Do not fail on qbootctl -m if device is not A/B

[Gelbpunkt]

It is common and makes sense to ship qbootctl -m as some kind of init service that runs at boot. However, not all devices are A/B partitioned and qbootctl -m will currently return an exit code of 1 on these devices, making the init service marked as failed. We should handle this case more gracefully and exit with exit code 0 if there was nothing to do.

[kcxt]

heya, thanks for this. I'm not sure how we get here on a non-A/B device since getCurrentSlot() should return an error (https://github.com/linux-msm/qbootctl/blob/main/qbootctl.c#L147)

I think we should fix whatever is wrong there. I'd also then suggest we add a --ignore-a-only flag to make qbootctl exit with 0 on A-only devices, since it might be used in an application where we /know/ the hardware is A/B and qbootctl thinking otherwise is something we'd want to know about (like some embedded device for example).

[Gelbpunkt]

heya, thanks for this. I'm not sure how we get here on a non-A/B device since getCurrentSlot() should return an error (https://github.com/linux-msm/qbootctl/blob/main/qbootctl.c#L147)

We get here because goofy me wrote this patch for an older version of qbootctl :)

I'd also then suggest we add a --ignore-a-only flag to make qbootctl exit with 0 on A-only devices

I'll update the PR to do do in the next days, sorry for taking so long to get back to this.