Re: [Inquire about plain dm-crypt]

Hi,

please do not send html mail to this list, just use text.

On 13/03/2020 05:25, Yoo-Seung Won (Dr) wrote:
> I want to encrypt single file using some commands as below.

Please note dm-crypt is block encryption, not a file encryption
so output will be BLOCK device not a file.

> ========================================================================================================================
> cryptsetup --verbose --cipher=aes-ecb --key-size=128 --key-file=/dev/example/aes_key.bin open --type=plain /dev/example/aes_data.bin enc
> 
> I know the meaning of each commands because I read the document.
> 
> aes_key.bin only has 16 bytes 0x00 (That is, 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00)
> 
> I means this file only consists of hexa.
> 
> And,
> 
> aes_data.bin has 0x800 bytes 0x00
> ========================================================================================================================
> 
> Finally, I got the enc in /dev/mapper
> 
> I opened the enc file. There is only 0x600 bytes. (I expected the 0x800 bytes because I use the AES-ECB)

As mentioned, it is block device, not a file!

I tried:
# dd if=/dev/zero of=/file bs=1 count=2048
# dd if=/dev/zero of=/key bs=1 count=16
# cryptsetup --verbose --cipher=aes-ecb --key-size=128 --key-file=/key open --type=plain /file enc

To get its size, try
# blockdev --getsize64 /dev/mapper/enc
2048

Note it must be always aligned to sector size (512 bytes in this case).

Also note there is a loopback allocated, that will align this for you,
striping the last incomplete sector.

# lsblk 
NAME   MAJ:MIN RM  SIZE RO TYPE  MOUNTPOINT
loop0    7:0    0    2K  0 loop  
└─enc  253:0    0    2K  0 crypt 


> Also, 0x600 bytes are composed of 96 16bytes (14 0f 0f 10 11 b5 22 3d 79 5 77 17 ff d9 ec 3a).
This is the correct *plaintext*.
 
> I expected that 128 16bytes (66 E9 4B D4 EF 8A 2C 3B 88 4C FA 59 CA 34 2B 2E) because I used the all zero keys and all zero data

You expected wrong, this is DECRYPTION of zero blocks, not encryption (you are reading the block device).

# hexdump -C /file
00000000  00 00 00 00 00 00 00 00  00 00 00 00 00 00 00 00  |................|
*
00000800

Try write zeroes as plaintext (use direct-io so it is not cached):

# dd if=/dev/zero of=/dev/mapper/enc oflag=direct

and now check ciphertext device:

# hexdump -C  /file

00000000  66 e9 4b d4 ef 8a 2c 3b  88 4c fa 59 ca 34 2b 2e  |f.K...,;.L.Y.4+.|
*
00000800


> I think that the encrypted data is wrong and file size also is wrong.
> 
> If you don't mind, could you tell me what I miss?

1) you have probably smaller file than 2048 bytes so the last sector is striped by loopback file (check ls -l)
2) you need to get block device size, not a file size
3) reading a device is AES decryption, not encryption

4) I guess this was a test, but never ever use ECB mode here.

So if tested properly, I get exactly what you expected.

Regards,
Milan
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