Bash Shell Script - Cut - linux

Question

How i can transform this in a while in bash shell script? Thank you.
a[4]=$(echo $c | cut -c1)
a[3]=$(echo $c | cut -c2)
a[2]=$(echo $c | cut -c3)
a[1]=$(echo $c | cut -c4)
a[0]=$(echo $c | cut -c5)
b[4]=$(echo $d | cut -c1)
b[3]=$(echo $d | cut -c2)
b[2]=$(echo $d | cut -c3)
b[1]=$(echo $d | cut -c4)
b[0]=$(echo $d | cut -c5)

Answer 1

You don't need cut also, following will work:
a=(); for ((i=0; i<5; i++)); do a+=( "${c:$i:1}" ); done
b=(); for ((i=0; i<5; i++)); do b+=( "${d:$i:1}" ); done

Answer 2

There are several ways of getting a reversed string into an array, here is one:
#!/bin/bash
STRING="test string"
# First, reverse the string into a new variable
REV_STRING=$(rev <<< $STRING)
# Declare the array
declare -a ARRAY
ARRAY_LEN=${#REV_STRING}
ARRAY_UPPER_BOUND=$((ARRAY_LEN - 1))
for i in $(seq 0 $ARRAY_UPPER_BOUND); do
# Here, ${REV_STRING:$i:1} is equivalent to say REV_STRING.charAtPosition(i)
ARRAY[$i]=${REV_STRING:$i:1}
done
# To print the array content
for i in $(seq 0 $ARRAY_UPPER_BOUND); do
echo ${ARRAY[$i]}
done
Also, you can use cut in the loop, like this:
#!/bin/bash
STRING="test string"
REV_STRING=$(rev <<< $STRING)
declare -a ARRAY
ARRAY_LEN=${#REV_STRING}
for i in $(seq 1 $ARRAY_LEN); do
# Here, ${REV_STRING:$i:1} is equivalent to say REV_STRING.charAtPosition(i)
# ARRAY[$i]=${REV_STRING:$i:1}
ARRAY[$i]=$(echo $REV_STRING | cut -c$i)
done
# To print the array content
for i in $(seq 1 $ARRAY_LEN); do
echo ${ARRAY[$i]}
done

Answer 3

If I'm understanding your end goal correctly, here's a quick one-liner (although it uses a couple external utilities, not just bash code):
b=( $( rev <<< "${c}" | sed -e 's/./& /g' ) )
That should take whatever ${c} contains, reverse it, split it into individual characters, and assign each character to an entry in the array b. Note that if the original ${c} contains spaces, those will get lost (i.e. there won't be any array elements containing a space).