Out of the results of ``` 0101 1002 0909 0210 1115 1511 spli kotlinlang #announcements

"0101-1002-0909-0210-1115-1511".split("-")
            .flatMap { it.chunked(2).map { it.toInt() } }
        	.map { if (it > 9) (it + 55).toChar() else (it + 48).toChar() }
        	.chunked(2)

I need:

11:A2:99:2A:BF:FB

, how could I do that idiomatically, also, is there a simpler algo for this? It's converting to hex..

elizarov

01/01/2018, 2:37 PM

"<tel:0101-1002-0909|0101-1002-0909>-0210-1115-1511".split("-")
        .map { it.chunked(2)
            .map { it.toInt().toString(16).toUpperCase() }
            .joinToString(separator = "")
        }.joinToString(separator = ":")

dave08

01/01/2018, 2:39 PM

Perfect 👍🏼, thanks!

Actually this gives the same:

"<tel:0101-1002-0909|0101-1002-0909>-0210-1115-1511".split("-")
				.joinToString(separator = ":") { it.chunked(2)
						.joinToString(separator = "") { it.toInt().toString(16).toUpperCase() }
				}

But your is maybe clearer.. is there a difference in performance? (I'm using this in a Vert.x api...)

karelpeeters

01/02/2018, 7:26 AM

If you need leading zeroes in the hexadecimal numbers make sure to use

"%02x".format(it)

dave08

01/02/2018, 7:35 AM

@karelpeeters I don't need those in the to hex algo, but I will need a way to convert the hex back to the number format, there I will need those zeros... my original code for that is not ideomatic at all (no

toInt(16)

, or is there?), but your hint might help, thanks!

karelpeeters

01/02/2018, 7:38 AM

There is

toInt(16)

, bith without leading zeroes. Good luck!

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