Re: Wrong input confuses bash's arithmetic unit permanently

[Chris F.A. Johnson]

On 2007-09-11, Enrico Scholz wrote:
> Configuration Information [Automatically generated, do not change]:
> Machine: i686
> OS: linux-gnu
> Compiler: gcc
> Compilation CFLAGS:  -DPROGRAM='bash' -DCONF_HOSTTYPE='i686' 
> -DCONF_OSTYPE='linux-gnu' -DCONF_MACHTYPE='i686-redhat-linux-gnu' 
> -DCONF_VENDOR='redhat' -DLOCALEDIR='/usr/share/locale' -DPACKAGE='bash' 
> -DSHELL -DHAVE_CONFIG_H   -I.  -I. -I./include -I./lib  -D_GNU_SOURCE 
> -D_LARGEFILE_SOURCE -D_FILE_OFFSET_BITS=64 -O2 -g -pipe -Wall 
> -Wp,-D_FORTIFY_SOURCE=2 -fexceptions -fstack-protector 
> --param=ssp-buffer-size=4 -m32 -march=i386 -mtune=generic 
> -fasynchronous-unwind-tables
> uname output: Linux kosh.bigo.ensc.de 2.6.22.3-ensc-1 #1 SMP PREEMPT Thu Aug 
> 16 23:36:51 CEST 2007 i686 i686 i386 GNU/Linux
> Machine Type: i686-redhat-linux-gnu
>
> Bash Version: 3.2
> Patch Level: 9
> Release Status: release
>
> Description:
>
> A syntactical wrong statement will confuse bash permanently so that 
> arithmetic operations return wrong results.
>
>
> Repeat-By:
> [copy from https://bugzilla.redhat.com/show_bug.cgi?id=286861]
> $ i=0
> $ let ++i
> $ echo $i
> 1
>
> $ let tmp="foo.a"+0
> bash: let: tmp=foo.a+0: syntax error: invalid arithmetic...
>
> $ i=0
> $ let ++i
> $ echo $i
> 0
>
> --> every further use of 'let' will generate bogus results. bash-3.1 returns
> expected results.

   I agree that it is a bug, but it is one that will never bother me
   because I always use the POSIX arithmetic syntax:

i=$(( $i + 1 ))

   What's the point of using let?

-- 
   Chris F.A. Johnson                      <http://cfaj.freeshell.org>
   ===================================================================
   Author:
   Shell Scripting Recipes: A Problem-Solution Approach (2005, Apress)
.