Variable from another function is not visible?

Question

I wrote the following script:

bar() {
        echo $(foo) >&2
        echo "bar" >&2
        echo $VARIABLE >&2
}

foo() {
        echo "foo" >&2
        VARIABLE="test"
        echo $VARIABLE >&2
}
bar

The output is

foo
test

bar

But if I wrote this:

bar() {
        foo
        echo "bar" >&2
        echo $VARIABLE >&2
}

foo() {
        echo "foo" >&2
        VARIABLE="test"
        echo $VARIABLE >&2
}
bar

It prints what I need:

foo
test
bar
test

The question is that I need to send foo's result to &2 yet I need the VARIABLE value that was set when calling to foo. How to do this in bash?

Answer 1

Bash runs $() in a subshell, so the assignment to VARIABLE inside $(foo) is thrown away when the $() ends. That is why the variable is not preserved.

Standard input/output/error are always available unless you change them. Therefore, you should be able to do:

bar() {
        foo >&2             # <-- redirect foo's output
        echo "bar" >&2
        echo $VARIABLE >&2
}

Explanation

You can often replace uses of echo $(something) and similar constructions, such as something | cat. In your case:

• foo produces output to stdout
• $() copies from stdout to the command line
• echo copies from the command line to stdout (!!)
• >&2 redirects from stdout to stderr

The echo $() sequence has the net effect of copying from stdout to stdout :) , so you can just eliminate it.^*

^{* But in other cases, you might want echo $(), e.g., if you are using $() for its side-effects, or if you are deliberately invoking a subshell to protect yourself from variables being changed by what you're executing. However, I think those cases are probably not common.}