modulus in bash

[abraxas334]

After googling many things i really cannot workout, where I am going wrong.

I amtrying to print out a variable if the number is even:
My very basic script looks like this:



#!/bin/bash

echo "this will be a T-jump experiment"

for traj in {1..10} #number of cycles loop
do
modulo = $traj%2
echo $modulo
if [ $traj%2==0 ]; then
echo $traj
for temp in {1..2} #number of temperatures
do
filename="nd_temp_($temp)_tran_($traj)"
echo $filename

done
else
for temp in {2..1} #number of temperatures
do
filename="nd_temp_($temp)_tran_($traj)"
echo $filename

done
fi

done


Modulus does exist in bash but I am apparently using it wrong. I don't really understand what is wrong tho.

Thanks for any help

[ofnuts]

You have to enclose your arithmetic in double parentheses, so:



modulo=$(( $traj % 2 ))

[LemursDontExist]

I don't think bash recognizes the modulo operator in that context - the shell needs to know you're doing arithmetic.


if [ $(($traj%2)) == 0 ]; then

should work. Alternatively


if (( $traj%2 == 0 )); then

should also work. Or, maybe best


if [ `expr $traj % 2` == 0 ]; then

The first two rely on bash syntax that I'm not positive works across all standard shells, but the last should work in any normal shell.

[sisco311]

Check out http://mywiki.wooledge.org/ArithmeticExpression

As LemursDontExist already mentioned it, the POSIX version is:

if test $((traj%2)) = 0; then ... #or
if [ $((traj%2)) = 0 ]; then ...

In BASH you can use let or ((


if ((traj%2 == 0)); then ...
if let "$traj%2 == 0"; then ...

[codemaniac]

You have to enclose your arithmetic in double parentheses, so:



modulo=$(( $traj % 2 ))

or the other way . : )


((modulo = $traj % 2 ))