copy array with array name inside string in bash

Question

In bash 4.3 script, I have variables:

 environment="local" config_local=("a" "b" "b" "d") copy_from="config_${environment}" 

I want to copy values from array with name stored in copy_from to another array named config.

I tried this and some variations:

 config=${!copy_from} echo "${config[@]}" config=${!copy_from[@]} echo "${config[@]}" config=("${!copy_from}") echo "${config[@]}" config=("${!copy_from[@]}") echo "${config[@]}" 

but I can't copy whole array to config and get only 0 or first element only:

 a 0 a 0 

How can I do it in bash?

Answer 1

There are ways to do this with eval, but they are insecure. There are some ways to do this with on bash 2.05b+:

$ foo=(a b c) $ name=foo $ temp_indirect=${name}[@] $ printf '<%s>\n' "${!temp_indirect}" <a> <b> <c> 

This is pretty ugly because it relies on (potentially undocumented?) implementation details of the bash parser, so use it at your own risk.

In general, if you want complex data structures, I'd suggest another language. In many cases there are workarounds to make them possible in bash, but the syntax is abstruse because the language is not optimised for them (it is mostly optimised for IPC and manipulation of files).