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In an HDLC frame, the address field can be 8 bits, or extended according to what the first bit is. I know that 11111111 is a broadcast address, and that there are 2^7 possible addresses in an 8-bit address.

My question is: How are those addresses are distributed? Is there a random addressing or something live DHCP provided by HDLC?

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    Aug 15, 2017 at 16:53

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First, HDLC is actually vendor-dependent, and the HDLC implementation from one vendor is not compatible with HDLC implementation from another vendor.

...there is 2^7 possible addresses in 8bits address.

In an 8-bit address, there are 2^8 possible addresses, from 0 to (2^8)-1, or 0 to 255.

my question is how those addresses are distributed, is there a random addressing or something live dhcp provided by hdlc?

You seem to be confusing things. DHCP is for layer-3 addressing, but HDLC is a layer-2 protocol. How each vendor assigns addresses is vendor-specific. For example, Cisco HDLC has two addresses: 10001111 and 00001111 for broadcast and unicast, respectively. The Cisco broadcast address only means that the layer-3 protocol has a broadcast address.

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