Question

How do I convert a ruby float/double to high endian order hex with high bytes and low bytes.

EXAMPLE:

start with 99.0

end up with

40 58 C0 00   00 00 00 00
high bytes    low bytes
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Solution

Well, like Patrick said, it doesn't take a lot to convert past using Array\#pack.

irb> [99.0].pack('G').split('').map { |ds| ds[0] }
#=> [64, 88, 192, 0, 0, 0, 0, 0]
irb> _.map { |d| "%02x" % d }
#=> ["40", "58", "c0", "00", "00", "00", "00", "00"]
irb> [99.0].pack('E').split('').map { |ds| ds[0] }
#=> [0, 0, 0, 0, 0, 192, 88, 64]
irb> _.map { |d| "%02x" % d }    
#=> ["00", "00", "00", "00", "00", "c0", "58", "40"]

So it depends whether you want to unpack it with the high-order byte in the zero index or the low order byte in the zero index:

      E     |  Double-precision float, little-endian byte order
      G     |  Double-precision float, network (big-endian) byte order

OTHER TIPS

The array class has a pack method:

a = [99.0]
s = a.pack("d")
s
=> "\000\000\000\000\000\300X@"

This gives you a byte string, but converting from that to hex for printing should be trivial.

If you want to go the other way, the string class has an unpack method:

s.unpack("d")
=>[99.0]
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