十六进制转储混乱 [英] hexdump confusion

问题描述

我正在使用unix hexdump实用程序.我的输入文件是UTF-8编码的，包含单个字符ñ，在十六进制UTF-8中为C3 B1.

I am playing with the unix hexdump utility. My input file is UTF-8 encoded, containing a single character ñ, which is C3 B1 in hexadecimal UTF-8.

hexdump test.txt
0000000 b1c3
0000002

嗯?这显示了B1 C3-与我期望的相反！有人可以解释吗?

Huh? This shows B1 C3 - the inverse of what I expected! Can someone explain?

为了获得预期的输出，我这样做:

For getting the expected output I do:

hexdump -C test.txt
00000000  c3 b1                                             |..|
00000002

我当时以为我了解编码系统.

I was thinking I understand encoding systems..

推荐答案

这是因为hexdump默认使用16位字，并且您使用的是Little-endian体系结构.因此，字节序列b1 c3被解释为十六进制字c3b1. -C选项强制hexdump使用字节而不是单词.

This is because hexdump defaults to using 16-bit words and you are running on a little-endian architecture. The byte sequence b1 c3 is thus interpreted as the hex word c3b1. The -C option forces hexdump to work with bytes instead of words.

这篇关于十六进制转储混乱的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！