Re: ((i++)) no longer supported?

[Chris F.A. Johnson]

On Thu, 3 May 2012, DJ Mills wrote:

> On Thu, May 3, 2012 at 2:53 AM, Pierre Gaston <pierre.gaston@gmail.com> wrote:
> > On Thu, May 3, 2012 at 9:34 AM, Pan ruochen <panruochen@gmail.com> wrote:
> > > Hi All,
> > >
> > > Suddenly I found that ((i++)) is not supported on bash.
> > > Just try the following simple case:
> > > $i=0; ((i++)); echo $?
> > > And the result is
> > > 1
> > > which means an error.
> > > I got the same result on GNU bash, version 4.1.2(1)-release
> > > (x86_64-redhat-linux-gnu) and GNU bash, version 4.1.10(4)-release
> > > (i686-pc-cygwin).
> > >
> > > - BR, Ruochen
> >
> > It has always been the case, and fits the documentation since i++
> > value is 0 and that is false in the arithmetic context.
> >
> > What changed is that it bash exits in this case if you use set -e.
> > Some Possible workarounds:
> > ((i++)) || :
> > ((i+=1))
> > i=$((i+1))
> > and a gazillon others.
>
> The most sane workaround would probably be to stop using set -e.

   Better still, use the POSIX standard method shown above: i=$((i+1))

--
   Chris F.A. Johnson, <http://cfajohnson.com/>
   Author:
   Pro Bash Programming: Scripting the GNU/Linux Shell (2009, Apress)
   Shell Scripting Recipes: A Problem-Solution Approach (2005, Apress)