How to convert hex to uint8_t using ruby

Question

I have a string

6e6de179a94a4b406efab31f29d216c0e2ff0000

which I am told is defined as uint8_t and unpacks as latitude [8], longitude [8] and altitude [4].

I think this hex should decode to 54.58335 -5.70542 -15.

How could I decode such a string using Ruby?

Answer 1

Fun question :)

The string is 40 hex chars long, so it represents 20 bytes.

Looking at your link, that would be 8 bytes for latitude, 8 bytes for longitude, and 4 bytes for the altitude. To be honest, I just tried the different parameters for pack and unpack until it looked like the desired floats:

hex = "6e6de179a94a4b406efab31f29d216c0e2ff0000"

lat_hex, lon_hex, alt_hex = hex[0,16], hex[16, 16], hex[32, 8]
lat_int, lon_int, alt_int = lat_hex.to_i(16), lon_hex.to_i(16), alt_hex.to_i(16)
p [lat_int].pack('q>').unpack('D').first
# 54.583297
p [lon_int].pack('q>').unpack('D').first
# -5.705235

Here's a shorter way:

hex.scan(/../).map{ |x| x.hex }.pack('C*').unpack('DDL')
# => [54.583297, -5.705235, 65506]

I'm not sure about the altitude. Since it's stored as an integer, it's probably modified later with some linear function to fit the GPS precision and the usual altitude range.