A gotcha for me originally and perhaps others is that while using ordering like

   $ ./outerr  >blah 2>&1

sends stdout and stderr to blah, imitating the order with pipe instead does not.

   $ ./outerr  | 2>&1 cat >blah
   err

This is because | is not a mere redirector but a statement terminator.

    (where outerr is the following...)
    echo out 
    echo err >&2

▲ time4tea 4 hours ago | parent | next [-]

Useless use of cat error/award

But also | isnt a redirection, it takes stdout and pipes it to another program.

So, if you want stderr to go to stdout, so you can pipe it, you need to do it in order.

bob 2>&1 | prog

You usually dont want to do this though.

	▲	kazinator 4 hours ago \| parent [-]
		The point is that the order in which that is processed is not left to right. First the \| pipe is established as fd [1]. And then 2>&1 duplicates that pipe into [2]. I.e. right to left: opposite to left-to-right processing of redirections. When you need to capture both standard error and standard output to a file, you must have them in this order: `bob > file 2>&1` It cannot be: `bob 2>&1 > file` Because then the 2>&1 redirection is performed first (and usually does nothing because stderr and stdout are already the same, pointing to your terminal). Then > file redirects only stdout. But if you change > file to \| process, then it's fine! process gets the combined error and regular output.

▲ inigyou 4 hours ago | parent | prev [-]

Why would that second one be expected to work?