Issue with program check if the number is divisible by 2 with no remainder BASH

Question

I tried to write a program to see if the count divisible by 2 without a remainder Here is my program

count=$((count+0))

while read line; do

if [ $count%2==0 ]; then
    printf "%x\n" "$line" >> file2.txt
else
    printf "%x\n" "$line" >> file1.txt
fi

count=$((count+1))
done < merge.bmp

This program doesnt work its every time enter to the true

Answer 1

In the shell, the [ command does different things depending on how many arguments you give it. See https://www.gnu.org/software/bash/manual/bashref.html#index-test

With this:

[ $count%2==0 ]

you give [ a single argument (not counting the trailing ]), and in that case, if the argument is not empty then the exit status is success (i.e. "true"). This is equivalent to [ -n "${count}%2==0" ]

You want

if [ "$(( $count % 2 ))" -eq 0 ]; then

or, if you're using bash

if (( count % 2 == 0 )); then

Answer 2

Some more "exotic" way to do this:

count=0
files=(file1 file2 file3)
num=${#files[@]}

while IFS= read -r line; do
    printf '%s\n' "$line" >> "${files[count++ % num]}"
done < input_file

This will put 1st line to file1, 2nd line to file2, 3rd line to file3, 4th line to file1 and so on.