Why does the array length function in Bash output a 1?

Question

i wonder why i get length 1 returned every time even though $test is empty, even though grep can't find anything.

#!/bin/bash

set -Eeuo pipefail
set -x

declare -a test

test=$(echo "$outputOfOtherCommand" | grep -i '>f\+\+\+\+\+\+\+\+\+' | cut -d' ' -f2) || true
echo "${#test[@]}"

for t in $test
do
    echo $t
done

The $outputOfOtherCommand includes the output of Rsync. The test_new_file was only created for clarity.

➜  ~ ./script.sh
.d..t...... ./
>f+++++++++ test_new_file

Number of files: 10 (reg: 9, dir: 1)
Number of created files: 1 (reg: 1)
Number of deleted files: 0
Number of regular files transferred: 1
Total file size: 0 bytes
Total transferred file size: 0 bytes
Literal data: 0 bytes
Matched data: 0 bytes
File list size: 0
File list generation time: 0.001 seconds
File list transfer time: 0.000 seconds
Total bytes sent: 280
Total bytes received: 38

The only thing I can explain is that an empty entry was created at array position 0. But i don't know. Maybe you can help me.

Your for loop does not make any sense. The loop body will be executed exactly once, with t having the literal string test. — user1934428
– user1934428, Commented Feb 27, 2020 at 11:51

user1934428 · Accepted Answer · 2020-02-27 11:47:20Z

3

test is an array only because you declared it using -a. An assignment of the form

test=$(.....)

assigns a single string to test, which ends up in the array element with index 0. Hence the length of your array is 1.

You can verify this by

declare -a arr
arr=$(echo a b c)
echo ${arr[0]}

answered Feb 27, 2020 at 11:47

user1934428

22.8k9 gold badges57 silver badges109 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

rosaLux161 Over a year ago

I see, then my question goes in a different direction. If I remove the "declare", I can go through the "array" with a for loop. Why is this possible when there is only an array with one element at position 0?

user1934428 Over a year ago

If you remove the 'declare', test is not an array anymore, but a string, but since you did not quote $test in your loop, word splitting occurs on white space and t will take the the values of the individual words. Whether this is or is not what you want, I don't know. You need to specify, what you want the array elements to contain.

Collectives™ on Stack Overflow

Why does the array length function in Bash output a 1?

1 Answer 1

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related