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From Максим Степачёв <maksim.stepac...@gmail.com>
Subject Re: Does KeystoreEncryptionSpi have problem with size calculation?
Date Fri, 01 Feb 2019 09:54:30 GMT
Nikolay,
Could you help again? I looked at the WalRecordCacheGroupAware, and found
that the DataRecord doesn't implement it.
This record type doesn't encrypt, does it? Is it a problem?

чт, 31 янв. 2019 г. в 18:23, Максим Степачёв <maksim.stepachev@gmail.com>:

> Nikolay,
> It's my mistake. You're right.
> Thanks for answering.
>
> чт, 31 янв. 2019 г. в 17:50, Nikolay Izhikov <nizhikov@apache.org>:
>
>> Maxim,
>>
>> > I suppose the AES algorithm work with blocks by 16 bytes for encryption
>> data + 2  bytes for padding in AES_WITH_PADDING mode
>>
>> Why do you make this conclusion?
>>
>> Actually, in AES_WITH_PADDING mode we should add 2 more *BLOCK*: for IV
>> and
>> for padding info.
>> You can find sanity check in KeystoreEncryptionSpiSelfTest.
>> Please, take a look into KeystoreEncryptionSpiSelfTest#testEncryptDecrypt
>> which check the case you described.
>>
>> чт, 31 янв. 2019 г. в 17:08, Максим Степачёв <maksim.stepachev@gmail.com
>> >:
>> >
>> > Nikolay,
>> > I see. We should close IGNITE-11129 as invalid. I'm going to ask the
>> > reporter, what he means.
>> >
>> > > Please, explain, why do you think so?
>> > I answered to Dmitriy Pavlov with explaining.
>> >
>> > In short, your method encryptedSize(int dataSize) returns the *"*wrong
>> > result*"* when I call it with dataSize = 20.
>> > I suppose the AES algorithm work with blocks by 16 bytes for
>> > encryption data + 2  bytes for padding in AES_WITH_PADDING mode.
>> > The current implementation works for dataSize = 20: ( (20 / 16) + 2 ) *
>> > 16.  " 20 / 16 = 1." , Are we lost one block for 4 bytes? Or is it fine?
>> >
>> >
>> >
>> >
>> >
>> >
>> > чт, 31 янв. 2019 г. в 16:15, Nikolay Izhikov <nizhikov@apache.org>:
>> >
>> > > Hello, Maxim.
>> > >
>> > > > IGNITE-11129
>> > >
>> > > Do we have reproducer for this ticket?
>> > > WalRecord will be encrypted only if record class
>> > > implements WalRecordCacheGroupAwarei.e it contains some cache data
>> that
>> > > should be protected with encryption.
>> > > Please, look into private boolean needEncryption(WALRecord rec).
>> > > SwitchSegmentRecord does not implement WalRecordCacheGroupAware, so
>> not
>> > > encryption would be applied for this types of records.
>> > >
>> > > I think we should close IGNITE-11129 as invalid.
>> > >
>> > > > Should we use this code:
>> > >
>> > > Please, explain, why do you think so?
>> > > Do you find some bug?
>> > > Can you send a reproducer?
>> > >
>> > > You can find details of encrypted data size calculation in AES
>> algorithm
>> > > description.
>> > >
>> > > чт, 31 янв. 2019 г. в 15:24, Максим Степачёв <
>> maksim.stepachev@gmail.com
>> >:
>> > >
>> > > > Dmitriy Pavlov,
>> > > >
>> > > > Your statement about page size is true. But our case about plainSize
>> of
>> > > > serialized record in bytes. This method calculates it:
>> > > plainSize(WALRecord
>> > > > record). For example, if you look in this method, you will see
>> > > > DATA_PAGE_UPDATE_RECORD section. A size of record calculates a sum
>> of: 4
>> > > +
>> > > > 8 + 2 + 4 +  uRec.payload().length where a length is an arbitrary
>> number.
>> > > >
>> > > > чт, 31 янв. 2019 г. в 14:54, Dmitriy Pavlov <dpavlov@apache.org>:
>> > > >
>> > > >> Hi Maxim, why do you think that data size can be divided to cipher
>> block
>> > > >> size with 0 remainder.
>> > > >>
>> > > >> I used to think that page size 4096 is always divisible by a usual
>> block
>> > > >> cipher block size, e.g 32, 16 or 8 bytes
>> > > >>
>> > > >> чт, 31 янв. 2019 г. в 13:11, Максим Степачёв
<
>> > > maksim.stepachev@gmail.com
>> > > >> >:
>> > > >>
>> > > >> > Hi, I have been trying to solve a problem with calculation
size
>> for
>> > > >> > encryption mode, it's ticket IGNITE-11129. But I found an
>> additional
>> > > >> place
>> > > >> > for wrong behavior. I'm confused, Is it fine or wrong? Look
at
>> > > >> > *KeystoreEncryptionSpi#encryptedSize*, the result calculation
>> works as
>> > > >> >
>> > > >> > (dataSize / BLOCK_SZ + cntBlocks) * BLOCK_SZ;
>> > > >> >
>> > > >> > But we don't have a guarantee that dataSize is multiple of
>> BLOCK_SZ.
>> > > >> > Should we use this code:
>> > > >> >
>> > > >> > ((dataSize + BLOCK_SZ - 1) / BLOCK_SZ + cntBlocks) * BLOCK_SZ;
>> > > >> >
>> > > >> > If yes, I'll fix it.
>> > > >> >
>> > > >>
>> > > >
>> > >
>>
>

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