網路工程考題

45000058000100001706.......

Rating
• FISH
Lv 4

表頭

4:0100 板本

5:0101 header length fiedl=5 * 32 =160bit

00 : 0000 0000

bits 0-2: precedence

bit 3: 0 = Normal Delay, 1 = Low Delay

bit 4: 0 = Normal Throughput, 1 = High Throughput

bit 5: 0 = Normal Reliability, 1 = High Reliability

bits 6-7: Reserved for future use

0058 : Total Length(datagram size)

0000 0000 0101 1000 = 64+16+8=58Bytes

0001 : Identification

0000 : (000) 000 0000 0000 =(Flags)+Fragment Offset

Fragment Offset

This allows a maximum offset of 65,528 () which would exceed the maximum IP packet length of 65,535 with the header length included.

17 : Time To Live (TTL)

0001 0111 = 32+4+2+1=39

所以剩16個count

06 : Protocol 0000 0110

說了一堆其實那些16進位要一個個化成二進位會更好懂

其實要看TTL就次從NO.64bit ~ 72 bit來看

化成二進位加總就好了最高是255

2008-03-08 10:22:45 補充：

更正內容 :

-----------------------------------

17 : Time To Live (TTL)

0001 0111 = 32+4+2+1=39

所以剩16個count

-----------------------------------

上面0001 0111的二進位加總是23

因此應該剩23個cunt

2008-03-08 12:08:43 補充：

好笑的是...我還寫成32+4+2+1......

應該是16+4+2+1......

一題兩笨點我應該要去po笨版了...

Source(s): 門市工程師