问题
I'm reading this code and I'm a tad confused as to what is going on. This code is using Ruby's OpenSSL library.
encrypted_message = cipher.update(address_string) + cipher.final
encrypted_message
=> "G\xCB\xE10prs\x1D\xA7\xD0\xB0\xCEmX\xDC@k\xDD\x8B\x8BB\xE1#!v\xF1\xDC\x19\xDD\xD0\xCA\xC9\x8B?B\xD4\xED\xA1\x83\x10\x1F\b\xF0A\xFEMBs'\xF3\xC7\xBC\x87\x9D_n\\z\xB7\xC1\xA5\xDA\xF4s \x99\\\xFD^\x85\x89s\e"
[3] pry(Encoder)> encrypted_message.unpack('H*')
=> ["47cbe1307072731da7d0b0ce6d58dc406bdd8b8b42e1232176f1dc19ddd0cac98b3f42d4eda183101f08f041fe4d427327f3c7bc879d5f6e5c7ab7c1a5daf47320995cfd5e8589731b"]
It seems that the H directive is this:
hex string (high nibble first)
How are the escaped characters in the encrypted_message transformed into letters and numbers?
I think the heart of the issue is that I don't understand this. What is going on?
['A'].pack('H')
=> "\xA0"
回答1:
Here is a good explanation of Ruby's pack and unpack methods.
According to your question:
> ['A'].pack('H')
=> "\xA0"
A byte consists of 8 bits. A nibble consists of 4 bits. So a byte has two nibbles. The ascii value of ‘h’ is 104. Hex value of 104 is 68. This 68 is stored in two nibbles. First nibble, meaning 4 bits, contain the value 6 and the second nibble contains the value 8. In general we deal with high nibble first and going from left to right we pick the value 6 and then 8.
In the above case the input ‘A’ is not ASCII ‘A’ but the hex ‘A’. Why is it hex ‘A’. It is hex ‘A’ because the directive ‘H’ is telling pack to treat input value as hex value. Since ‘H’ is high nibble first and since the input has only one nibble then that means the second nibble is zero. So the input changes from ['A'] to ['A0'] .
Since hex value A0 does not translate into anything in the ASCII table the final output is left as it and hence the result is \xA0. The leading \x indicates that the value is hex value.
来源：https://stackoverflow.com/questions/48492676/what-is-this-unpack-doing-can-someone-help-me-understand-just-a-few-letters