Parsing byte string into a native scala byte

Question

I'm trying to parse a string like this into a List[Byte].

Here is the string

"0x4e 0x01 0x09"

How can you instantiate a byte from a string representation?

What's the expected output for that example?

Chris Martin
– Chris Martin

2016-01-20 02:44:44 +00:00
Commented Jan 20, 2016 at 2:44 — Chris Martin
– Chris Martin, Commented Jan 20, 2016 at 2:44
List(0x4e,0x010000000, 0x09)

Chris Stewart
– Chris Stewart

2016-01-20 02:46:03 +00:00
Commented Jan 20, 2016 at 2:46 — Chris Stewart
– Chris Stewart, Commented Jan 20, 2016 at 2:46
That's a List[Int].

Chris Martin
– Chris Martin

2016-01-20 02:46:50 +00:00
Commented Jan 20, 2016 at 2:46 — Chris Martin
– Chris Martin, Commented Jan 20, 2016 at 2:46
You're right, simplifying the OP to use other data.

Chris Stewart
– Chris Stewart

2016-01-20 02:48:41 +00:00
Commented Jan 20, 2016 at 2:48 — Chris Stewart
– Chris Stewart, Commented Jan 20, 2016 at 2:48

Chris Martin · Accepted Answer · 2016-01-20 03:00:30Z

3

Here's one solution using a regex and parseInt.

def parseBytes(s: String): List[Byte] =
  (raw"\b0x([0-9a-f]{2})\b".r
    .findAllMatchIn(s)
    .map(g => Integer.parseInt(g.group(1), 16).toByte)
    .toList)

Test:

scala> parseBytes("0x4e 0x01 0x09 0xff")
0: List[Byte] = List(78, 1, 9, -1)

answered Jan 20, 2016 at 3:00

Chris Martin

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