Arithmetic operation on all elements in BASH array

Question

Say we have a BASH array with integers:

declare -a arr=( 1 2 3 )

and I want to do an arithmetic operation on each element, eg add 1. Is there an altenative to a for loop:

for (( i=0 ; i<=${#arr[@]}-1 ; i++ )) ; do
    arr[$i]=$(( ${arr[$i]} + 1 ))
done

I have tried a few options:

arr=$(( ${arr[@]} + 1 ))

fails, while

arr=$(( $arr + 1 ))

results in

echo ${arr[@]}
2 2 3

thus only the first (zeroth) element being changed.

I read about awk solutions, but would like to know if BASH arithmetics support such batch operations on each array element.

Answer 1

I know your question is not fresh new but you can accomplish what you want by declaring your array as integer and then applying a substitution:

declare -ia arr=( 1 2 3 )
value=1

declare -ia 'arr_added=( "${arr[@]/%/+$value}" )'
echo "arr_added: ${arr_added[*]}"

value=42
declare -ia 'arr_added=( "${arr[@]/%/+$value}" )'
echo "arr_added: ${arr_added[*]}"

It outputs:

arr_added: 2 3 4
arr_added: 43 44 45

You can perform other math operations as well:

value=3
declare -ia 'arr_multd=( "${arr[@]/%/*$value}" )'
echo "arr_multd: ${arr_multd[*]}"

Outputs:

arr_multd: 3 6 9

Answer 2

you can use eval to have the feeling of lambda function (not sure about the syntax but this should be the main idea ) :

eval "function add1 { x=$1; y=$((x+1)) ; return $y; }"
for (( i=0 ; i<=${#arr[@]}-1 ; i++ )) ; do
    add1 ${arr[i]}
done