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The Bash man page describes use of ${!a} to return the contents of the variable whose name is the contents of a (a level of indirection).

I'd like to know how to return all elements in an array using this, i.e.,

a=(one two three)
echo ${a[*]}


one two three

I would like for:

echo ${!b[*]}

to return the same. Unfortunately, it doesn't, but returns 0 instead.


Given the replies, I now realise that my example was too simple, since of course, something like:


Will achieve exactly what I said I needed.

So, here's what I was trying to do:

LIST_lys=(lys1 lys2)
LIST_diaspar=(diaspar1 diaspar2)

whichone=$1   # 'lys' or 'diaspar'


Of course, carefully reading the Bash man page shows that this won't work as expected because the last line simply returns the indices of the "array" $_LIST (not an array at all).

In any case, the following should do the job (as pointed out):

LIST=($(eval echo \${$_LIST[*]}))

or ... (the route that I went, eventually):

LIST_lys="lys1 lys2"

Assuming, of course, that elements don't contain whitespace.

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3 Answers

up vote 2 down vote accepted

I think the use of indirect reference of bash variable should be treated literally.

Eg. For your original example:

a=(one two three)
echo ${a[*]} # one two three
echo ${!b[*]} # this would not work, because this notation 
              # gives the indices of the variable b which
              # is a string in this case and could be thought
              # as a array that conatins only one element, so
              # we get 0 which means the first element
echo ${!c} # this will do exactly what you want in the first
           # place

For the last real scenario, I believe the code below would do the work.

LIST_lys=(lys1 lys2)
LIST_diaspar=(diaspar1 diaspar2)

whichone=$1   # 'lys' or 'diaspar'

LIST=( "${!_LIST}" ) # Of course for indexed array only 
                     # and not a sparse one

It is better to use notation "${var[@]}" which avoid messing up with the $IFS and parameter expansion. Here is the final code.

LIST_lys=(lys1 lys2)
LIST_diaspar=(diaspar1 diaspar2)

whichone=$1   # 'lys' or 'diaspar'

LIST=( "${!_LIST}" ) # Of course for indexed array only 
                     # and not a sparse one
                     # It is essential to have ${!_LIST} quoted
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You need to copy the elements explicitly. For an indexed array:


For an associative array (note that a is the name of the array variable, not a variable whose value is the name of an array variable):

typeset -A b
for k in "${!a[@]}"; do b[$k]=${a[$k]}; done

If you have the variable name in an array, you can use the element-by-element method with an extra step to retrieve the keys.

eval "keys=(${!$name[@]})"
for k in "${keys[@]}"; do eval "b[\$k]=${$name[\$k]}"; done

(Warning, the code in this post was typed directly in a browser and not tested.)

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You're quite right, but unfortunately, that doesn't solve my problem (my example was too simplistic). I've provided an update to be more clear on my intention. –  Eric Smith Sep 6 '11 at 8:52
@Eric I think in ksh/bash you need eval at that stage. See my edit. –  Gilles Sep 6 '11 at 10:21
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${!b[*]} expands to the indices used in array b.

What you would like has to be done in two steps, so eval will help: eval echo \${$b[*]}. (Note the \ which ensures that the first $ will pass the first step, the variable expansion, and will be only expanded in the second step by eval.)

According to Parameter Expansion ! is both used for indirect expansion ({!a}), Names matching prefix (${!a*}) and List of array keys (${!a[*]}). Because List of array keys has the same syntax as your intended indirect expansion+array element expansion, the later is not supported as is.

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${!a} does expand to the value of the variable whose name is $a. This is rather tersely described in the manual, in the parragraph that begins with “If the first character of parameter is an exclamation point (!), a level of variable indirection is introduced.” –  Gilles Sep 6 '11 at 7:35
Yep - @Gilles is right, but @manatwork, on second reading, I noticed that ${! is kinda ambigious since if it's an array you're dealing with, the behaviour is different. –  Eric Smith Sep 6 '11 at 8:50
@Gilles you are right on that sentence, but sadly it not applies as "The exceptions to this are the expansions of ${!prefix*} and ${!name[@]} described below." But my reply is certainly an ambiguous mess, so I will edit it. –  manatwork Sep 6 '11 at 9:02
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