Lesson 03 · Packing the flag bits - Build a DNS Resolver

Packing the flag bits

The header's flags word is sixteen bits packing eight separate fields. Today you build the high flag byte - QR, Opcode, AA, TC, and RD - and pin the RD bit to its exact position, the bit a standard query sets to ask for recursion.

The goal

Pack the high-byte flag fields into the top byte of the flags word.

Start here - the target

TO DO

Scenario: The high flag byte packs QR, Opcode, AA, TC, and RD

Giventhe flag layout where the high byte holds QR (bit 7), Opcode (bits 6-3), AA (bit 2), TC (bit 1), and RD (bit 0)

Whena query sets only RD (recursion desired)

Thenthe high flag byte is 0x01 and the flags word is 0x0100

Andsetting QR, AA, and RD together makes the high flag byte 0x85 (0x80 | 0x04 | 0x01)

Background

The flags word looks like one 16-bit number but is really eight fields packed into its bits. The high byte, reading from the top bit down, is QR (0 for a query, 1 for a response), a 4-bit Opcode, AA (authoritative answer), TC (truncated), and RD (recursion desired) in the lowest bit. Because a DNS message is stored big-endian, this high byte is the third byte of the whole message, right after the two ID bytes.

The bit that matters most for a client is RD: set it and you are asking the server to do the recursive work of chasing the answer for you. RD is bit 0 of the high byte, so a query that sets only RD has a high flag byte of 0x01 and a flags word of 0x0100. Build the packing as an OR of the individual bit masks so you can combine any set of flags; the low byte (RA, Z, RCODE) is tomorrow’s job.

Make it work

// high byte bit positions (a query with recursion desired sets RD)
const (
  flagRD uint16 = 1 << 8  // bit 0 of the high byte
  flagTC uint16 = 1 << 9
  flagAA uint16 = 1 << 10
  flagQR uint16 = 1 << 15 // top bit of the whole word
)
// Opcode occupies bits 11-14; OR the flags you want set together

CheckpointDONE

The high flag byte packs correctly, with RD in the right place. Commit and stop here.