CODEWITHSUNDEEP
protocol-buffers
Brad Mcquaid
Brad Mcquaid

Reputation: 1

What is the relation between length of data block and file length in varint?

In proto stream when decoding the file size and data block length in varint, the data block length is bigger than the file size? Please Help

Upvotes: 0

Views: 73

Answers (2)

Ruman Ahmed Rizvi
Ruman Ahmed Rizvi

Reputation: 11

00 E4 02 00 A5 0A F0 3E 1E 08 C9 3F 12 19 08 05 12 03 01 00 05 18 75 2A 0E AE 44 93 42 AF 44 BF 3F B2 44 B1 44 B0 44 0A 03 08 93 42 22 33 12 31 42 2F 0A 0A 54 72 61 6E 73 69 74 69 6F 6E 12 04 6E 6F 6E 65 19 00 00 01 02 08 F0 3F 29 01 07 01 01 54 30 00 58 EB D4 FF AA 07 80 01 00 2A 03 08 AF 44 32 03 08 AE 44 3A 0D 05 78 AF 44 8A 01 03 08 BF 3F 98 01 01 A2 01 03 08 B0 44 DA 01 03 08 B2 44 A2 02 03 08 B1 44 D2 02 01 26 01 06 28 AF 44 1F 08 AE 44 12 1A 08 07 12 05 94 A0 C3 01 22 08 0A 04 0A 02 01 06 18 01 2A 04 B3 44 DC 3B 0A BE 01 0A AF 01 0A 3E 0A 1F 0A 0A 0D 00 00 C8 42 15 00 00 5A 43 12 01 0C E0 4E 44 15 00 00 E1 43 18 03 25 00 00 00 00 12 03 08 C9 3F 1A 12 08 04 10 02 18 01 25 00 00 40 41 2D 00 00 00 3F 30 00 28 00 38 00 12 03 08 DC 3B 1A 68 08 00 10 00 2A 62 12 1D 3D 18 1A 54 0A 0E 08 01 12 01 12 08 00 00 15 01 48 0C 0A 0E 08 02 05 10 05 5F 00 00 3A 10 00 04 E1 43 15 20 05 30 14 E1 43 0A 02 08 05 3E 44 00 14 12 03 08 B3 44 22 01 05 30 30 01 10 03 15 08 B3 44 12 10 08 D1 0F 09 E4 F0 4A 6C 2A 04 83 42 C7 41 12 03 08 A8 3F 1A 1F 41 66 74 65 72 20 4B 65 79 6E 6F 74 65 20 38 3F 29 0A 62 75 6C 6C 65 74 20 77 6F 6D 62 61 74 2A 0D 0A 07 08 00 12 03 08 83 42 0A 02 08 12 32 08 0A 06 08 00 10 00 18 00 3A 09 0A 07 08 01 69 10 C7 41 50 01 72 15 17 04 9A 01 05 0B 10 12 02 65 6E C2 09 0B 18 10 00 18 00 1F 08 AF 62 65 01 0C C0 41 B5 44 4E 65 01 04 A0 41 39 06 00 40 9E 65 01 08 C0 41 1A 42 65 01 00 40 C2 65 01 00 40 3A 65 01 00 40 5E 65 01 00 B5 21 65 00 B5 21 65 0C 02 15 08 B5 2E 65 01 18 4F 2A 04 AE 3B A2 40 29 65 08 06 57 6F 25 4C 31 33 08 A2 40 32 25 26 21 1B 00 3A 11 15 04 AE 3B 86 48 01 18 12 08 B2 44 12 0D 08 2D C9 0C 05 2A 02 B4 61 35 10 B4 44 15 08 B4 2E 7D 00 00 3E 01 7D 0C AC 3B 08 04 05 7F 15 77 04 AC 3B 82 77 00 3D B4 20 0F 08 B1 44 12 0A 08 E7 17 49 36 0C 00 1F 08 B0 62 C4 01 08 DB 40 B6 3E C4 01 18 00 44 15 00 80 3D 44 25 C4 20 48 42 15 00 00 E8 41 18 00 92 29 03 04 DB 40 32 C4 01 0D 3D 6A 29 03 05 22 7D 29 05 10 04 E8 41 36 29 03 04 E8 41 5A 29 03 00 B6 21 C4 00 B6 21 C4 18 01 17 08 B6 44 12 12 75 29 20 4C 2A 06 AE 3B C0 40 B7 44 25 47 10 1A 03 EF BF BC 35 4C 00 C0 56 C3 01 00 4A 31 CE 04 B7 44 2E CE 01 32 57 01 00 B7 21 57 00 FB 2D C4 44 11 0A 04 0A 00 10 00 10 00 22 07 64 65 63 69 6D 61 6C

Decoding the first six bytes

Upvotes: 0

Marc Gravell
Marc Gravell

Reputation: 1062630

the data block length is bigger than the file size

That sounds unlikely, and is most likely the result of (any one of):

  • a broken encoder (very unlikely if you're using any of the established implementations, but not impossible)
  • corruption of the data-stream in transit (the most common variant of this being: using a text-encoding such as UTF-8 backwards to try to get a string - when if a string is really needed, something like base-16 (hex) or base-64 should be used)
  • accidentally truncating a stream prematurely
  • a broken decoder

It sounds like you're parsing the stream manually, so frankly my guess is the last option. If you can post the bytes you're decoding we can probably advise on whether your interpretation is correct. Alternatively, you can try pushing your data through https://protogen.marcgravell.com/decode which will pull apart the data-stream and show how it has interpreted the bytes.

Upvotes: 1

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