Round a divided number in Bash

How would I round the result from two divided numbers, e.g.

3/2

As when I do

testOne=$((3/2))

$testOne contains "1" when it should have rounded up to "2" as the answer from 3/2=1.5

Solution 1:

To do rounding up in truncating arithmetic, simply add (denom-1) to the numerator.

Example, rounding down:

N/2
M/5
K/16

Example, rounding up:

(N+1)/2
(M+4)/5
(K+15)/16

To do round-to-nearest, add (denom/2) to the numerator (halves will round up):

(N+1)/2
(M+2)/5
(K+8)/16

Solution 2:

Good Solution is to get Nearest Round Number is

var=2.5
echo $var | awk '{print int($1+0.5)}'

Logic is simple if the var decimal value is less then .5 then closest value taken is integer value. Well if decimal value is more than .5 then next integer value gets added and since awk then takes only integer part. Issue solved

Solution 3:

bash will not give you correct result of 3/2 since it doesn't do floating pt maths. you can use tools like awk

$ awk  'BEGIN { rounded = sprintf("%.0f", 3/2); print rounded }'
2

or bc

$ printf "%.0f" $(echo "scale=2;3/2" | bc)
2