How do I make this bitwise code more concise? - bash

Question

My bitwise code which works is:
#!/bin/bash -e
random=$((RANDOM % 32));
bitWiseAnd() {
local IFS='&'
printf "%s\n" "$(( $* ))"
}
echo "random=${random}";
if [ $(bitWiseAnd ${random} "0x10") -ne 0 ]; then
echo "1"
else
echo "0"
fi
if [ $(bitWiseAnd ${random} "0x8") -ne 0 ]; then
echo "1"
else
echo "0"
fi
if [ $(bitWiseAnd ${random} "0x4") -ne 0 ]; then
echo "1"
else
echo "0"
fi
if [ $(bitWiseAnd ${random} "0x2") -ne 0 ]; then
echo "1"
else
echo "0"
fi
if [ $(bitWiseAnd ${random} "0x1") -ne 0 ]; then
echo "1"
else
echo "0"
fi
A sample output from the above code:
random=15
0
1
1
1
1
I have no idea how this code works. Can this code be made more concise?

Answer 1

You could use a for loop instead of repeated if blocks. Also, the for loop spells out your purpose more clearly, like this:
#!/bin/bash -e
bitWiseAnd() {
local IFS='&'
printf "%s\n" "$(( $* ))"
}
random=$((RANDOM % 32));
echo "random=${random}";
for i in 10 8 4 2 1; do
if [ $(bitWiseAnd ${random} "0x$i") -ne 0 ]; then
echo "1"
else
echo "0"
fi
done
Output:
random=10
0
1
0
1
0
Also, the function below
bitWiseAnd() {
local IFS='&'
printf "%s\n" "$(( $* ))"
}
could be made more explicit this way, as long as we are looking at just two input arguments:
bitWiseAnd() {
printf "%s\n" $(($1 & $2))
}

Answer 2

This won't produce the exact same output as you need, because you asked for an optimization and it doesn't involve multiple useless echo statements.
Another approach would be using a while-loop and check if bit-positions are still valid. I have removed the useless echo statements, assuming you don't need any of that and just want to improve the code.
bitMask="0x10"
while [ $(bitWiseAnd "31" "$bitMask") -ne 0 ] && ((bitMask))
do
bitMask=$((bitMask >> 1))
done
Just replace the hard-coded number 31 with the variable value $random and bitMask of your choice.