Piping both stdout and stderr in bash?

Question

It seems that newer versions of bash have the &> operator, which (if I understand correctly), redirects both stdout and stderr to a file (&>> appends to the file instead, as Adrian clarified).

What's the simplest way to achieve the same thing, but instead piping to another command?

For example, in this line:

cmd-doesnt-respect-difference-between-stdout-and-stderr | grep -i SomeError

I'd like the grep to match on content both in stdout and stderr (effectively, have them combined into one stream).

Note: this question is asking about piping, not redirecting - so it is not a duplicate of the question it's currently marked as a duplicate of.

Answer 1

(Note that &>>file appends to a file while &> would redirect and overwrite a previously existing file.)

To combine stdout and stderr you would redirect the latter to the former using 2>&1. This redirects stderr (file descriptor 2) to stdout (file descriptor 1), e.g.:

$ { echo "stdout"; echo "stderr" 1>&2; } | grep -v std
stderr
$

stdout goes to stdout, stderr goes to stderr. grep only sees stdout, hence stderr prints to the terminal.

On the other hand:

$ { echo "stdout"; echo "stderr" 1>&2; } 2>&1 | grep -v std
$

After writing to both stdout and stderr, 2>&1 redirects stderr back to stdout and grep sees both strings on stdin, thus filters out both.

You can read more about redirection here.

Regarding your example (POSIX):

cmd-doesnt-respect-difference-between-stdout-and-stderr 2>&1 | grep -i SomeError

or, using >=bash-4:

cmd-doesnt-respect-difference-between-stdout-and-stderr |& grep -i SomeError

Answer 2

Bash has a shorthand for 2>&1 |, namely |&, which pipes both stdout and stderr (see the manual):

cmd-doesnt-respect-difference-between-stdout-and-stderr |& grep -i SomeError

This was introduced in Bash 4.0, see the release notes.