Confused with `bash -c` command - does not work with global variables

Solution 1:

Your double quoted string is being expanded BEFORE the commands in the string are being interpreted by bash -c so that $? is likely a previous command's exit code.

You can see this with an example where we also use the -x flag to see what's happening under the hood:

$ set -x
$ bash -c "v=5; echo $v"
+ bash -c 'v=3; echo '

You can see that there is no value of $v to echo when execution hits that line as $v was already replaced before the command was interpreted by bash -c.

For further expirementation; to get nearly the same output as your issue, do the following:

1. Run cat with no parameters and ctrl+c to interupt.
2. Run bash -c "echo 'hi'; echo $?"

You should get hi and 130 as the output, which was the exit code of your sigint'd cat command. (note that it's not the exit 0 of the echo "hi" as you might expect).

Thankfully for your command, you can just swap out your quotes to avoid this issue:

/bin/bash -c '(while true; do \
sleep 1; \
ping -c 1 "www.google.com"; \
if [ $? -eq 0 ]; then break; fi; \
done); \
exit 0'