Why doesn't Bash overwrite my array with a new one?

Question

I stripped my use case to these 3 lines (plus the output):

A=(foo bar)
A=$(echo "spam egg")
echo ${A[@]}
spam egg bar

It creates an array with two element written by hand. Then some time later I want to replace my array with the output from a command line tool (e.g. ls *.vhd). Instead of replacing the array, Bash only replaces the first element, so bar gets "attached" at the end.

If I use another handwritten array, then I can't reproduce this behavior.

A=(foo bar)
A=(spam egg)
echo ${A[@]}
spam egg

So I suspect it's related to using $(). How can I solve my problem?

`A=$(echo "spam egg")` is equivalent to `A[0]=$(echo "spam egg")`. — chepner, Nov 10 '16 at 18:05

Joshua Goldberg · Accepted Answer · 2016-11-10T19:30:05.557

5

Use parentheses in the assignment, so that it is treated as an array, rather than one long string with spaces:

A=$(echo "spam egg")
echo ${A[0]}

-> spam egg

A=($(echo "spam egg"))
echo ${A[0]}

-> spam

(The parentheses for $() that tell bash to run a command don't also count for telling it to collect the result as an array.)

edited Nov 10 '16 at 19:30

answered Nov 10 '16 at 17:48

Joshua Goldberg

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Indexing does not work on regular strings. `${A[1]}` would not be `s` in your first example. – chepner Nov 10 '16 at 18:07
Ah, sorry! I wrote the examples in `zsh`, though the main point of needing parens still stands. Change to: `A=($(echo "spam egg")); echo ${A[0]} -> spam` for `bash` – Joshua Goldberg Nov 10 '16 at 18:16
1

It would be nice to update your answer to show relevant `bash` code. – chepner Nov 10 '16 at 18:20
Thanks. I've spent half of the day debugging this ... – VladSavitsky Aug 03 '20 at 14:44

Why doesn't Bash overwrite my array with a new one?

1 Answers1