Bash - Math: Returns integer instead of float

Question

Bash returns value 4 instead of 4.2 when dividing 21/5. And returns 36 when doing 9 * (21/5) instead of 37.8.

echo "$(( 21/5 ))"
4
    
echo "21/5" | bc
4

Similarly 9 * ( 21/5 ) returns 36 but it should be 37.8;

y = 9;
num_lines = 21;
w = 5;

let value="$y * ($num_lines/$w)"

echo $value
36

Bash's arithmetic expansion does not support floating-point calculations. Look at [this question](https://unix.stackexchange.com/q/40786/381289) for other possibilities. — Piotr P. Karwasz, Mar 28 '20 at 11:21

asktyagi · Accepted Answer · 2020-03-28T10:18:54.343

3

try something like this

$ echo "21/5" | bc -l
4.20000000000000000000

$ echo "$y * ($num_lines/$w)" |bc -l
37.80000000000000000000

For setting up limit of decimal point use like below, For 2 decimal point

 $echo "scale=2; $y * ($num_lines/$w)" | bc -l
 37.80

For 3 decimal point, just change the scale number.

$ echo "scale=3; $y * ($num_lines/$w)" | bc -l
37.800

edited Mar 28 '20 at 10:18

answered Mar 28 '20 at 10:14

asktyagi

nice. Is there a way to limit decimal points? – Robert Mar 28 '20 at 10:15
found it. echo "scale=2; 21/5" | bc -l – Robert Mar 28 '20 at 10:16
is I possible to do something like this? `while [ "$y" -le (( $y * ($num_lines/$w) | bc -l )) ]` – Robert Mar 28 '20 at 10:21
use like this `while [ $y -le $(( $y * $num_lines/$w | bc -l )) ];do echo 1; sleep 3;done` – asktyagi Mar 28 '20 at 10:35

1 Answers1