You can also use let, which may make more sense
     let "myVar=$myVar+1"
 As for floating point you need something like bc to handle that
     myVar=5.4
     myOtherVar=3.1
     myVar=$(echo "scale=2; ($myVar + 3.14) / $myOtherVar" | bc)
     echo $myVar
If ran the above would print 2.75

On Mon, Mar 2, 2015 at 7:15 PM, der.hans <PLUGd@lufthans.com> wrote:

> Am 02. Mär, 2015 schwätzte Michael Havens so:
>
> moin moin Mike,
>
> see the Arithmetic Expansion portion of the bash man page for details.
>
> $((expression))
>
> That says to evaluate the expression and substitute the result. Remember
> that the shell only does interger math, no decimals.
>
> :) ~$ echo $(( 4 / 2 ))
> 2
> :) ~$ echo $(( 4 / 3 ))
> 1
> :) ~$ echo $(( 4 * 3 ))
> 12
> :) ~$
>
> Note that you don't need to escape the * when using arithmetic expansion
> because the $(( )) is already quoting.
>
> ciao,
>
> der.hans
>
>
>  just starting with this so please, bear with me....
>>
>> Anyways I am looking at this beginner's script:
>>
>>   #! /bin/bash
>>   myvar=0
>>   while [ $myvar -ne 20 ]
>>   do
>>           echo $myvar
>>           myvar=$(( $myvar + 1 ))
>>   done
>>
>> Now my mind can wrap itself around everything in this script except for
>> tis
>> line:
>>
>>   myvar=$(( $myvar + 1 ))
>>
>> Where my difficulty arises is why the '$' before the '(('?
>> (it was geast fun running that script with a minus sign instead!)
>> :-)~MIKE~(-:
>>
>>
> --
> #  http://www.LuftHans.com/        http://www.PhxLinux.org/
> #  "The first requisite of a good citizen in this republic of ours is that
> #  he should be able and willing to pull his weight."  -- Theodore
> Roosevelt
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