Bash function local array variable is not accessible in called place

Question

There is a 2 bash script file. First file b.sh is mentioned below.

#!/bin/bash
declare -a arr1=()
func() {
    var_a=12
    arr1[0]=20
    arr1[1]=30
    declare -a arr2=()
    arr2[0]=40
    arr2[1]=50
}

Second file a.sh is mentioned below.

#!/bin/bash
source b.sh
func
echo $var_a
echo ${arr1[1]}
echo ${arr2[1]}

Output is

12
30

My doubt is, why the local array variable (arr2) in func is not accessible in a.sh. But the local variable var_a is accessible.

BTW, this question could be simplified. There's no need to have two separate scripts and a source command to reproduce this problem -- it could easily be created with just one. — Charles Duffy
– Charles Duffy, Commented Jan 25, 2018 at 21:12
Ya true, It is a simplified version of the issue faced in an another script based test framework, which constains 2 script file. — rashok
– rashok, Commented Mar 4, 2018 at 17:59

Barmar · Accepted Answer · 2018-01-25 20:48:54Z

5

arr2 is a local variable because it was created using declare. As stated in the Bash Manual:

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

Since you didn't create var_a with declare, the assignment creates a global variable, so it's accessible outside the funtion. If you'd written:

declare var_a=12

or

local var_a=12

inside the function then it would have been local.

answered Jan 25, 2018 at 20:31

Barmar

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