大家幫忙看看這道題答案

x9858627899

<interface>
$ g  \  B3 \6 m& `  <loopback>
) ~7 E! ]. T  ?/ j4 m- a/ y     <name>100</name>
. J+ C( U( z; z& j2 V     <enable>true</enable>
8 C4 f6 ^! o  r8 u* T  </loopback>
# R+ T' F( T/ K, d1 ?) j. s

<interface>
* `' p7 Z* O9 t8 }6 O. n; V: x) J
8 q) M6 Z/ S) N1 K& s- w( a/ O

What is achieved by this code?
A. It displays the loopback interface
B. It renames the loopback interface
C. It deletes the loopback interface
7 K; A) d: ?! S# Q2 u0 I* I5 Q% pD. It unshuts the loopback interface
4 `" D+ c* \: ~/ r

0 _; F8 a6 D8 s. L. H% S查不到啊...有些說D有些說A..沒有程式背景
請各位幫忙看看唄
/ x1 _( g6 b+ L/ d& p

lincc

D:
. @8 s6 j' n9 ^' j" s% kenable就是noshut
3 r: h2 R' R. O" r: H- z$ V0 ?

thamky

D. It unshuts the loopback interface. Enable=true means unshut the port.

朱大帅爱皇马

  {6 s) H4 W0 Y: ~+ I7 }: WD:
8 ^( W* ?* a: @* u  Yenable就是noshut:

jycert1

it should be D: enable the loopback 100

shangd

这不编程么？

tiang

D enable就是noshut