问题
Even after reading the standard documentation, I still can't understand how Ruby's Array#pack and String#unpack exactly work. Here is the example that's causing me the most trouble:
irb(main):001:0> chars = ["61","62","63"]
=> ["61", "62", "63"]
irb(main):002:0> chars.pack("H*")
=> "a"
irb(main):003:0> chars.pack("HHH")
=> "```"
I expected both these operations to return the same output: "abc". Each of them "fails" in a different manner (not really a fail since I probably expect the wrong thing). So two questions:
- What is the logic behind those outputs?
- How can I achieve the effect I want, i.e. transforming a sequence of hexadecimal numbers to the corresponding string. Even better - given an integer n, how to transform it to a string identical to the text file that when is considered as a number (say, in a hex editor) equals n?
回答1:
We were working on a similar problem this morning. If the array size is unknown, you can use:
ary = ["61", "62", "63"]
ary.pack('H2' * ary.size)
=> "abc"
You can reverse it using:
str = "abc"
str.unpack('H2' * str.size)
=> ["61", "62", "63"]
回答2:
The 'H' String directive for Array#pack says that array contents should be interpreted as nibbles of hex strings.
In the first example you've provided:
irb(main):002:0> chars.pack("H*")
=> "a"
you're telling to pack the first element of the array as if it were a sequence of nibbles (half bytes) of a hex string: 0x61 in this case that corresponds to the 'a' ASCII character.
In the second example:
irb(main):003:0> chars.pack("HHH")
=> "```"
you're telling to pack 3 elements of the array as if they were nibbles (the high part in this case): 0x60 corresponds to the '`' ASCII character. The low part or second nibble (0x01) "gets lost" due to missing '2' or '*' modifiers for "aTemplateString".
What you need is:
chars.pack('H*' * chars.size)
in order to pack all the nibbles of all the elements of the array as if they were hex strings.
The case of 'H2' * char.size only works fine if the array elements are representing 1 byte only hex strings.
It means that something like chars = ["6161", "6262", "6363"] is going to be incomplete:
2.1.5 :047 > chars = ["6161", "6262", "6363"]
=> ["6161", "6262", "6363"]
2.1.5 :048 > chars.pack('H2' * chars.size)
=> "abc"
while:
2.1.5 :049 > chars.pack('H*' * chars.size)
=> "aabbcc"
回答3:
The Array#pack method is pretty arcane. Addressing question (2), I was able to get your example to work by doing this:
> ["61", "62", "63"].pack("H2H2H2")
=> "abc"
See the Ruby documentation for a similar example. Here is a more general way to do it:
["61", "62", "63"].map {|s| [s].pack("H2") }.join
This is probably not the most efficient way to tackle your problem; I suspect there is a better way, but it would help to know what kind of input you are starting out with.
The #pack method is common to other languages, such as Perl. If Ruby's documentation does not help, you might consult analogous documentation elsewhere.
回答4:
I expected both these operations to return the same output: "abc".
The easiest way to understand why your approach didn't work, is to simply start with what you are expecting:
"abc".unpack("H*")
# => ["616263"]
["616263"].pack("H*")
# => "abc"
So, it seems that Ruby expects your hex bytes in one long string instead of separate elements of an array. So the simplest answer to your original question would be this:
chars = ["61", "62", "63"]
[chars.join].pack("H*")
# => "abc"
This approach also seems to perform comparably well for large input:
require 'benchmark'
chars = ["61", "62", "63"] * 100000
Benchmark.bmbm do |bm|
bm.report("join pack") do [chars.join].pack("H*") end
bm.report("big pack") do chars.pack("H2" * chars.size) end
bm.report("map pack") do chars.map{ |s| [s].pack("H2") }.join end
end
# user system total real
# join pack 0.030000 0.000000 0.030000 ( 0.025558)
# big pack 0.030000 0.000000 0.030000 ( 0.027773)
# map pack 0.230000 0.010000 0.240000 ( 0.241117)
来源：https://stackoverflow.com/questions/13729475/rubys-pack-and-unpack-explained