Reputation: 81
When I run a tcpdump – what exactly does the number reported between the () after the lowest and highest sequence number mean?
I.e: Typical tcpdump on the sender:
11:47:18.278352 IP 1.1.1.36.41034 > 1.1.1.37.60960: . 3263771:3272459(8688) ack 1 win 54 11:47:18.278371 IP 1.1.1.36.41034 > 1.1.1.37.60960: . 3272459:3282595(10136) ack 1 win 54 11:47:18.278620 IP 1.1.1.36.41034 > 1.1.1.37.60960: . 3282595:3298523(15928) ack 1 win 54 11:47:18.278727 IP 1.1.1.36.41034 > 1.1.1.37.60960: . 3298523:3301419(2896) ack 1 win 54 11:47:18.278731 IP 1.1.1.36.41034 > 1.1.1.37.60960: P 3301419:3301719(300) ack 1 win 54 11:47:18.777160 IP 1.1.1.36.41034 > 1.1.1.37.60960: P 3301719:3301723(4) ack 1 win 54 11:47:18.777175 IP 1.1.1.36.41034 > 1.1.1.37.60960: . 3301723:3303171(1448) ack 1 win 54
And the receiver:
11:47:18.277895 IP 1.1.1.36.41034 > 1.1.1.37.60960: P 990413:990417(4) ack 1 win 54 11:47:18.277948 IP 1.1.1.36.41034 > 1.1.1.37.60960: . 990417:991865(1448) ack 1 win 54 11:47:18.277953 IP 1.1.1.36.41034 > 1.1.1.37.60960: . 991865:993313(1448) ack 1 win 54 11:47:18.277958 IP 1.1.1.36.41034 > 1.1.1.37.60960: . 993313:994761(1448) ack 1 win 54 11:47:18.278028 IP 1.1.1.36.41034 > 1.1.1.37.60960: . 994761:996209(1448) ack 1 win 54
Upvotes: 0
Views: 4824
Reputation: 43688
That is the length of the data, as you can see if you substract the two sequence numbers (after accounting +1 for SYN and FIN flags, if they are present in the packet).
As to why it's greater than the MTU, either your netword card can send jumbo-frames (quite normal on 1Gbps cards), or your card can accept large segments and divide them on hardware (called LSO/GSO or some similar acronym I cannot seem to recall).
Edit: See Large Segment Offload on Wikipedia.
Upvotes: 1