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https://issues.apache.org/jira/browse/HADOOP-11343?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel&focusedCommentId=14233337#comment-14233337
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Mike Yoder commented on HADOOP-11343:
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{quote}
Just described as above, combination of the current caculateIV and other Cipher
counter increment will cause problem if these two are not consistent.
{quote}
Yeah, you're right. This is a good catch. Let me see if I can state this
problem differently.
If the underlying java (or openssl) ctr calculation is different than
calculateIV, there is a problem IF
- assume an initial IV of 00 00 00 00 00 00 00 00 ff ff ff ff ff ff ff ff
- the file is 32 bytes
- File A is written, all 32 bytes at once (one call to calculateIV with counter
of 0)
- File B is written, the first 16 bytes and then the second 16 bytes (two calls
to calculateIV with counter of 0 and 1)
- Then the last 16 bytes of files A and B will be different
This actually isn't a problem *if* the files are read back _exactly_ as they
are written. But if you try to read file A in two steps, or read file B in one
step, the second block will look corrupted. It seems possible to construct a
test case for this.
The code in the patch looks reasonable, although I haven't sat down with paper
and pencil to work through the math. The test cases are convincing. Have you
tested with both the openssl and java crypto implementations?
I do believe that you still need to provide an upgrade path. This means
defining a new crypto SUITE and make it the default. Existing files will use
the old SUITE; the upgrade path is to simply copy all the files in an EZ; when
writing the new files the new SUITE will be used and everything will work out.
> Overflow is not properly handled in caclulating final iv for AES CTR
> --------------------------------------------------------------------
>
> Key: HADOOP-11343
> URL: https://issues.apache.org/jira/browse/HADOOP-11343
> Project: Hadoop Common
> Issue Type: Bug
> Components: security
> Affects Versions: trunk-win, 2.7.0
> Reporter: Jerry Chen
> Assignee: Jerry Chen
> Priority: Blocker
> Attachments: HADOOP-11343.patch
>
>
> In the AesCtrCryptoCodec calculateIV, as the init IV is a random generated 16
> bytes,
> final byte[] iv = new byte[cc.getCipherSuite().getAlgorithmBlockSize()];
> cc.generateSecureRandom(iv);
> Then the following calculation of iv and counter on 8 bytes (64bit) space
> would easily cause overflow and this overflow gets lost. The result would be
> the 128 bit data block was encrypted with a wrong counter and cannot be
> decrypted by standard aes-ctr.
> /**
> * The IV is produced by adding the initial IV to the counter. IV length
> * should be the same as {@link #AES_BLOCK_SIZE}
> */
> @Override
> public void calculateIV(byte[] initIV, long counter, byte[] IV) {
> Preconditions.checkArgument(initIV.length == AES_BLOCK_SIZE);
> Preconditions.checkArgument(IV.length == AES_BLOCK_SIZE);
>
> System.arraycopy(initIV, 0, IV, 0, CTR_OFFSET);
> long l = 0;
> for (int i = 0; i < 8; i++) {
> l = ((l << 8) | (initIV[CTR_OFFSET + i] & 0xff));
> }
> l += counter;
> IV[CTR_OFFSET + 0] = (byte) (l >>> 56);
> IV[CTR_OFFSET + 1] = (byte) (l >>> 48);
> IV[CTR_OFFSET + 2] = (byte) (l >>> 40);
> IV[CTR_OFFSET + 3] = (byte) (l >>> 32);
> IV[CTR_OFFSET + 4] = (byte) (l >>> 24);
> IV[CTR_OFFSET + 5] = (byte) (l >>> 16);
> IV[CTR_OFFSET + 6] = (byte) (l >>> 8);
> IV[CTR_OFFSET + 7] = (byte) (l);
> }
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