java.lang.Integer.parseInt fails with hex strings hat use the highest bit

Question

I was surprised that the following forward backward conversion from 32-bit ints to hex strings fails:

Integer.parseInt(-2028332484.toHexString, 16)

Gives:

java.lang.NumberFormatException: For input string: "871a1a3c"

Obviously a workaround is

java.lang.Long.parseLong(-2028332484.toHexString, 16).toInt

But I wonder if there is not a better (and possibly more Scala'ish) solution?

score 0 · Accepted Answer · edited May 23 '17 at 11:56

0

It's been answered for Java already here.

Unfortunately there is no additional treatment for that transformation in scala AFAIK.

Scala defines in RichInt:

def toHexString: String = java.lang.Integer.toHexString(self)

and in StringLike:

def toInt: Int         = java.lang.Integer.parseInt(toString)

apart from import java.lang.{Long => JLong} and using JLong I don't know a more scala-ish solution than yours.

edited May 23 '17 at 11:56

Community

answered Apr 07 '13 at 05:48

michael_s

Thanks for the link. I see that the symmetric counterpart of `parseInt(_, 16)` is indeed `toString(_, 16)`, and there is no counterpart for `toHexString`. – 0__ Apr 07 '13 at 11:58

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