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Redirect stderr and stdout in Bash

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Original source (stackoverflow.com)
Tags: bash shell-scripting streams file-descriptors stderr stdout
Clipped on: 2017-12-19

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29 upvote
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This redirects stderr to the original stdout, not to the file where stdout is going. Put '2>&1' after '>file.log' and it works. – user25148 Mar 12 '09 at 9:25
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What is the advantage of this approach over some_command &> file.log? – ubermonkey May 27 '09 at 14:04
4 upvote
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If you want to append to a file then you must do it this way: echo "foo" 2>&1 1>> bar.txt AFAIK there's no way to append using &> – SlappyTheFish Jun 8 '10 at 10:58
6 upvote
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Argh, sorry, echo "foo" 1>> bar.txt 2>&1 – SlappyTheFish Jun 8 '10 at 11:17
1 upvote
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I think the interpretation that 2>&1 redirects stderr to stdout is wrong; I believe it is more accurate to say it sends stderr to the same place that stdout is going at this moment in time. Thus place 2>&1 after the first redirect is essential. – jdg Aug 7 '15 at 17:47
# Close STDOUT file descriptor
exec 1<&-
# Close STDERR FD
exec 2<&-

# Open STDOUT as $LOG_FILE file for read and write.
exec 1<>$LOG_FILE

# Redirect STDERR to STDOUT
exec 2>&1

echo "This line will appear in $LOG_FILE, not 'on screen'"

Now, simple echo will write to $LOG_FILE. Useful for daemonizing.

To the author of the original post,

It depends what you need to achieve. If you just need to redirect in/out of a command you call from your script, the answers are already given. Mine is about redirecting within current script which affects all commands/built-ins(includes forks) after the mentioned code snippet.


Another cool solution is about redirecting to both std-err/out AND to logger or log file at once which involves splitting "a stream" into two. This functionality is provided by 'tee' command which can write/append to several file descriptors(files, sockets, pipes, etc) at once: tee FILE1 FILE2 ... >(cmd1) >(cmd2) ...

exec 3>&1 4>&2 1> >(tee >(logger -i -t 'my_script_tag') >&3) 2> >(tee >(logger -i -t 'my_script_tag') >&4)
trap 'cleanup' INT QUIT TERM EXIT


get_pids_of_ppid() {
    local ppid="$1"

    RETVAL=''
    local pids=`ps x -o pid,ppid | awk "\\$2 == \\"$ppid\\" { print \\$1 }"`
    RETVAL="$pids"
}


# Needed to kill processes running in background
cleanup() {
    local current_pid element
    local pids=( "$$" )

    running_pids=("${pids[@]}")

    while :; do
        current_pid="${running_pids[0]}"
        [ -z "$current_pid" ] && break

        running_pids=("${running_pids[@]:1}")
        get_pids_of_ppid $current_pid
        local new_pids="$RETVAL"
        [ -z "$new_pids" ] && continue

        for element in $new_pids; do
            running_pids+=("$element")
            pids=("$element" "${pids[@]}")
        done
    done

    kill ${pids[@]} 2>/dev/null
}

So, from the beginning. Let's assume we have terminal connected to /dev/stdout(FD #1) and /dev/stderr(FD #2). In practice, it could be a pipe, socket or whatever.

  • Create FDs #3 and #4 and point to the same "location" as #1 and #2 respectively. Changing FD #1 doesn't affect FD #3 from now on. Now, FDs #3 and #4 point to STDOUT and STDERR respectively. These will be used as real terminal STDOUT and STDERR.
  • 1> >(...) redirects STDOUT to command in parens
  • parens(sub-shell) executes 'tee' reading from exec's STDOUT(pipe) and redirects to 'logger' command via another pipe to sub-shell in parens. At the same time it copies the same input to FD #3(terminal)
  • the second part, very similar, is about doing the same trick for STDERR and FDs #2 and #4.

The result of running a script having the above line and additionally this one:

echo "Will end up in STDOUT(terminal) and /var/log/messages"

...is as follows:

$ ./my_script
Will end up in STDOUT(terminal) and /var/log/messages

$ tail -n1 /var/log/messages
Sep 23 15:54:03 wks056 my_script_tag[11644]: Will end up in STDOUT(terminal) and /var/log/messages

If you want to see clearer picture, add these 2 lines to the script:

ls -l /proc/self/fd/
ps xf
answered Dec 13 '13 at 10:30
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Explanation:

Consider the following code which prints the word "stdout" to stdout and the word "stderror" to stderror.

$ (echo "stdout"; echo "stderror" >&2)
stdout
stderror

Note that the '&' operator tells bash that 2 is a file descriptor (which points to the stderr) and not a file name. If we left out the '&', this command would print stdout to stdout, and create a file named "2" and write stderror there.

By experimenting with the code above, you can see for yourself exactly how redirection operators work. For instance, by changing which file which of the two descriptors 1,2, is redirected to /dev/null the following two lines of code delete everything from the stdout, and everything from stderror respectively (printing what remains).

$ (echo "stdout"; echo "stderror" >&2) 1>/dev/null
stderror
$ (echo "stdout"; echo "stderror" >&2) 2>/dev/null
stdout

Now, we can explain why the solution why the following code produces no output:

(echo "stdout"; echo "stderror" >&2) >/dev/null 2>&1

To truly understand this, I highly recommend you read this webpage on file descriptor tables. Assuming you have done that reading, we can proceed. Note that Bash processes left to right; thus Bash sees >/dev/null first (which is the same as 1>/dev/null), and sets the file descriptor 1 to point to /dev/null instead of the stdout. Having done this, Bash then moves rightwards and sees 2>&1. This sets the file descriptor 2 to point to the same file as file descriptor 1 (and not to file descriptor 1 itself!!!! (see this resource on pointers for more info) . Since file descriptor 1 points to /dev/null, and file descriptor 2 points to the same file as file descriptor 1, file descriptor 2 now also points to /dev/null. Thus both file descriptors point to /dev/null, and this is why no output is rendered.


To test if you really understand the concept, try to guess the output when we switch the redirection order:

(echo "stdout"; echo "stderror" >&2)  2>&1 >/dev/null

stderror

The reasoning here is that evaluating from left to right, Bash sees 2>&1, and thus sets the file descriptor 2 to point to the same place as file descriptor 1, ie stdout. It then sets file descriptor 1 (remember that >/dev/null = 1>/dev/null) to point to >/dev/null, thus deleting everything which would usually be send to to the standard out. Thus all we are left with was that which was not send to stdout in the subshell (the code in the parentheses)- i.e. "stderror". The interesting thing to note there is that even though 1 is just a pointer to the stdout, redirecting pointer 2 to 1 via 2>&1 does NOT form a chain of pointers 2 -> 1 -> stdout. If it did, as a result of redirecting 1 to /dev/null, the code 2>&1 >/dev/null would give the pointer chain 2 -> 1 -> /dev/null, and thus the code would generate nothing, in contrast to what we saw above.


Finally, I'd note that there is a simpler way to do this:

From section 3.6.4 here, we see that we can use the operator &> to redirect both stdout and stderr. Thus, to redirect both the stderr and stdout output of any command to \dev\null (which deletes the output), we simply type $ command &> /dev/null or in case of my example:

$ (echo "stdout"; echo "stderror" >&2) &>/dev/null

Key takeaways:

  • File descriptors behave like pointers (although file descriptors are not the same as file pointers)
  • Redirecting a file descriptor "a" to a file descriptor "b" which points to file "f", causes file descriptor "a" to point to the same place as file descriptor b - file "f". It DOES NOT form a chain of pointers a -> b -> f
  • Because of the above, order matters, 2>&1 >/dev/null is != >/dev/null 2>&1. One generates output and the other does not!

Finally have a look at these great resources:

Bash Documentation on Redirection, An Explanation of File Descriptor Tables, Introduction to Pointers

answered Nov 19 at 1:55

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