Environment variable passed to bash script is empty inside function

Question

I have bash script: myscript.sh. Here is the content:

#!/bin/bash

function my_func {
    echo "my_func start"
    echo $MY_VARIABLE  # THIS PRINTS NOTHING
    echo "my_func end"
}

function func2 {
}

echo "after my_func"
echo $MY_VARIABLE # this prints "YES"

I want to call only function my_func from command line with environment variable "MY_VARIABLE". I launch script with this command:

MY_VARIABLE=YES source myscript.sh; my_func

Here is the output:

after my_func
YES
my_func start

my_func end

Variable $MY_VARIABLE inside my_func() is empty. I want that variable $MY_VARIABLE in my_func() were equal to "YES".

What documentation leads you to the expectation that your original code would work as described? — Charles Duffy
– Charles Duffy, Commented Nov 22, 2019 at 14:46

chepner · Accepted Answer · 2019-11-22 14:31:07Z

3

You are only setting MY_VARIABLE for the environment in which source runs. Once that complete, my_func runs in your original environment. If you want my_func to have access, use

export MY_VARIALBE=YES; source my script.sh; my_func

bash doesn't have closures; when you define my_func, the shell isn't using the current value of MY_VARIABLE in the definition. It just creates a function that, when called, looks in the calling scope for the value.

answered Nov 22, 2019 at 14:31

chepner

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