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  1. Hadoop HDFS
  2. HDFS-15186

Erasure Coding: Decommission may generate the parity block's content with all 0 in some case



    • Bug
    • Status: Resolved
    • Critical
    • Resolution: Fixed
    • 3.0.3, 3.2.1, 3.1.3
    • 3.3.0
    • datanode, erasure-coding
    • None
    • Reviewed
    • Patch, Important


      1. I can find some parity block's content with all 0 when i decommission some DataNode(more than 1) from a cluster. And the probability is very big(parts per thousand).This is a big problem.You can think that if we read data from the zero parity block or use the zero parity block to recover a block which can make us use the error data even we don't know it.

      There is some case in the below:

      B: Busy DataNode, 

      D:Decommissioning DataNode,

      Others is normal.

      1.Group indices is [0, 1, 2, 3, 4, 5, 6(B,D), 7, 8(D)].

      2.Group indices is [0(B,D), 1, 2, 3, 4, 5, 6(B,D), 7, 8(D)].


      In the first case when the block group indices is [0, 1, 2, 3, 4, 5, 6(B,D), 7, 8(D)], the DN may received reconstruct block command and the liveIndices=[0, 1, 2, 3, 4, 5, 7, 8] and the targets's(the field which  in the class StripedReconstructionInfo) length is 2.

      The targets's length is 2 which mean that the DataNode need recover 2 internal block in current code.But from the liveIndices we only can find 1 missing block, so the method StripedWriter#initTargetIndices will use 0 as the default recover block and don't care the indices 0 is in the sources indices or not.

      When they use sources indices [0, 1, 2, 3, 4, 5] to recover indices [6, 0] use the ec algorithm.We can find that the indices [0] is in the both the sources indices and the targets indices in this case. The returned target buffer in the indices [6] is always 0 from the ec  algorithm.So I think this is the ec algorithm's problem. Because it should more fault tolerance.I try to fixed it .But it is too hard. Because the case is too more. The second is another case in the example above(use sources indices [1, 2, 3, 4, 5, 7] to recover indices [0, 6, 0]). So I changed my mind.Invoke the ec  algorithm with a correct parameters. Which mean that remove the duplicate target indices 0 in this case.Finally, I fixed it in this way.



        1. HDFS-15186.005.patch
          6 kB
          Yao Guangdong
        2. HDFS-15186.004.patch
          6 kB
          Yao Guangdong
        3. HDFS-15186.003.patch
          5 kB
          Yao Guangdong
        4. HDFS-15186.002.patch
          5 kB
          Yao Guangdong
        5. HDFS-15186.001.patch
          4 kB
          Yao Guangdong

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              yaoguangdong Yao Guangdong
              yaoguangdong Yao Guangdong
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