0

I want to for example divide all integers from 1 to 5 by 8 and put them in an array. I was thinking something similar to the code below but I'm sure it's wrong and dirty:

 COUNTER=0
 until [  $COUNTER -lt 8 ]; do
        ${steps[$COUNTER]}=1+(5-1)/8*$COUNTER
        let COUNTER+=1
 done

Is there anyway to do this cleanly if my code is correct at all?

Improve this question
1

bash can only do integer arithmetic. Use bc for floats:

from=1
to=5
steps=8
for c in $(seq 0 $steps) ; do
    ar[c]=$(echo $(bc -l <<< "$from + ($to - $from) / $steps * $c"))
done
echo "${ar[@]}"
Improve this answer
5
  • The OP wants to divide all integers from 1 to 5 by 8 and store that in the array. – terdon Oct 7 '15 at 13:29
  • @terdon: I understood the question differently before you edited it. – choroba Oct 7 '15 at 13:54
  • I know, it really wasn't clear, hence the comment and lack of downvote :) – terdon Oct 7 '15 at 13:56
  • I get really massive values with a lot of zeros in front of them any fixes? – Raymond Ghaffarian Shirazi Oct 7 '15 at 16:21
  • 1
    @RaymondGhaffarianShirazi: You can shorten the output of bc by declaring scale: bc -l <<< "scale=3; $from + ... – choroba Oct 7 '15 at 19:15
2

The seq command can increment by a float:

declare -a steps=($(seq 1 .5 5))

You can get the increment with echo 'scale=1;(5-1)/8'|bc

Improve this answer
2

Yes, but since bash doesn't do floating point arithmetic, you'll need to use another tool and save its output in the bash array. For example:

$ perl -le 'print $_/8 for 1..5' 
0.125
0.25
0.375
0.5
0.625

Save in an array with:

array=( $(perl -le 'print $_/8 for 1..5') )
Improve this answer

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.