Bash array declaration and appendation

Question

I'm trying to declare and append to an array in a bash script, after searching i resulted in this code.

list=()
list+="string"

But when i echo this out it results in nothing. I have also tried appending to the array like this

list[$[${#list[@]}+1]]="string"

I don't understand why this is not working, anyone have any suggestions?

EDIT: The problem is list is appended to inside a while loop.

list=()

git ls-remote origin 'refs/heads/*' | while read sha ref; do
    list[${#list[@]}+1]="$ref"
done

declare -p list

see stackoverflow.com/q/16854280/1126841

I have tried ' echo "${list}" ', ' echo "$list" ' and ' echo $list ' — A.Jac, Feb 03 '17 at 11:47
@A.Jac Cannot reproduce your first error; `$list` and `${list[0]}` are effectively equivalent. `list+="string"` won't add `string` to the array, but it will append `string` to the end of the first element of the array (creating said element if necessary). — chepner, Feb 03 '17 at 12:10
`$[...]` is obsolete (use `$((...))` instead) and unnecessary; the index of a regular array is automatically evaluated in an arithmetic context, so `list[${#list[@]}+1]="string"` would be sufficient. — chepner, Feb 03 '17 at 12:11
By the way, you can see what bash thinks about that variable by running `declare -p list`. If it is an array will be printed like `declare -a list=.` — George Vasiliou, Feb 03 '17 at 12:14
ok, so i tried `list[${#list[@]}+1]="string"` and `declare -p list` it results in `declare -a list='()'` — A.Jac, Feb 03 '17 at 12:22
Not sure what is going on: I get `declare -a list='([1]="string")'`. — chepner, Feb 03 '17 at 12:30
The only thing i can think of is incompatibility between bash and zsh (which im using). But shouldn't be an issue since i have shebang added — A.Jac, Feb 03 '17 at 12:38
The problem seems to be my lack of shell knowledge, since the appending is happening inside a while loop it is not remembered after loop is done, need to find a workaround for it. — A.Jac, Feb 03 '17 at 12:57
The loop wouldn't happen to be in a pipeline, would it? This is a *completely* different problem, and this is why your question needs to provide an example that actually reproduces your problem. — chepner, Feb 03 '17 at 13:34

score 7 · Answer 1 · edited Feb 03 '17 at 13:33

7

You can append new string to your array like this:

#!/bin/bash

mylist=("number one")

#append "number two" to array    
mylist=("${mylist[@]}" "number two")

# print each string in a loop
for mystr in "${mylist[@]}"; do echo "$mystr"; done

For more information you can check http://wiki.bash-hackers.org/syntax/arrays

edited Feb 03 '17 at 13:33

chepner

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answered Feb 03 '17 at 12:43

Ali Okan Yüksel

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The OP is already using the correct operator to append to an array; this won't make a difference. – chepner Feb 03 '17 at 13:35

Jdamian · Answer 2 · 2017-07-08T11:21:54.007

Ali Okan Yüksel has written down an answer for the first method you mentioned about adding items in an array.

list+=( newitem  another_new_item ··· )

The right way of the second method you mentioned is:

list[${#list[@]}]="string"

Assuming that a non-sparse array has N items and because bash array indexes starts from 0, the last item in the array is N-1th. Therefore the next item must be added in the N position (${#list[@]}); not necessarily in N+1 as you wrote.

Instead, if a sparse array is used, it is very useful the bash parameter expansion which provides the indexes of the array:

${!list[@]}

For instance,

$ list[0]=3
$ list[12]=32
$ echo ${#list[@]}
2
$ echo ${!list[@]}
0 12

Bash array declaration and appendation

2 Answers2