Shell Programming and Scripting

Input :-
array1=(2 6 8 9 10 12)
array2=(5 4 6 8 12 14)
Output :-
array3=(7 1 0 1 4 1 7 2 2 2 6)

Explanation :- 5+2 = 7   (less then 10 so no action required)
                    6+4 = 10 (greater then 9 so it gets split into 1,0)

#!/bin/bash
welcome()
{
echo "Welcome to first array progg."
}
main_logic()
{
 declare -a arr1=(5 7 9 12 14 19)
 declare -a arr2=(7 4 3 8 12 17)
 declare -a arr3=()
 sum=0
 len=${#arr1[@]}
 for ((i=0; i<$len; i++))
 do
     sum=$(( ${arr1[$i]}+${arr2[$i]} )) 
     #echo $sum
     arr3=("${arr3[@]}" "$sum")
 done 
echo ${arr3[@]}
}
##Entry Point of the Progg.
welcome
main_logic

     sum=$(( ${arr1[$i]}+${arr2[$i]} ))       #echo $sum
     if [ $sum -gt 9 ]
     then
       arr3=("${arr3[@]}" "$sum/10") #sum divided by 10
       arr3=("${arr3[@]}" "$sum%10") #mod 10 sum
     else
       arr3=("${arr3[@}" "$sum")
      fi

for i in ${!array1[@]}; do printf "%s" $((array1[i] + array2[i])); done | od -An -c
   7   1   0   1   4   1   7   2   2   2   6

od -An -c