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set -e ignored in subshell if part of command list
From: |
Shaun Crampton |
Subject: |
set -e ignored in subshell if part of command list |
Date: |
Wed, 13 Nov 2019 10:24:23 +0000 |
Configuration Information [Automatically generated, do not change]:
Machine: x86_64
OS: linux-gnu
Compiler: gcc
Compilation CFLAGS: -g -O2
-fdebug-prefix-map=/build/bash-LQgi2O/bash-5.0=.
-fstack-protector-strong -Wformat -Werror=format-security -Wall
-Wno-parentheses -Wno-format-security
uname output: Linux shauns-laptop 5.1.16-050116-generic #201907031232
SMP Wed Jul 3 12:35:21 UTC 2019 x86_64 x86_64 x86_64 GNU/Linux
Machine Type: x86_64-pc-linux-gnu
Bash Version: 5.0
Patch Level: 3
Release Status: release
Description:
I was trying to get a function to return early if a command
fails by putting
the body of the function in a subshell and using set -e inside
the subshell.
If I run a subshell on its own, this works, but when I try to combine it
into a larger program, the set -e gets ignored.
Repeat-By:
Managed to boil it down to this smaller example:
# On its own, subshell behaves as expected:
$ ( set -ex; false; echo here )
+ false
# In a list, behaviour changes, "echo here" gets executed:
$ ( set -ex; false; echo here ) && echo there
+ false
+ echo here
here
there
# If the subshell is executed in the background, it works
$ ( set -e; false; echo here ) & pid=$!; wait $pid && echo there
[1] 26219
[1]+ Exit 1 ( set -e; false; echo here )