Source: Vaneyckt: Safer bash scripts with 'set -euxo pipefail'
Shortcut: set -Eeuox pipefail
Makes the script exit immediately when a command fails.
Before:
#!/bin/bash
# 'foo' is a non-existing command
foo
echo "bar"
# output
# ------
# line 4: foo: command not found
# bar
#
# Note how the script didn't exit when the foo command could not be found.
# Instead it continued on and echoed 'bar'.
After:
#!/bin/bash
set -e
# 'foo' is a non-existing command
foo
echo "bar"
# output
# ------
# line 5: foo: command not found
#
# This time around the script exited immediately when the foo command wasn't
# found. Such behavior is much more in line with that of higher-level languages.
Makes the script exit immediately when a command in a pipeline fails instead of looking only at the exit code of the pipeline.
Before:
#!/bin/bash
set -e
# 'foo' is a non-existing command
foo | echo "a"
echo "bar"
# output
# ------
# a
# line 5: foo: command not found
# bar
#
# Note how the non-existing foo command does not cause an immediate exit, as
# it's non-zero exit code is ignored by piping it with '| echo "a"'.
After:
#!/bin/bash
set -o pipefail
# 'foo' is a non-existing command
foo | echo "a"
echo "bar"
# output
# ------
# a
# line 5: foo: command not found
#
# This time around the non-existing foo command causes an immediate exit, as
# '-o pipefail' will prevent piping from causing non-zero exit codes to be
# ignored.
Treat unset variables as an error and exit immediately.
Before:
#!/bin/bash
echo "$a
echo "bar"
# output
# ------
#
# bar
#
# The default behavior will not cause unset variables to trigger an immediate
# exit. In this particular example, echoing the non-existing $a variable will
# just cause an empty line to be printed.
After:
#!/bin/bash
set -u
echo "$a"
echo "bar"
# output
# ------
# line 5: a: unbound variable
#
# Notice how 'bar' no longer gets printed. We can clearly see that '-u' did
# indeed cause an immediate exit upon encountering an unset variable.
For cases where you want to deal with unset variables, check the ${a:-b}
Makes the bash script print each command before executing it.
#!/bin/bash
set -x
a=5
echo $a
echo "bar"
# output
# ------
# + a=5
# + echo 5
# 5
# + echo bar
# bar
Traps are fired when a bash script catches certain signals (e.g. SIGINT or SIGTERM).
However using -e without -E will cause an ERR trap to not fire in certain conditions:
Before:
#!/bin/bash
set -e
trap "echo ERR trap fired!" ERR
myfunc()
{
# 'foo' is a non-existing command
foo
}
myfunc
echo "bar"
# output
# ------
# line 9: foo: command not found
#
# Notice that while '-e' did indeed cause an immediate exit upon trying to execute the non-existing
# foo command, it did not case the ERR trap to be fired.
After:
#!/bin/bash
set -Ee
trap "echo ERR trap fired!" ERR
myfunc()
{
# 'foo' is a non-existing command
foo
}
myfunc
echo "bar"
# output
# ------
# line 9: foo: command not found
# ERR trap fired!
#
# Not only do we still have an immediate exit, we can also clearly see that the
# ERR trap was actually fired now.
thanks @Hologos, finally corrected :)