[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