On 12/30/14 7:03 AM, Ville Oikarinen wrote:
> Hello.
>
> I often use "set -eu" to make my scripts more robust by exiting when
> there are errors.
>
> But one of my scripts just kept on running after failures. I found out
> that the keyword "local" does not obey it. Here is a script to
> reproduce the problem:
This is not a bug.
> #!/bin/bash
>
> set -eu
>
> failure() {
> echo "going to fail"
> false
> }
>
> cmd-subst-failure-with-local() {
> echo "calling failure without failing because of local"
> local FAILUREOUT=$(failure)
> echo "Bug: set -eu wasn't effective: failure output was '$FAILUREOUT'"
> }
>
> cmd-subst-failure-without-local() {
> echo "calling failure so it fails (without local)"
> FAILUREOUT=$(failure)
> # here set -eu works so we will fail instead of getting here
> echo "Internal error, set -eu didn't work even without local"
> }
>
> cmd-subst-failure-with-local
The `local' builtin succeeds because the assignment is performed succesfully.
> cmd-subst-failure-without-local
In the absence of a command name, the assignment statement's return
status is dependent on whether or not there is a command substitution
in the assignment statement, and that command substitution's exit
status.
--
``The lyf so short, the craft so long to lerne.'' - Chaucer
``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, ITS, CWRU c...@case.edu http://cnswww.cns.cwru.edu/~chet/
