Re: [PATCH v1 2/2] zram: remove init_lock in zram_make_request

[Sergey Senozhatsky]

On (01/30/15 15:52), Ganesh Mahendran wrote:
> >> When I/O operation is running, that means the /dev/zram0 is
> >> mounted or swaped on. Then the device could not be reset by
> >> below code:
> >>
> >>     /* Do not reset an active device! */
> >>     if (bdev->bd_holders) {
> >>         ret = -EBUSY;
> >>         goto out;
> >>     }
> >>
> >> So the zram->init_lock in I/O path is to check whether the device
> >> has been initialized(echo xxx > /sys/block/zram/disk_size).
> >>
> 
> Thanks for your explanation.
> 
> >
> > for mounted device (w/fs), we see initial (well, it goes up and down
> 
> What does "w/" mean?

'with fs'

> > many times while we create device, but this is not interesting here)
> > ->bd_holders increment in:
> >   vfs_kern_mount -> mount_bdev -> blkdev_get_by_path -> blkdev_get
> >
> > and it goes to zero in:
> >   cleanup_mnt -> deactivate_super -> kill_block_super -> blkdev_put
> >
> >
> > after umount we still have init device. so, *theoretically*, we
> > can see something like
> >
> >         CPU0                            CPU1
> > umount
> > reset_store
> > bdev->bd_holders == 0                   mount
> > ...                                     zram_make_request()
> > zram_reset_device()
> 
> In this example, the data stored in zram will be corrupted.
> Since CPU0 will free meta while CPU1 is using.
> right?
> 

with out ->init_lock protection in this case we have 'free' vs. 'use' race.

> 
> >
> > w/o zram->init_lock in both zram_reset_device() and zram_make_request()
> > one of CPUs will be a bit sad.
> what does "w/o" mean?

'with out'


	-ss
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