I have an array of integers that range from 0 to 255, each representing two hexadecimal digits. I want to convert this array into one hexadecimal string using Ruby. How would I do that?
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Cisplatin
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Does something like this help? http://stackoverflow.com/questions/84421/converting-an-integer-to-a-hexadecimal-string-in-ruby?rq=1 – Antonio Nov 18 '16 at 17:15
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`arr.map { |n| n.to_s(16) }.join('')` – Dave Newton Nov 18 '16 at 17:17
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@Antonio I saw that, but it would convert characters like `1` to `1` instead of `01`, which is an issue because each element represents two hexadecimal characters. – Cisplatin Nov 18 '16 at 17:22
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2`arr.map { |n| sprintf('%02x', n) }.join('')` – Dave Newton Nov 18 '16 at 17:28
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...or a variant of @Dave's #2: `[0, 128, 255].each_with_object('') { |n,s| s << "%02x" % n } #=> "0080ff"` or `..(."%02x" % n).upcase..` if `=> "0080FF"` is desired. – Cary Swoveland Nov 18 '16 at 19:58
2 Answers
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With pack and unpack: (or unpack1 in Ruby 2.4+)
[0, 128, 255].pack('C*').unpack('H*')[0]
#=> "0080ff"
[0, 128, 255].pack('C*').unpack1('H*')
#=> "0080ff"
The actual binary hexadecimal string is already returned by pack('C*'):
[0, 128, 255].pack('C*')
#=> "\x00\x80\xFF"
unpack('H*') then converts it back to a human readable representation.
A light-weight alternative is sprintf-style formatting via String@% which takes an array:
'%02x%02x%02x' % [0, 128, 255]
#=> "0080ff"
x means hexadecimal number and 02 means 2 digits with leading zero.
Stefan
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This is roughly twice as fast as the other answer, if speed is important. – zetetic Nov 18 '16 at 19:55
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I would do something like this:
array = [0, 128, 255]
array.map { |number| number.to_s(16).rjust(2, '0') }.join
#=> "0080ff"
spickermann
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