BASH: Capturing both the output and the exit code of a process

Of course I found an answer to this in a collection of Stack Overflow questions, but to make things easier for anybody who might stumble into this post (and mainly for Future Me), here’s the answer.

Seriously, don’t do this.

Basically I have a command that sometimes produces an error, but usually just re-running it produces the correct output. When it encounters an error, it dutifully sets its error code to something other than 0 (success). But how to I capture both the command’s output and its exit code? This way.

exit_code=$? # This HAS to be exeuted right after the command above

So I made a little wrapper script that repeatedly calls the first one until it gets an answer, with a set maximum number of retries to avoid infinite loops.

max_retries=10 # Change this appropriately
# The "2> /dev/null" silences stderr redirecting it to /dev/null
# The command must be first executed outside of the while loop:
# bash does not have a do...while construct
output=$(myCommandWhichSometimesDoesntWork 2> /dev/null)
while [[ $? -gt 0 ]]
        if [[ retries -gt max_retries ]]
                exit 1
        output=$(myCommandWhichSometimesDoesntWork 2> /dev/null)
echo $output