编码的十六进制字符串中的字节数
这可能是一个简单的问题，但我正在尝试检查编码和字节（我有一段时间没有看过）来实现二进制协议。
看起来普通字符是1个字节。但是，当您将它们编码为十六进制时，字节数会减半。
ruby-1.9.2-p180 :001 > "abcd".bytesize
=> 4
ruby-1.9.2-p180 :002 > ["abcd"].pack("H*")
=> "\xAB\xCD"
ruby-1.9.2-p180 :003 > ["abcd"].pack("H*").bytesize
=> 2
我还期望十六进制编码能出现字符 0-9 和 AF
有人可以帮助澄清这里发生了什么吗？另外，如果您能为我指出在线的一般编码的良好评论，我很乐意温习。我还没有看到任何对此的简单概述，这会很棒。
谢谢！
This is probably a simple question, but I'm trying to review encodings and bytes (which I haven't looked at in a while) to implement a binary protocol.
It looks like normal characters are 1 byte. But when you encode them in Hex, it halves the number of bytes.
ruby-1.9.2-p180 :001 > "abcd".bytesize
=> 4
ruby-1.9.2-p180 :002 > ["abcd"].pack("H*")
=> "\xAB\xCD"
ruby-1.9.2-p180 :003 > ["abcd"].pack("H*").bytesize
=> 2
I was also expecting the Hex encoding to come out with characters 0-9 and A-F
Can someone help clarify what is going on here? And also if you can point me to a good review of encodings in general that is online I would love to brush up. I haven't seen any simple overviews of this yet which would be great.
Thanks!
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#pack读取给定的字符串并根据给定的格式将其转换为二进制。pack('H*')表示您给出的字符串表示十六进制表示法，因此它将从十六进制转换字节AB和CD转换为二进制 (1010 1011 1100 1101)，即两个字节。尝试
["g"].pack("H*")，其中g不是有效的十六进制字符...#packreads the given string and transforms it into binary according to the given format.pack('H*')means that the string you are giving represents hexadecimal notation, so it will convert the bytesABandCDfrom hex into binary (1010 1011 1100 1101), which is two bytes.Try
["g"].pack("H*"), wheregis not a valid hexadecimal character...