Just can't find a way to transform an Hex String to a number (Int, Long, Short) in Scala.
Is there something like "A".toInt(base)?
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5 Answers
You can use the Java libs:
val number = Integer.parseInt("FFFF", 16)
> number: Int = 65535
Or if you are feeling sparky :-):
implicit def hex2int (hex: String): Int = Integer.parseInt(hex, 16)
val number: Int = "CAFE" // <- behold the magic
number: Int = 51966
Also, if you aren't specifically trying to parse a String parameter into hex, note that Scala directly supports hexadecimal Integer literals. In this case:
val x = 0xCAFE
> x: Int = 51966
Isn't Scala wonderful? :-)
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2Hahah, mind block! You are right. Anyway, with Integer scala 2.9 resolves the parseInt method but with Short and Long don't. Any ideas?– rsanCommented May 26, 2012 at 5:10
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2I think "Short" is resoling to Scala's Short. This works: val x = java.lang.Short.parseShort("FF", 16) // x: Short = 255– 7zark7Commented May 26, 2012 at 5:15
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5Came here when googling for the opposite, converting int to str. For the record: That is
Integer.toString(number, base)orBigInt(number).toString(base), if you happen to have a bigint. (but doesn't work with bases above 36, very annoying)– BeniBelaCommented Sep 30, 2012 at 10:20 -
How do you go to Long (since that's in the question Google takes you here)? Commented Jan 9, 2017 at 20:08
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1
7zark7 answer is correct, but I want to make some additions.
Implicit from String to Int can be dangerous. Instead you can use implicit conversion to wrapper and call parsing explicitly:
class HexString(val s: String) {
def hex = Integer.parseInt(s, 16)
}
implicit def str2hex(str: String): HexString = new HexString(str)
val num: Int = "CAFE".hex
answered May 26, 2012 at 6:04
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1Indeed, that implicit conversion sounded error prone. But this way you have fexibility and clarity at the same time. Thanx– rsanCommented May 26, 2012 at 23:34
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1You could just do the same with an implicit class:
implicit class StringToHexStringAugmenter(val s: String) { def hex(): Int = Integer.parseInt(s, 16) }Less code for the same thing Commented Aug 31, 2016 at 14:03
What about a one-liner?
def hexToInt(s: String): Int = {
s.toList.map("0123456789abcdef".indexOf(_)).reduceLeft(_ * 16 + _)
}
scala> hexToInt("cafe")
res0: Int = 51966
And to answer your second item:
Is there something like "A".toInt(base)?
Yes, still as a one-liner:
def baseToInt(s: String, base: String): Int = {
s.toList.map(base.indexOf(_)).reduceLeft(_ * base.length + _)
}
scala> baseToInt("1100", "01")
res1: Int = 12
Anyone wanting to convert a UUID from hex to a decimal number can borrow from Benoit's answer and use BigDecimal for the job:
scala> "03cedf84011dd11e38ff0800200c9a66".toList.map(
| "0123456789abcdef".indexOf(_)).map(
| BigInt(_)).reduceLeft( _ * 16 + _)
res0: scala.math.BigInt = 5061830576017519706280227473241971302
Or more generally:
def hex2dec(hex: String): BigInt = {
hex.toLowerCase().toList.map(
"0123456789abcdef".indexOf(_)).map(
BigInt(_)).reduceLeft( _ * 16 + _)
}
def uuid2dec(uuid: UUID): BigInt = {
hex2dec(uuid.toString.replace("-",""))
}
Then:
scala> import java.util.UUID
scala> val id = UUID.fromString("3CEDF84-011D-D11E-38FF-D0800200C9A66")
id: java.util.UUID = 03cedf84-011d-d11e-38ff-0800200c9a66
scala> uuid2dec(id)
res2: BigInt = 5061830576017519706280227473241971302
One practical application for this is encoding the UUID in a barcode, where Code128 produces a shorter barcode for all digits than it does with alphanumeric strings. See notes about subtype "128A" on http://en.wikipedia.org/wiki/Code128#Subtypes.
For Long and Short, it is also possible to use the Java methods directly like
Long2long(java.lang.Long.valueOf(hexString, 16))
where Long2long can be even be omitted in some cases.
