Bash loop until a certain command stops failing
<p>I would like to write a loop in bash which executes until a certain command stops failing (returning non-zero exit code), like so:</p>

<pre><code>while ! my_command; do
    # do something
done
</code></pre>

<p>But inside this loop I need to check which exit code <code>my_command</code> returned, so I tried this:</p>

<pre><code>while ! my_command; do
    if [ $? -eq 5 ]; then
        echo "Error was 5"
    else
        echo "Error was not 5"
    fi
    # potentially, other code follows...
done
</code></pre>

<p>But then the special variable <code>?</code> becomes <code>0</code> inside the loop body.
The obvious solution is:</p>

<pre><code>while true; do
    my_command
    EC=$?
    if [ $EC -eq 0 ]; then
        break
    fi
    some_code_dependent_on_exit_code $EC
done
</code></pre>

<p>How can I check the exit code of <code>my_command</code> (called in loop header) inside loop body without rewriting this example using a <code>while true</code> loop with a break condition as shown above?</p>