On 12/10/10 12:17 PM, Marc Herbert wrote:
> Le 10/12/2010 16:05, Andreas Schwab a écrit :
>>>
>>> This is a design mistake: it trades a few characters for a lot of confusion.
>>
>> You can always choose to ignore the exit status. The converse is not
>> true.
>
> Agreed, but that does not imply an
Le 10/12/2010 16:19, Dominic Raferd a écrit :
> Thanks Greg (and also Eric and Andreas). Your FAQ makes it very clear;
> that is to say, it makes it clear how darned complicated it is. Seems
> best to avoid using 'set -e' altogether, as you say (except perhaps at
> an early stage for debugging):
On Thu, Dec 09, 2010 at 05:52:49PM +, Dominic Raferd wrote:
> $ val=0; let val++; echo $val,$?; unset val
> 1,1
>
> see the error code 1. Setting any other start value (except
> undefined) for val does not produce this error, the problem occurs
> for let val++ and let val-- if the start value
Le 10/12/2010 16:05, Andreas Schwab a écrit :
>>
>> This is a design mistake: it trades a few characters for a lot of confusion.
>
> You can always choose to ignore the exit status. The converse is not
> true.
Agreed, but that does not imply any command should try to be creative
and throw random
On 10/12/2010 15:15, Greg Wooledge wrote:
On Thu, Dec 09, 2010 at 05:52:49PM +, Dominic Raferd wrote:
Why does this happen? Is it 'by design'? It makes arithmetic with bash
let very dangerous because it can throw unexpected errors (and break
scripts running with set -e).
http://mywiki.wool
Marc Herbert writes:
>> let intentionally returns status 1 if the value was 0; and status > 1 if
>> there was an error. Why? So you can do loops such as:
>>
>> countdown=10
>> while let countdown--; do ... ; done
>>
>>> Why does this happen? Is it 'by design'?
>>
>> Yes. The same as for 'ex
On 12/10/2010 08:49 AM, Marc Herbert wrote:
>> let intentionally returns status 1 if the value was 0; and status > 1 if
>> there was an error. Why? So you can do loops such as:
>>
>> countdown=10
>> while let countdown--; do ... ; done
>>
>>> Why does this happen? Is it 'by design'?
>>
>> Yes. T
> let intentionally returns status 1 if the value was 0; and status > 1 if
> there was an error. Why? So you can do loops such as:
>
> countdown=10
> while let countdown--; do ... ; done
>
>> Why does this happen? Is it 'by design'?
>
> Yes. The same as for 'expr' which is standardized by POS
Dominic Raferd writes:
> Why does this happen? Is it 'by design'?
$ help let
[...]
Exit Status:
If the last ARG evaluates to 0, let returns 1; let returns 0 otherwise..
Andreas.
--
Andreas Schwab, sch...@linux-m68k.org
GPG Key fingerprint = 58CA 54C7 6D53 942B 1756 01D3 44D5 214B 827
On Thu, Dec 09, 2010 at 05:52:49PM +, Dominic Raferd wrote:
> Why does this happen? Is it 'by design'? It makes arithmetic with bash
> let very dangerous because it can throw unexpected errors (and break
> scripts running with set -e).
http://mywiki.wooledge.org/BashFAQ/105 -- Why doesn't s
On 12/09/2010 10:52 AM, Dominic Raferd wrote:
> Description:
>
> $ val=0; let val++; echo $val,$?; unset val
> 1,1
Not a bug.
>
> see the error code 1. Setting any other start value (except undefined)
> for val does not produce this error, the problem occurs for let val++
> and let val-- if the
Configuration Information [Automatically generated, do not change]:
Machine: i486
OS: linux-gnu
Compiler: gcc
Compilation CFLAGS: -DPROGRAM='bash' -DCONF_HOSTTYPE='i486'
-DCONF_OSTYPE='linu
x-gnu' -DCONF_MACHTYPE='i486-pc-linux-gnu' -DCONF_VENDOR='pc'
-DLOCALEDIR='/usr/
share/locale' -DPACKAGE=
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