Bash 后增量中止脚本执行

Question

Running with bash -e :

round=0
((round++))
echo "Done"

Will not show Done . Why? How can I use post-increment when -e is set?

Answer 1

Please take a look of some examples:

round=0
((round++))
echo $?

round=1
((round-1))
echo $?

((0))
echo $?

In all cases $? returns 1.

Bash manpage states:

((expression))
The expression is evaluated according to the rules described below under ARITHMETIC EVALUATION. If the value of the expression is non-zero, the return status is 0; otherwise the return status is 1. This is exactly equivalent to let "expression".

If the -e option is set, then the script exits because:

-e
Exit immediately if a pipeline (which may consist of a single simple command), a list, or a compound command (see SHELL GRAMMAR above), exits with a non-zero status. ...

You can avoid the termination by assigning a variable to the result of arithmetic evaluation:

set -e
round=0
dummy=$((round++))
echo "Done"

Hope this helps.