Date: Mon, 13 Oct 2014 07:14:31 -0500 From: "William A. Mahaffey III" <wam@hiwaay.net> To: freebsd-questions@freebsd.org Subject: Re: sh man page .... Message-ID: <543BC227.50004@hiwaay.net> In-Reply-To: <20141013124649.4082d94f@gumby.homeunix.com> References: <5437FB8B.9080008@hiwaay.net> <20141010183814.3ae32a05@gumby.homeunix.com> <5438755B.2000108@hiwaay.net> <20141013124649.4082d94f@gumby.homeunix.com>
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On 10/13/14 06:46, RW wrote: > On Fri, 10 Oct 2014 19:10:03 -0500 > William A. Mahaffey III wrote: > > >> Straight out of the script which is failing. Under linux, if I call >> the script w/ no '-s #' option, the variable 'slept' is not set, & >> linux (or more accurately linux bash) evaluates that to the value oif >> zero (0). >> >> >> [wam@kabini1, ~, 7:07:22pm] 386 % sh >> $ if [ 0 -lt $(($slept)) ] ; then echo -n "$cmd: sleeping $slept secs >> ...." ; sleep $(($slept)) ; echo " done." ; fi >> arithmetic expression: expecting primary: "" >> [wam@kabini1, ~, 7:07:45pm] 387 % > > The problem here is that you have: > > [ 0 -lt $(($slept)) ] > > If you change it to the normal usage > > [ 0 -lt $((slept)) ] > > it works as expected. > > Is there any particular reason for the extra "$"? > > I guess the difference is not in the handling of uninitialised > variables, but specifically in the handling of $(()) which is an error > in sh, but not is bash. > _______________________________________________ > freebsd-questions@freebsd.org mailing list > http://lists.freebsd.org/mailman/listinfo/freebsd-questions > To unsubscribe, send any mail to "freebsd-questions-unsubscribe@freebsd.org" > Good question. I am *not* a bash/shell wiz, but I think the extra "$" was needed to get bash to behave (Linux, FC14 64-bit, i.e. a bit dated). Not 100% on that, but pretty sure .... -- William A. Mahaffey III ---------------------------------------------------------------------- "The M1 Garand is without doubt the finest implement of war ever devised by man." -- Gen. George S. Patton Jr.
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