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kotlinlang

Out of the results of:
```        "0101-1002-0909-0210-1115-1511".split("-")
            .flatMap { it.chunked(2).map { it.toInt() } }
        	.map { if (it &gt; 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..

```
"<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 = ":")
```

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...)

If you need leading zeroes in the hexadecimal numbers make sure to use `"%02x".format(it)`.

<@U5UU34LPK> 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!

There is `toInt(16)`, bith without leading zeroes. Good luck!