How to convert Java into Kotlin

Question

I don't know if somebody has already answered this question. If it's the case, then, please for the redondancy.

I would like to transform this function into Kotlin function. Thanks for the help.

private final static String HEX = "XXXXXXXXX";
private static void appendHex(StringBuffer sb, byte b) {
    sb.append(HEX.charAt((b>>4)&0x0f)).append(HEX.charAt(b&0x0f));
}

could it be that HEX actually rather contains something like 0123456789ABCDEF? if so... what about: String.format("%02X", b) instead? — Roland
– Roland, Commented Mar 21, 2019 at 9:29
@Roland. Yes it contains. But it is the content of the function that is very important to me! — K. Donon
– K. Donon, Commented Mar 21, 2019 at 10:00
I am not sure whether I understood you correctly... sb.append(HEX.charAt((b>>4)&0x0f)).append(HEX.charAt(b&0x0f)); is what you want to translate to Kotlin right? But if it is already simplifiable to sb.append(String.format("%02X", b)), that's not ok for you? — Roland
– Roland, Commented Mar 21, 2019 at 10:11
There's a chance that you want to keep this static function in Java so that you can still keep using Byte with >>. — EpicPandaForce
– EpicPandaForce, Commented Mar 21, 2019 at 11:36

EpicPandaForce · Accepted Answer · 2019-03-21 11:35:15Z

4

I suggest using an extension function for that:

private const val HEX = "XXXXXXXXX"
fun StringBuffer.appendHex(b: Byte): StringBuffer {
    val i = b.toInt()
    append(HEX[i shr 4 and 0x0f])
    append(HEX[i and 0x0f])
    return this
}

I also made the function return this again so you can even call it in a chain:

myStringBuffer
   .append(someThing)
   .appendHex(myByte)
   .append(someThingElse)

edited Mar 21, 2019 at 11:35

EpicPandaForce

81.7k28 gold badges269 silver badges451 bronze badges

answered Mar 21, 2019 at 8:54

Lino

20k6 gold badges55 silver badges73 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Roland · Accepted Answer · 2019-03-21 11:04:50Z

2

If it's just a hex string representation you need (and that's also what I get from your comment), you may just want to use the following instead:

fun StringBuffer.appendHex(b : Byte) = append("%02X".format(b))

Usage is then the same as also Lino has shown:

StringBuffer()
    .append("some text")
    .appendHex(someByte)

or if you need to keep your current signature:

companion object {
  @JvmStatic
  fun appendHex(s : StringBuffer, b : Byte) {
    s.append("%02X".format(b))
  }
}

edited Mar 21, 2019 at 11:04

answered Mar 21, 2019 at 10:22

Roland

23.5k5 gold badges62 silver badges91 bronze badges

2 Comments

Lino Over a year ago

You could also use "%02X".format(b), which does the exact same as String.format()

Roland Over a year ago

@Lino thanks! didn't know that one ;-) but to be honest, I didn't even care so far as I used a static import for the String.format and that was already short enough ;-) but I will remember now ;-)

EpicPandaForce · Accepted Answer · 2019-03-21 08:39:43Z

1

companion object {
    private const val HEX = "XXXXXXXXX"

    @JvmStatic
    private fun appendHex(sb: StringBuffer, b: Int) {
        sb.append(HEX[b shr 4 and 0x0f]).append(HEX[b and 0x0f])
    }
}

Please note that I had to change the Byte to Int, because Byte doesn't seem to have shr and and defined for some unknown reason.

No, I really don't know why bitwise operators would not be defined over byte.

answered Mar 21, 2019 at 8:39

EpicPandaForce

81.7k28 gold badges269 silver badges451 bronze badges

Collectives™ on Stack Overflow

How to convert Java into Kotlin

3 Answers 3

Comments

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related