了解TCP数据包重新排序

时间:2016-06-17 08:56:52

标签: tcp pcap

我正在尝试从TCP嗅探(使用pcap)过滤重复/丢失的数据包,但我停止了对seq / ack的理解。这是我的日志与相对seq / ack:

CLIENT->SERVER/Seq=0;Ack=0/SYN/P.size:0; No data in TCP. Size: 66/66 -> 20 E7 1E 61 15 5B 4E 1D 00 00 00 00 80 02 F7 D3 4D 03 00 00 02 04 05 B4 01 03 03 06 01 01 04 02 | No payload
SERVER->CLIENT/Seq=0;Ack=1/ACK+SYN/P.size:0; No data in TCP. Size: 58/58 -> 1E 61 20 E7 C4 9D 5B 6B 15 5B 4E 1E 60 12 20 00 2D D1 00 00 02 04 05 B4 | No payload
CLIENT->SERVER/Seq=1;Ack=1/ACK/P.size:0; No data in TCP. Size: 54/54 -> 20 E7 1E 61 15 5B 4E 1E C4 9D 5B 6C 50 10 01 6D 64 21 00 00 | No payload
SERVER->CLIENT/Seq=1;Ack=268/ACK/P.size:0; No data in TCP. Size: 54/54 -> 1E 61 20 E7 C4 9D 5B 6C 15 5B 4F 29 50 10 5B 40 09 43 00 00 | No payload
CLIENT->SERVER/Seq=1;Ack=1/ACK+PSH/P.size:267; 20 E7 1E 61 15 5B 4E 1E C4 9D 5B 6C 50 18 01 6D AF 0B 00 00 | 0B 01 00 EA 02 00 00 09 07 54 56 03 09 0B 01 07 02 54 54 56 07 00 02 55 56 00 51 00 53 57 04 07 55 08 54 01 07 01 53 00 56 55 56 01 06 05 04 51 03 08 51 08 51 56 04 54 06 55 08 02 09 51 56 01 53 06 55 04 53 00 56 56 53 01 09 02 09 01 51 54 51 09 55 56 09 03 04 07 05 55 04 06 55 04 06 09 04 51 01 08 08 06 05 52 06 04 01 07 54 03 06 52 55 06 55 55 51 01 02 04 54 03 55 54 01 57 51 55 05 52 05 54 07 51 51 55 07 02 53 53 00 52 05 52 07 01 54 00 03 05 05 08 06 05 05 06 03 00 0D 08 01 07 09 03 51 03 07 53 09 51 06 07 54 0A 50 56 02 52 04 05 55 51 02 53 00 08 54 04 52 56 06 02 09 00 08 03 53 56 01 05 00 55 06 08 56 04 0D 06 07 52 06 07 04 0A 06 01 04 54 04 00 05 02 04 54 00 09 52 53 05 04 01 04 05 05 01 52 51 52 0D 06 51 08 09 54 53 00 0D 01 02 03 54 53 01 05 03 08 56 54 07 02 54 0B 06 DC 4F 61 4F
CLIENT->SERVER/Seq=267;Ack=1/ACK/P.size:0; No data in TCP. Size: 54/54 -> 20 E7 1E 61 15 5B 4F 28 C4 9D 5B 6C 50 10 01 6D 63 17 00 00 | No payload
SERVER->CLIENT/Seq=1;Ack=268/ACK+PSH/P.size:20; 1E 61 20 E7 C4 9D 5B 6C 15 5B 4F 29 50 18 5B 40 3A C6 00 00 | 14 00 00 01 E0 41 9A F0 98 F5 A4 37 01 00 00 00 01 00 00 00
CLIENT->SERVER/Seq=268;Ack=21/ACK/P.size:0; No data in TCP. Size: 54/54 -> 20 E7 1E 61 15 5B 4F 29 C4 9D 5B 80 50 10 01 6D 63 02 00 00 | No payload
SERVER->CLIENT/Seq=21;Ack=305/ACK/P.size:0; No data in TCP. Size: 54/54 -> 1E 61 20 E7 C4 9D 5B 80 15 5B 4F 4E 50 10 5B 40 09 0A 00 00 | No payload
CLIENT->SERVER/Seq=268;Ack=21/ACK+PSH/P.size:37; 20 E7 1E 61 15 5B 4F 29 C4 9D 5B 80 50 18 01 84 6B AF 00 00 | 25 00 32 DF 4C C6 2A 51 18 85 82 AC 27 D8 7A 06 44 DF F7 27 BD FC 59 43 3B E7 19 53 33 37 78 7B 93 81 38 51 CB
CLIENT->SERVER/Seq=304;Ack=21/ACK/P.size:0; No data in TCP. Size: 54/54 -> 20 E7 1E 61 15 5B 4F 4D C4 9D 5B 80 50 10 01 84 62 C7 00 00 | No payload
SERVER->CLIENT/Seq=21;Ack=305/ACK+PSH/P.size:328; 1E 61 20 E7 C4 9D 5B 80 15 5B 4F 4E 50 18 5B 40 AD 89 00 00 | 48 01 F3 B3 29 D9 41 E1 45 1B D3 98 0B 6E CF CC FD 18 F8 B9 23 3B 66 93 37 62 AA E9 7A 43 E2 B9 88 1F FF 77 80 70 E8 1D B9 8E 46 61 F2 F3 52 3E 0F 98 78 3B A1 51 C9 1E BA 8D 45 63 F0 F1 50 F9 F1 67 87 9E 3A C8 50 9D CB 03 34 63 CD C6 B0 FF 7A 4D ED 9F 36 F5 5E 98 43 FC 74 5A 8D 9E 3F 07 BC 10 F3 B2 28 D8 40 81 25 12 DA FD 6E 6F CE A2 93 04 E4 A5 3F CF 57 A2 06 31 F9 DE 4D 4C ED 81 B0 27 C7 86 1C EC 74 81 25 12 DA FD 6E 6F CE A2 93 04 E4 A5 3F CF 57 A2 06 31 F9 DE 4D 4C ED 81 B0 27 C7 86 1C EC 74 81 25 12 DA FD 6E 6F CE A2 93 04 E4 A5 3F CF 57 33 86 71 B9 8A 17 C7 66 1E 21 67 87 7A 95 B4 2C D9 7D 4A 82 A5 36 37 96 8B DB 85 65 24 BE 4E D6 23 87 B0 78 5F CC CD 6C 00 31 A6 46 07 9D 6D F5 00 A4 93 5B 7C EF EE 4F 23 12 85 65 24 BE 4E D6 23 87 B0 78 3F AE AF 0E 5E 6F F8 18 65 FF 0F 97 22 86 B1 79 5E CD CC 6D 01 30 A7 47 F3 76 86 1E EB 4F 78 B0 93 00 01 A0 CC FD 6A 8A C9 53 A3 3B B...
CLIENT->SERVER/Seq=305;Ack=349/ACK/P.size:0; No data in TCP. Size: 54/54 -> 20 E7 1E 61 15 5B 4F 4E C4 9D 5C C8 50 10 01 84 61 7E 00 00 | No payload
CLIENT->SERVER/Seq=305;Ack=349/ACK/P.size:0; No data in TCP. Size: 54/54 -> 20 E7 1E 61 15 5B 4F 4E C4 9D 5C C8 50 10 01 9B 61 67 00 00 | No payload
CLIENT->SERVER/Seq=304;Ack=349/ACK/P.size:0; No data in TCP. Size: 54/54 -> 20 E7 1E 61 15 5B 4F 4D C4 9D 5C C8 50 10 01 B2 61 51 00 00 | No payload
SERVER->CLIENT/Seq=349;Ack=305/ACK+PSH/P.size:7; 1E 61 20 E7 C4 9D 5C C8 15 5B 4F 4E 50 18 5B 40 05 3F 00 00 | 07 00 67 24 BE 4E D6 

第5行

SERVER->CLIENT/Seq=1;Ack=268/ACK/P.size:0; No data in TCP. Size: 54/54 -> 1E 61 20 E7 C4 9D 5B 6C 15 5B 4F 29 50 10 5B 40 09 43 00 00 | No payload

服务器使用268发送ACK,而客户端仅在下一行发送这些267长度的数据。为什么订单在这里打破?

据我所知,fisrt客户端应该发送seq1 / ack1 / L = 267,然后服务器应该使用seq1 / Ack268。

或者是否意味着,我必须在TCP协议中实现整个逻辑以进行数据包交换(包括选择性ACK?)

0 个答案:

没有答案