How do I do if statement arithmetic in bash?

Question

I want to do something like this:

if [ $1 % 4 == 0 ]; then
...

But this does not work.

What do I need to do instead?

Answer 1

If you want something a bit more portable - for example, something that works in sh as well as in bash - use

if [ $(echo "$1 % 4" | bc) -eq 0 ]; then
...

An example

#!/bin/bash
#@file: trymod4.bash

if [ $(echo "$1 % 4" | bc) -eq 0 ]; then
  echo "$1 is evenly divisible by 4"
else
  echo "$1 is NOT evenly divisible by 4"
fi

$ chmod +x trymod4.bash
$ ./trymod4.bash 224
224 is evenly divisible by 4
$ ./trymod4.bash 223
223 is NOT evenly divisible by 4

I put this in, because you used the single [ ... ] conditional, which I usually associate with sh-compatible programs.

---

Check that this works in sh.

#!/bin/sh
#@file: trymod4.sh

if [ $(echo "$1 % 4" | bc) -eq 0 ]; then
  echo "$1 is evenly divisible by 4"
else
  echo "$1 is NOT evenly divisible by 4"
fi

$ chmod +x trymod4.sh
$ ./trymod4.sh 144
144 is evenly divisible by 4
$ ./trymod4.sh 19
19 is NOT evenly divisible by 4

All right, it works with sh.

Note the "theoretical" (but not always implemented as such) differences between [ ... ] and [[ ... ]] from this site (archived).