[ixp1200] WWBump Code in 'Network System Design'

[Lim Boon Ping]

Hi all,

This is a question regarding WWBump example code in 'Network System Design, using network processors', from Prof. Douglas Comer.

Trace through the code in WWBump.uc, i have some questions on the section regarding extracting destination port from SDRAM transfer register. I might have misunderstood the overall concept behind these codes, and hope if someone could point out.

Thanks in advance.

....

/* 'dpoff' contains the offset for destination port in TCP packet,*/
/* that is 14 bytes Ethernet header + n bytes IP header + 2     */ 
/* bytes source port, where n has been calculated earlier.      */

/* Use lower three bits of the byte offset to determine which */
/* byte the destination port will be in.  If value >= 4, dest. */
/* port is in the 2nd longword; otherwise it's in the first. */
 alu[ dpoff, dpoff, AND, 0x7 ]  ; Get lowest three bits
 alu[ --, dpoff, -, 4]   ; Test and conditional
 br>=0[SecondWord#]   ;   branch if value >=4

FirstWord#: /* Load upper two bytes of register $$hdr0 */
 ld_field_w_clr[dport, 0011, $$hdr0, >>16] ; Shift before mask
 br[GotDstPort#]    ; Check port number

SecondWord#: /* Load lower two bytes of register $$hdr1 */
 ld_field_w_clr[dport, 0011, $$hdr1, >>16] ; Shift before mask

.....


May i know why the author uses lower 3 bits of the byte offset to determine byte the destination port will be in? 

Isnt' dpoff has pointed to the offset of destination port, extracting 8 bytes from the offset onwards will always start from destination port until the first 2 bytes of ACK number, (means it will always be in $$hdr0) ? 

Then why must the code check for condition as it will never branch to SecondWord#?

If it will branch to SecondWord#, under what condition?

I have done some calculation, please kindly indicate if i have misunderstood the concept.

Case I: (minimum IP header length)
 
Assume IP HLEN = 20, dpoff = 14 + 20 + 2 = 36 bytes
That is destination port offset will be at 37 & 38 bytes of the packet.

36 bytes -> 288 bits -> 100100000
100100000 & 000000111 = 000000000  /* dpoff AND 0x7 */
result: branch to FirstWord#


Case II: (40 bytes IP header length)
 
Assume IP HLEN = 40, dpoff = 14 + 40 + 2 = 56 bytes
That is destination port offset will be at 57 & 58 bytes of the packet.

56 bytes -> 448 bits -> 111000000
111000000 & 000000111 = 000000000  /* dpoff AND 0x7 */
result: branch to FirstWord#


Case III: (60 bytes IP header length)
 
Assume IP HLEN = 60, dpoff = 14 + 60 + 2 = 76 bytes
That is destination port offset will be at 77 & 78 bytes of the packet.

76 bytes -> 608 bits -> 1001100000
1001100000 & 000000111 = 000000000  /* dpoff AND 0x7 */
result: branch to FirstWord#

 



Best regards,
Lim Boon Ping
---------------------------------------
Post-Graduate Student
Multimedia University
Cyberjaya, Malaysia

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