes the output sent to a file descriptor 2 (by default, standard error) and sends it instead to file descriptor 1 (by default a standard output).

But what does 2<&1 redirect do?

Does it send stderr to stdin?

My theory was that it was sending stdin to stderr (e.g. same as 1>&2) but experimentally, that is NOT the case:

$ perl -e 'print "OUT\n"; print STDERR "ERR\n"; \
  while (<>) { print "IN WAS $_\n";}'           \
  > out3 2<&1
df
$ cat out3
ERR
OUT
IN WAS df

Note that standard out AND standard error both went to file out3 where stdout was redirected.

Answer 1

The <& operator duplicates an “input” file descriptor. According to IEEE Std 1003.1-2001 (aka Single Unix Specification v3, the successor to POSIX), it’s supposed to be an error to say 2<&1 if 1 is not a file descriptor open for input. However, it appears that bash is lazy and doesn’t care if the file descriptor is open for input or for output.

So both 2<&1 and 2>&1 simply perform the system call dup2(1, 2), which copies file descriptor 1 to file descriptor 2.

You can check by running a command like this, since redirections are performed left-to-right:

sleep 99999 1>/dev/null 2<&1

Then in another window, run lsof on the sleep process. You’ll see that both file descriptors 1 and 2 point to /dev/null. Example (on my Mac):

:; ps axww | grep sleep
 8871 s001  R+     0:00.01 grep sleep
 8869 s003  S+     0:00.01 sleep 99999
:; lsof -p 8869 | tail -2
sleep   8869 mayoff    1w   CHR    3,2       0t0       316 /dev/null
sleep   8869 mayoff    2w   CHR    3,2       0t0       316 /dev/null