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

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Hello Sergey

2015-01-29 23:12 GMT+08:00 Sergey Senozhatsky <sergey.senozhatsky@xxxxxxxxx>:
> On (01/29/15 21:48), Ganesh Mahendran wrote:
>> > Admin could reset zram during I/O operation going on so we have
>> > used zram->init_lock as read-side lock in I/O path to prevent
>> > sudden zram meta freeing.
>>
>> 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?

> 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?


>
> 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?

Thanks

>
>         -ss

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