On Mon, Nov 9, 2009 at 4:49 AM, Chet Ramey <chet.ra...@case.edu> wrote:
> Ciprian Dorin, Craciun wrote:
>> Shouldn't any of the following scripts print `error`? (Bash
>> 4.0.35(2)-release on ArchLinux.)
>>
>> Or I've miss-interpreted the documentation...
>>
>> Thanks,
>> Ciprian.
>>
>>
>> ~~~~
>> set -e -o pipefail
>> ( false ; echo ok ; ) || echo error
>> ~~~~
>>
>> ~~~~
>> set -e -o pipefail
>> ( false ; echo ok ; ) | true || echo error
>> ~~~~
>>
>> ~~~~
>> set -e -o pipefail
>> { false ; echo ok ; } || echo error
>> ~~~~
>>
>> ~~~~
>> set -e -o pipefail
>> { false ; echo ok ; } | true || echo error
>> ~~~~
>
> No. Since `set -e' has no effect on the left side of the || or &&
> operators, all of the commands preceding the || exit with status 0.
>
> Chet
> --
> ``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/
Sorry, but I don't understand at all... So please bare with me and
make me understand.
So I've interpreted `set -e` as a way to tell bash to treat any
process exiting with non-zero (and not succeeded by a || ), as an
error and end the current shell / sub-shell.
Thus if I say: `set -e ; { false ; true ; }` it works, but when I
put the `||`, it doesn't...
So my question is how can I solve this problem? (And obtain the
needed behaviour.) (I see `()` and `{}` as blocks in normal
programming languages (of course with some particularities), and
non-zero exit codes as exceptions. And this is very helpful to write
robust bash scripts.)
Thanks,
Ciprian.
