capture exit code and output of a command

Question

There are similar questions, but mine is a little bit different. Or so I think. What I'd like to do is:

1.sh:

#!/usr/bin/env bash
set -eu
r=0
a=$(./2.sh || r=$?)
echo "$a"
echo "$r"

2.sh:

#!/usr/bin/env bash
echo output
exit 2

But it outputs:

$ ./1.sh
output
0   # I'd like to have `2` here

Since $(...) runs separate shell. So, how do I capture both, exit code and output?

a=$(./2.sh); r=$?; ## doesn't work? – Jeff Schaller Mar 4 '16 at 11:34 — Jeff Schaller, Mar 4 '16 at 11:34

Jeff Schaller · Accepted Answer · 2016-03-04 12:10:11Z

up vote 5 down vote accepted

The exit code of a process calling another process is the one of the called process.

$($($($($(exit 2)))))
echo $?
2

Here there are 5 levels of calling.

In your case:

r=0
a=$(./2.sh)
r=$?
echo "$a"
echo "$r"

Jeff Schaller

11.9k62242

answered Mar 4 '16 at 11:36

Emmanuel

2,3371614

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