dividing in bash script with digits

Question

I want to dive integers but my results would be a digit bash can't do digits:

from=69;to=64;steps=$(( 12-1 )); ssize=$(( (to - from) / steps ))
echo $ssize $steps

results

0 11

Edit:

bc works with multiplication:

echo "5.94*11.14" | bc

gives 66.17

but division

echo "5.94/11.14" | bc

gives 0

What did you expect that echo to produce? ssize is zero because bash only does integer arithmetic (-5/11 = 0) — glenn jackman, Oct 07 '15 at 18:33
I see, I thought bc can solve the issue but also bc can only multiply with digits — Raymond Ghaffarian Shirazi, Oct 07 '15 at 18:45
You can set the precision for bc. Either echo "scale=3; 5.94/11.14" | bc or echo "5.94/11.14" | bc -l — glenn jackman, Oct 07 '15 at 19:30
Teach bash integer division with floating point results: http://stackoverflow.com/a/24431665/3776858 — Cyrus, Oct 08 '15 at 05:20
Related: Bash only gives integer as output regardless of input when performing calculations, Why does division in bash arithmetic compute most percentages as 0? — Eliah Kagan, Sep 09 '19 at 19:33

score 1 · Accepted Answer · answered Oct 07 '15 at 19:30

As was already pointed out, bash cannot handle floating point calculation by itself. You're doing a floor-division

(69-64)//11 = 5//11 = 0

Instead use bc

steps=12;ssize=$(bc -l <<< "from=69;to=64;(to - from)/$steps;"); echo $ssize $steps

The option -l will load the mathlib and with it set scale=20 telling bc how many digits after the period to handle. The default is 0, that's why you're also seeing only integer division when invoking bc without -l. Alternatively set scale manually

steps=12;ssize=$(bc <<< "scale=20;from=69;to=64;(to - from)/$steps;"); echo $ssize $steps

dividing in bash script with digits

1 Answers1