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, e.g. 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.

I don't think there is an alternative. As a side note, none of the programming languages I use can do that natively. Also, you can simplify arr[$i]=$(( ${arr[$i]} + 1 )) to arr[$i]++ — Aserre
– Aserre, Commented Jul 30, 2015 at 7:58
I do not know about your preferred languages, but e.g. FORTRAN does support arr(:) = arr(:) + 1 or arr3(:) = arr1(:) * arr2(:) , so I had the hope that BASH would be capable. Sorry, I cannot reproduce the arr[$i]++ thing. Care to explain? — FelixJN
– FelixJN, Commented Jul 30, 2015 at 8:06
forgot the parenthesis. correct syntax is (( $arr[$i]++ )) — Aserre
– Aserre, Commented Jul 30, 2015 at 8:19
The shell would only be looping in the background anyway if there were a function. So if it bothers you just write your own function to do it. — 123
– 123, Commented Jul 30, 2015 at 9:02
@Plutox sorry, still cannot reproduce, could you explicitly write the for loop start to end? Thanks in advance. — FelixJN
– FelixJN, Commented Jul 30, 2015 at 9:17

j4x · Accepted Answer · 2017-09-15 14:41:06Z

4

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

answered Sep 15, 2017 at 14:41

j4x

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2 Comments

FelixJN Over a year ago

Thank you - in my tests also works without using quotes, e.gg declare -ia new_arr=( ${arr[@]}/%/+$value }. Is there any documentation regarding this? I didn't see it in man bash.

j4x Over a year ago

Yes, there is no direct mention for -i in array documentation neither in any man page nor TLDP or the like. You can check -i, though in "Typing variables: declare or typeset" [tldp.org/LDP/abs/html/declareref.html]. It works for any variable, including arrays.

Zohar81 · Accepted Answer · 2015-07-30 08:18:06Z

-1

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

answered Jul 30, 2015 at 8:18

Zohar81

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2 Comments

Aserre Over a year ago

I don't think this answers the question. OP wants to avoid a for loop altogether

chepner Over a year ago

It also is a pointless use of eval, since the parameter expansions will be expanded before the function is ever defined. return is used to return an exit status, not a computation.

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