Cannot get bash variable changed in sourced script after pipe?

Question

I constructed a simple example to illustrate the problem:

caller.sh:

#!/usr/bin/env bash
state="loaded"
source "sh/callee.sh" 2>&1
echo "$state"

callee.sh:

#!/usr/bin/env bash
state="integrated"

when I run caller.sh, it gives the result I want:

integrated

But if I add a pipe after the source command:

caller.sh:

#!/usr/bin/env bash
state="loaded"
source "sh/callee.sh" 2>&1 | cat
echo "$state"

The result becomes:

loaded

Question: How can I preserve/retrieve the changed value of $state in caller.sh?

pipe-line forks a sub-shell, the value is lost, once the sub shell terminates — Inian
– Inian, Commented Mar 2, 2017 at 18:05

chepner · Accepted Answer · 2017-03-02 18:03:52Z

3

You can't use a pipe. The source command runs in a subshell that exits after the pipe completes, so state is never set in the shell that calls echo "$state". You can use process substitution to "invert" the pipeline:

state="loaded"
source "sh/callee.sh" 2>&1 > >(cat)
echo "$state"

Now source executes in the same shell.

answered Mar 2, 2017 at 18:03

chepner

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3 Comments

Inian Over a year ago

But why is cat needed at all here?

Jasper_Li Over a year ago

what does the first > do?

Jasper_Li Over a year ago

cat is just something I came up with in this example. In my program I want to direct stdin and stderr to a log file

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