XXD二进制转储问题 [英] xxd binary dump problems

问题描述

是应该从XXD预期或做在右边的人物的bizzare的presence建议我做错了什么格式上面的输出？

Is the above output in the format that should be expected from xxd or does the presence of the bizzare characters on the right suggest i've done something wrong?

我试图连载简单的链表，这就是输出我得到。将未能从序列中删除前哨字符\\ 0会导致错误？

I'm attempting to serialise a simple linked list and that's the output i get. Would failing to remove the sentinal character "\0" from the serialisation cause the error?

推荐答案

我猜你是作为序列的二进制的没有的字符串的，所以这是正常的。左边每两个十六进制数字对应右边一个字符。这是的一个字节的，即八位。有些字符打印（见ASCII表波纹管），有些则不是（显示为点）。

I'm guessing you are serializing as binary not as strings, so this is normal. Each two hex digits on the left correspond to a one character on the right. This is one byte, i.e. eight bits. Some characters are printable (see ASCII table bellow), some are not (shown as dots).

ASCII Table - | hex value - name/char |

| 00 nul| 01 soh| 02 stx| 03 etx| 04 eot| 05 enq| 06 ack| 07 bel|
| 08 bs | 09 ht | 0a nl | 0b vt | 0c np | 0d cr | 0e so | 0f si |
| 10 dle| 11 dc1| 12 dc2| 13 dc3| 14 dc4| 15 nak| 16 syn| 17 etb|
| 18 can| 19 em | 1a sub| 1b esc| 1c fs | 1d gs | 1e rs | 1f us |
| 20 sp | 21  ! | 22  " | 23  # | 24  $ | 25  % | 26  & | 27  ' |
| 28  ( | 29  ) | 2a  * | 2b  + | 2c  , | 2d  - | 2e  . | 2f  / |
| 30  0 | 31  1 | 32  2 | 33  3 | 34  4 | 35  5 | 36  6 | 37  7 |
| 38  8 | 39  9 | 3a  : | 3b  ; | 3c  < | 3d  = | 3e  > | 3f  ? |
| 40  @ | 41  A | 42  B | 43  C | 44  D | 45  E | 46  F | 47  G |
| 48  H | 49  I | 4a  J | 4b  K | 4c  L | 4d  M | 4e  N | 4f  O |
| 50  P | 51  Q | 52  R | 53  S | 54  T | 55  U | 56  V | 57  W |
| 58  X | 59  Y | 5a  Z | 5b  [ | 5c  \ | 5d  ] | 5e  ^ | 5f  _ |
| 60  ` | 61  a | 62  b | 63  c | 64  d | 65  e | 66  f | 67  g |
| 68  h | 69  i | 6a  j | 6b  k | 6c  l | 6d  m | 6e  n | 6f  o |
| 70  p | 71  q | 72  r | 73  s | 74  t | 75  u | 76  v | 77  w |
| 78  x | 79  y | 7a  z | 7b  { | 7c  | | 7d  } | 7e  ~ | 7f del|

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