Convert Integer in Hex String Notation into a signed 8-Bit Integer

Question

I'd like to convert a hex string into an 8-bit signed integer, as shown in the following example:

irb> "0xff".unpack(X)
=> [-1]

The correct value for X is missing.

By reding https://ruby-doc.org/core-2.3.0/String.html#method-i-unpack and experimenting I do not find the correct value for X.

I could code that myself, like also posted here: https://www.ruby-forum.com/topic/138200#615299 but I am convinced, there is a more elegant concise ruby way.

irb> length = 8
irb> mid = 2**(length-1)
irb> max_unsigned = 2**length
irb> to_signed = proc {|n| (n>=mid) ? n - max_unsigned : n}
irb> to_signed["0xFF".to_i(16)]
=> -1

Note: For the unsinged case, this is easy:

irb> "e0".hex
=> 224

cremno · Accepted Answer · 2018-03-21 12:23:11Z

5

The c directive comes very close but it expects a different input: a single character representing the signed 8-bit integer. This requires the input "0xff" to be "\xff". So this conversion must take place first. At least one method known by you can be used here; the other one is Integer#chr:

"0xff".hex.chr # => "\xFF"

And the complete solution:

"0xff".hex.chr.unpack1(?c) # => -1

answered Mar 21, 2018 at 12:23

cremno

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