问题
This outputs before\n:
#!/usr/bin/env bash
set -e
echo before
((0))
echo after
Removing set -e or changing ((0)) to ((1)) makes the program output before\nafter\n as expected.
Why does ((0)) trigger the set -e exit condition?
回答1:
This will explain:
((0))
echo $?
1
((1))
echo $?
0
So it is due to non-zero return status of arithmetic expression evaluation in (( and )) your script is exiting when set -e is being used.
As help set says this:
-e Exit immediately if a command exits with a non-zero status.
回答2:
Line
set -e
means:
-e Exit immediately if a command exits with a non-zero status. (see: https://www.gnu.org/software/bash/manual/html_node/The-Set-Builtin.html)
((0)) is an expression that evaluates to 1. That's why the script exits.
来源：https://stackoverflow.com/questions/24148146/why-does-0-cause-a-bash-script-to-exit-if-set-e-directive-is-present