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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.
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>

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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