[Solved] Append elements to an array in bash

I tried the += operator to append an array in bash but do not know why it did not work

#!/bin/bash


i=0
args=()
while [ $i -lt 5 ]; do

    args+=("${i}")
    echo "${args}"
    let i=i+1

done

expected results

0
0 1
0 1 2
0 1 2 3
0 1 2 3 4

actual results

0
0
0
0
0

Any help would be appreciated.

Enquirer: Steve J

||

Solution #1:

It did work, but you’re only echoing the first element of the array. Use this instead:

echo "${args[@]}"

Bash’s syntax for arrays is confusing. Use ${args[@]} to get all the elements of the array. Using ${args} is equivalent to ${args[0]}, which gets the first element (at index 0).

See ShellCheck: Expanding an array without an index only gives the first element.

Also btw you can simplify let i=i+1 to ((i++)), but it’s even simpler to use a C-style for loop. And also you don’t need to define args before adding to it.

So:

#!/bin/bash
for ((i=0; i<5; ++i)); do
    args+=($i)
    echo "${args[@]}"
done
Respondent: wjandrea

The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 .

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