How to redirect stdout+stderr to one file while keeping streams separate?

Question

Redirecting stdout+stderr such that both get written to a file while still outputting to stdout is simple enough:

cmd 2>&1 | tee output_file

But then now both stdout/stderr from cmd are coming on stdout. I'd like to write stdout+stderr to the same file (so ordering is preserved assuming cmd is single threaded) but then still be able to also separately redirect them, something like this:

some_magic_tee_variant combined_output cmd > >(command-expecting-stdout) 2> >(command-expecting-stderr)

So combined_output contains the both with order preserved, but the command-expecting-stdout only gets stdout and command-expecting-stderr only gets stderr. Basically, I want to log stdout+stderr while still allowing stdout and stderr to be separately redirected and piped. The problem with the tee approach is it globs them together. Is there a way to do this in bash/zsh?

Answer 1

From what I unterstand this is what you are looking for. First I made a litte script to write on stdout and stderr. It looks like this:

$ cat foo.sh 
#!/bin/bash

echo foo 1>&2
echo bar

Then I ran it like this:

$ ./foo.sh 2> >(tee stderr | tee -a combined) 1> >(tee stdout | tee -a combined)
foo
bar

The results in my bash look like this:

$ cat stderr
foo
$ cat stdout 
bar
$ cat combined 
foo
bar

Note that the -a flag is required so the tees don't overwrite the other tee's content.

Answer 2

{ { cmd | tee out >&3; } 2>&1 | tee err >&2; } 3>&1

Or, to be pedantic:

{ { cmd 3>&- | tee out >&3 2> /dev/null; } 2>&1 | tee err >&2 3>&- 2> /dev/null; } 3>&1

Note that it's futile to try and preserve order. It is basically impossible. The only solution would be to modify "cmd" or use some LD_PRELOAD or gdb hack,