Workshop
本页目录· 17 条
- Quiz & Assignment 3会用到这次课的知识
- d) 从Ethernet帧起始处算起，ASCII码的“G”（即HTTP GET里的G）出现在第几字节？
- 解题步骤（现在基于你这张正确截图）
- e) Datalink、Network和Transport层各自头部的大小是多少？
- 1. Datalink Layer（Ethernet II Header）大小
- 2. Network Layer（IPv4 Header）大小
- 3. Transport Layer（TCP Header）大小
- Activity B - Question D
- Activity B - Question E
- Activity B - Final Report
- Question A
- Question B
- Question C
- Question D
- Question E
- 总结
- MAC地址变化表
Quiz & Assignment 3会用到这次课的知识
- Answer the following questions, based on the contents of the Ethernet frame containing the HTTP GET message (GET /HTTP/1.1..). a) What is the value of the Ethernet source address? Hint: Example Ethernet Address (or MAC Address): 12:40:63:46:EF:9D b) What is the destination address in the Ethernet frame? Is this the Ethernet address of http://www.bom.gov.au? Hint: It is the MAC address (i.e. 52:54:00:12:35:02) of VBox playing the role of the default gateway of the VM. Again, if you are using the Machine’s O.S then it will be the Mac address of the modem router. c) Give the hexadecimal value for the two-byte Frame type field. Hint: See under the Ethernet II, Source Type. d) How many bytes from the very start of the Ethernet frame does the ASCII “G” in GET appear in the Ethernet frame? e) What is the total size of each header for Datalink, Network and Transport layers?
a) Ethernet源地址是多少？
第一个Frame详细信息里Ethernet II部分可以看到
b) Ethernet目的地址是多少？它是 bom.gov.au 的MAC地址吗？
- 不是。
这是 虚拟机默认网关（VBox网关或路由器）的MAC地址。因为在局域网内访问互联网时，包首先发送到默认网关，再由网关路由到外部网站。
c) Frame Type字段的十六进制值是多少？
- 含义：这表示 上层封装的是IPv4协议数据包（不是ARP，也不是IPv6）。
问题c，Frame Type字段的十六进制值 0x0800 是怎么从Wireshark里看出来的？
点开（小三角形展开）之后，会出现更多细节，包括：