In bash how to modify an external variable from within a function that is piped?

Question

I have two scripts that call each other. I need to modify a given variable in both. It doesn't work.

A.sh :

#! /bin/sh
funcA() {
    echo "var B in funcA (1) $__varB"
    __varB="xxx"
    echo "var B in funcA (2) $__varB"
}
. B.sh

B.sh :

! /bin/sh
__varB="asdf"
funcA | tee -a out.txt 2>&1 #if no pipe then it works
echo "var B in B.sh $__varB"

Execution

./A.sh
var B in funcA (1) asdf
var B in funcA (2) xxx
var B in B.sh asdf

If in B I do not pipe funcA into tee then it works as expected and __varB is modified. My problem is that in the real case I can't modify B.sh.

What can I do in A.sh so that I modify __varB ?

choroba · Accepted Answer · 2013-04-30 09:18:10Z

1

The redirected function is apparently run in a subshell. To avoid it, you have use the process substitution:

funcA > >( tee -a out.txt 2>&1 )

This change must appear in B.sh. If you cannot modify it, you are doomed.

answered Apr 30, 2013 at 9:18

choroba

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