Bash background process still blocking script flow in subshell

I am trying to do something that involves scripts calling other scripts in subshells to capture their output.

One of the scripts needs to have a side effect of starting a background process. This all works when executed directly, but it blocks when called within a subshell.

As a self-contained example consider the following 2 scripts:

test1.sh

#!/bin/bash
echo $(./test2.sh)

test2.sh

#!/bin/bash
(yes > /dev/null ; echo 'yes killed') &
echo success

When I run test2.sh by itself, I get the expected result of "success" on the terminal, and yes running in the background. Killing yes prints "yes killed" to the terminal as expected.
When I run test1.sh I expect to get the same behavior, but what actually happens is that the terminal hangs until I kill yes after which "success yes killed" is printed to the terminal.

What do I change to these scripts so that I can get the same behavior from calling either one?

The premise here is that the subshell evaluation in test1.sh will actually be stored in a variable for later use. The background process started in test2.sh should live past the execution of either script.

Solution 1:

Command substitution $(...) turns output into arguments. It first needs the whole output, because it's not possibly to supply arguments dynamically one by one.

Solution 2:

As @choroba and @GordonDavisson suggested, command substitution $( ... ) will not return until all of its stdout has disconnected ^[1].

The trick here is that even if you redirect all stdout of the commands, the subshell itself eg. ( ... ) still has its stdout attached. This means that the following will not work:

#!/bin/bash
(yes > /dev/null ; echo 'yes killed' > /dev/null) &
echo success

But this will:

#!/bin/bash
(yes ; echo 'yes killed' ) > /dev/null &
echo success

Note: You will no longer get any output to the terminal or stdout like this, but for my purpose this is not an issue. If necessary you could always redirect to a file or something if the output is needed

^[1] See also https://stackoverflow.com/questions/16874043/bash-command-substitution-forcing-process-to-foreground

Solution 3:

I think bash's disown is the answer. See help disown.

Use it like this:

yes &>/dev/null </dev/zero &
disown -h $!

Redirect the output or closing terminal would result in SIGHUP signal or a broken pipe for output (SIGPIPE) eventually blocking or killing the program.

If you care about output redirect it to a file. Or use nohup like this: