how to use looping variables to populate an array

Question

when I populate an array using explicit range value it get populated correctly using the code below:

arr=()
for i in {0..10}
do
    arr+=( 1 )
done
echo ${arr[@]}

and I got the following output (nothing surprising)

However, when storing the range in a variable and calling it as in the code below it only shows one element:

range=10
arr=()
for i in {0..$range}
do
    arr+=( 1 )
done
echo ${arr[@]}

I got the following output

Make it `for i in $(seq 0 $range)` as explained in the linked question. — 0stone0, Jan 05 '22 at 12:21
Read the chapter [brace expansion](https://www.gnu.org/software/bash/manual/bash.html#Brace-Expansion) in the Bash manual. — ceving, Jan 05 '22 at 12:34
ugly solution but works: `range=10; eval "arr=( \$(printf '1%.0s ' {0..$range}) )"` — Fravadona, Jan 05 '22 at 12:50

score 1 · Answer 1 · answered Jan 05 '22 at 12:22

1

Variable expansion does not occur before brace expansion so you only get a literal {0..1} as a single iterated argument.

Use the old school method instead:

for (( i = 0; i <= range; ++i )); do

answered Jan 05 '22 at 12:22

konsolebox

Just out of curiosity, why did you decide to answer this instead of marking it as a duplicate since your explanation is also mentioned in the question I linked. – 0stone0 Jan 05 '22 at 12:27
@0stone0 I didn't look at it before I answered and it's trivial enough that I can simply give a solution specific to the example in the question. – konsolebox Jan 05 '22 at 12:30

1 Answers1