How to rename an associative array in Bash?

I need to loop over an associative array and drain the contents of it to a temp array (and perform some update to the value).

The leftover contents of the first array should then be discarded and i want to assign the temp array to the original array variable.

Sudo code:

declare -A MAINARRAY declare -A TEMPARRAY ... populate $MAINARRAY[...] ... while something; do #Drain some values from MAINARRAY to TEMPARRAY $TEMPARRAY["$name"]=(($MAINARRAY["$name"] + $somevalue)) done ... other manipulations to TEMPARRAY ... unset MAINARRAY #discard left over values that had no update declare -A MAINARRAY MAINARRAY=$TEMPARRAY[@] #assign updated TEMPARRAY back to MAINARRAY (ERROR HERE)

See here: stackoverflow.com/questions/3112687/…

– user840803
Jul 12 '11 at 13:28

Yeh, that's what I've got now, was thinking there should be some more efficient way to rename the variable.

– David Parks
Jul 12 '11 at 14:15

The solution by [FlorianFeldhaus] would work if just variable renaming was needed. Infact, it could be a single statement with a little help from sed like this: eval $(declare -p old_name |sed 's/old_name=/new_name=/')

– Samveen
Jun 6 '13 at 23:12

eval $(declare -p old_name |sed 's/old_name=/new_name=/')

8 Answers
8

With associative arrays, I don't believe there's any other method than iterating

for key in "$!TEMPARRAY[@]" # make sure you include the quotes there do MAINARRAY["$key"]="$TEMPARRAY["$key"]" # or: MAINARRAY+=( ["$key"]="$TEMPARRAY["$key"]" ) done

And how do you do that in a function? Where you only get $1 and $2 as variable names (because there's now way to pass the actual variable)?

– cprn
May 22 '17 at 12:39

Copying associative arrays is not directly possible in bash. The best solution probably is, as already been pointed out, to iterate through the array and copy it step by step.

There is another solution which I used to pass variables to functions. You could use the same technique for copying associative arrays:

# declare associative array declare -A assoc_array=(["key1"]="value1" ["key2"]="value2") # convert associative array to string assoc_array_string=$(declare -p assoc_array) # create new associative array from string eval "declare -A new_assoc_array="$assoc_array_string#*= # show array definition declare -p new_assoc_array

The eval seems to be necessary. Why is that?

– Michael Hoffman
Oct 31 '12 at 22:36

eval

The ``eval'' is necessary because it declares the new array. The problem is, that declare -A expect a statement in the assignement and not a string. Thus you need to create a string and evaluate it as a statement.

– Florian Feldhaus
Nov 1 '12 at 10:52

More important is that declare -A is absolutely necessary — without it (for example, inside a function working with arrays from parent environment rather than creating local variables) the whole trick does not work, as it will only create a single item with key 0 and value (["key1"]="value1" ["key2"]="value2" ), because escaping rules seem incompatible between declare and assignment.

– Anton Samsonov
Jun 7 '16 at 17:46

declare -A

0

(["key1"]="value1" ["key2"]="value2" )

declare

This should be selected as the answer. It works, and it's WAY faster that iterating.

– Fmstrat
May 16 '18 at 15:17

if you want to understand better the '#*=', it's well explained here : tldp.org/LDP/abs/html/string-manipulation.html ('Substring Removal' section)

– Gremi64
Sep 12 '18 at 9:02

This one-liner does an associative array copy: MAINARRAY=TEMPARRAY

eval $(typeset -A -p TEMPARRAY|sed 's/ TEMPARRAY=/ MAINARRAY=/')

Following both the suggestions of glenn jackman and ffeldhaus, you can build a function which might become handy:

function cp_hash local original_hash_name="$1" local copy_hash_name="$2" local __copy__=$(declare -p $original_hash_name); eval declare -A __copy__="$__copy__:$(expr index "$__copy__" =)"; for i in "$!__copy__[@]" do eval $copy_hash_name[$i]=$__copy__[$i] done

Usage:

declare -A copy_hash_name cp_hash 'original_hash_name' 'copy_hash_name'

Example:

declare -A hash hash[hello]=world hash[ab]=cd declare -A copy cp_hash 'hash' 'copy' for i in "$!copy[@]" do echo "key : $i | value: $copy[$i]" done

Will output

key : ab | value: cd key : hello | value: world

upvoted even though (eval declare -A __copy__="$__copy__:$(expr index "$__copy__" =)";) explodes for me - still a good starting point.

– keen
Aug 9 '16 at 22:05

eval declare -A __copy__="$__copy__:$(expr index "$__copy__" =)";

If your array contains regular expression like 's/([a-zA-Z]*).*/1/p', this will not work.

– fab
Apr 3 '18 at 19:20

expanding on Luca Borrione's cp_hash - which didn't work for me, and I gave up trying to track down the eval expansion issue - I ran into differences before and after bash 4.2. after 4.2(something) this gets a lot easier... but that's not backwards compatible. See 1 and 2

so my variation tested on 4.1.2(1) and 4.3.46(1):

#!/bin/bash ## bash4 due to associative arrays! function cp_hash() ## REQUIRES you to declare -A $2 in advance. local original_hash_name="$1" local copy_hash_name="$2" # # sadly we have no way to identify if you have already declared it, so bull ahead. # ## store the definition of the old array local __copy__=$(declare -p $original_hash_name) ## rename the array inside the definition __copy__=$__copy__/$original_hash_name=/__copy__= ## for bash 4.2 > we could end here. ## declare -A creates local scope variables by default, so add -g ## this DOES NOT work prior to 4.2, even w/o -g and w/ a declare outside. # __copy__=$__copy__/$original_hash_name=/$copy_hash_name= # eval $__copy__/-A/-g -A ## for bash4 where we can't do -g, then: ## local associative array based on the definition we stored and modified eval $__copy__ ## loop through the local copy, and store it in the declared-outside copy. for i in "$!__copy__[@]" do eval $copy_hash_name[$i]=$__copy__[$i] done declare -A hash hash[hello]=world hash[ab]=cd #not required for 4.2+ if you use -g, neither helps nor hinders declare -A copy cp_hash 'hash' 'copy' echo hash: $hash[@] echo copy: $copy[@] echo "copy result loop" for i in "$!copy[@]" do echo "key : $i | value: $copy[$i]" done

If your array contains regular expression like 's/([a-zA-Z]*).*/1/p', this will not work.

– fab
Apr 3 '18 at 19:19

@fab do you have any suggested improvements that might help that case?

– keen
Apr 11 '18 at 20:38

In my case Florian Feldhaus solution works. You can use also n parameter for declare. declare -n params=HISTORY_PARAM_MAP

– fab
Apr 19 '18 at 14:48

How about this one (Doesn't create a real copy, just a link to source variable):

#!/bin/bash declare -A my_array=(["key1"]="value1" ["key2"]="value2") declare -n arr=my_array arr['LOG_FILE']=/tmp/log.txt echo $arr['key1'] echo $arr['LOG_FILE']

Will print:

value1 /tmp/log.txt

The nameref attribute cannot be applied to array variables. see gnu.org/software/bash/manual/html_node/…

– Feng
Sep 18 '18 at 0:43

You're right but this solution works in my bash version: 4.3.48(1)

– fab
Sep 18 '18 at 19:03

Here is a small Copy-Function for bash-Variables of any kind

- normal scalar variables

- indexed arrays

- associative arrays

### Function vcp -VariableCoPy- # $1 Name of existing Source-Variable # $2 Name for the Copy-Target vcp() local var=$(declare -p $1) var=$var/declare /declare -g eval "$var/$1=/$2="

Usage, Examples:

# declarations var=" 345 89 " ind_array=(Betty " 345 89 ") declare -A asso_array=([one]=Harry [two]=Betty [some_signs]=" +*.<$~,'/ ") # produce the copy vcp var varcopy vcp ind_array ind_array_copied vcp asso_array asso_array_2 # now you can check the equality between original and copy with commands like # declare -p <name>

The results

MAINARRAY=( "$TEMPARRAY[@]" )

That's what I thought, but I get an error on that line: "must use subscript when assigning associative array"

– David Parks
Jul 12 '11 at 6:27

This won't work for associative arrays

– Luca Borrione
Aug 27 '12 at 21:17

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