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

```
while ! my_command; do
    # do something
done

```

But inside this loop I need to check which exit code `my_command` returned, so I tried this:

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

```

But then the special variable `?` becomes `0` inside the loop body.
The obvious solution is:

```
while true; do
    my_command
    EC=$?
    if [ $EC -eq 0 ]; then
        break
    fi
    some_code_dependent_on_exit_code $EC
done

```

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

Answer: 
In addition to the well-known `while` loop, POSIX provides an `until` loop that eliminates the need to negate the exit status of `my_command`.

```
# To demonstrate
my_command () { read number; return $number; }

until my_command; do
    if [ $? -eq 5 ]; then
        echo "Error was 5"
    else
        echo "Error was not 5"
    fi
    # potentially, other code follows...
done

```
You can get the status of a negated command from the `PIPESTATUS` built-in variable:

```
while ! my_command ; do
    some_code_dependent_on_exit_code "${PIPESTATUS[0]}"
done

```

chepner's solution is better in this case, but `PIPESTATUS` is sometimes useful for similar problems.