How to divide numbers and return as float in #!/bin/sh

Question

Hi I have to divide two numbers in #!/bin/sh

$x = 5678
$x/1000

but calculations return only integer: 5.

I need 5,678.

I've tried

$x/1000.0 
$x/1000.0 | bc

on first method expect integer

second method: I don't have bc.

Answer 1

You can use awk:

#! /bin/sh -
x=5678 y=1000
awk -- 'BEGIN{printf "%.3f\n", ARGV[1]/ARGV[2]}' "$x" "$y"

Whether you'll get 5.678 or 5,678 will depend on the locale ($LC_ALL, $LC_NUMERIC, $LANG variables) and the implementation of awk (some of which will use . regardless of the locale).

With the GNU implementation of awk, you need to pass a POSIXLY_CORRECT variable into its environment for it to honour the locale for the decimal radix character.

That also applies to input numbers:

$ LC_ALL=fr_FR.UTF-8 POSIXLY_CORRECT=1 gawk -- 'BEGIN{
   printf "%.3f\n", ARGV[1]/ARGV[2]}' 1,2 10
0,120

$ LC_ALL=fr_FR.UTF-8 POSIXLY_CORRECT=1 gawk -- 'BEGIN{
   printf "%.3f\n", ARGV[1]/ARGV[2]}' 1.2 10
0,100

(in that latter case, the 1.2 was interpreted as 1, because when doing an explicit conversion of string to number (like when using the / arithmetic operator here), awk parses as much of the string as it can as long as it makes a valid number and ignores the rest. That .2 is not understood in that French locale as the decimal radix is , there, not .).

The --, to mark the end of options is needed in the busybox implementation of awk which you seem to be using as it accepts options after non-option arguments, which makes it non-POSIX compliant. Without it, the above wouldn't work if $x or $y were negative number.

Alternatvely, you could again set $POSIXLY_CORRECT, which would force it to parse its options the standard way. Note that in any case, busybox awk always uses . as the decimal radix character regardless of the locale and regardless of $POSIXLY_CORRECT.

Answer 2

you can use the following dc snippet to get the result:

$ dc -e "3k $x 1000/p"