Bash array declared in a function is not available outside the function

Question

on bash (v4.3.11) terminal type this:

function FUNCtst() { declare -A astr; astr=([a]="1k" [b]="2k" ); declare -p astr; };FUNCtst;declare -p astr

(same thing below, just to be easier to read here)

function FUNCtst() { 
  declare -A astr; 
  astr=([a]="1k" [b]="2k" ); 
  declare -p astr; 
};
FUNCtst;
declare -p astr

will output this (outside of the function the array looses its value, why?)

declare -A astr='([a]="1k" [b]="2k" )'
bash: declare: astr: not found

I was expecting it output this:

declare -A astr='([a]="1k" [b]="2k" )'
declare -A astr='([a]="1k" [b]="2k" )'

how to make it work?

ilkkachu · Accepted Answer · 2023-02-01 10:30:27Z

12

From the man page:

When used in a function, declare makes each name local, as with the local command, unless the -g option is used.

Example:

FUNCtst() { 
    declare -gA astr
    astr=([a]="1k" [b]="2k" )
    declare -p astr
}
FUNCtst
declare -p astr

prints

declare -A astr=([a]="1k" [b]="2k" )
declare -A astr=([a]="1k" [b]="2k" )

ilkkachu

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answered Jun 12, 2014 at 7:23

Nykakin

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