Bash declare associative array through function gives non-associative array

Question

I tried to declare an associative array through a function, and I found that the associative array becomes a plain array. The test code goes like this:

bash-5.0$ cat test.sh
#!/bin/bash

createArr(){ declare -A "$1"; }

# create array using `createArr'
name=array1
createArr $name
array1[1]=1
echo "${!array1[@]}"
array1[a]=1
echo "${!array1[@]}"
declare -p array1

# create array directly
name=array2
declare -A $name
array2[1]=1
echo "${!array2[@]}"
array2[a]=1
echo "${!array2[@]}"
declare -p array2

And executing the code gives the following result:

bash-5.0$ ./test.sh
1
0 1
declare -a array1=([0]="1" [1]="1")
1
1 a
declare -A array2=([1]="1" [a]="1" )

I'd like to know why the results are different, thanks! (My bash version is 5.0.11)

oguz ismail · Accepted Answer · 2020-01-30 12:17:27Z

1

help declare says

When used in a function, declare makes NAMEs local, as with the local command.

That means outside of createArr, array1 is not declared. And an assignment like array1[1]=1 where array1 is unset, implies that array1 is a regular, indexed array.

Using the -g flag in your function fixes this though

createArr() { declare -gA "$1"; }

answered Jan 30, 2020 at 12:17

oguz ismail

51.8k16 gold badges64 silver badges83 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Bash declare associative array through function gives non-associative array

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related