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?

Accepted Answer

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.

How to convert hex to uint8_t using ruby

Solution

Problem