[mpls] (no subject)

loa@pi.nu Sun, 10 March 2024 09:26 UTC

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Date: Sun, 10 Mar 2024 10:26:17 +0100
From: loa@pi.nu
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Subject: [mpls] (no subject)
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Working Group, Authors.

I have a question on draft-ietf-mpls-mna-hdr@ietf.org, in section 13.3 the
document says>

   There are at most 15 Format D LSEs
   per opcode, so there are at most 20 + 15 * 30 = 470 bit positions.
   The Bit Position is an integer with value 0-469.

The encoding looks like:

    0                   1                   2                   3
    0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1
   +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
   |      MNA-Label=bSPL (TBA)             | TC  |S|    TTL        | A
   +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
   |   Opcode    |        Data             |R|IHS|S| Res |U|  NASL | B
   +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
   |   Opcode    |             Data              |S|  Data |  NAL  | C
   +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
   ~1|                   Data                    |S|     Data      ~ D
   +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+

The NASL and NAS are defined as.

   *  NASL (4 bits) : The Network Action Sub-Stack Length (NASL).  The
      number of additional LSEs in the sub-stack, not including the
      leading Format A LSE and the Format B LSE.

   *  NAL (4 bits): Network Action Length.  The number of LSEs of
      additional data, encoded in LSE Format D (Section 4.4) following
      this LSE.

The highest value you can have in the NASL is 15, and that would include
the Format C LSE, how can there be 15 Format D LSEs, if NAL is =15, the
last Format D LSE would "stick beyond" the length of the NAS, right?
Isn't the maximum number of Format D LSE 14?


/Loa