Long to hex kotlin

[Solved]-android, kotlin convert Long to hex string-kotlin

So if you don’t plan on having to format values over 2^63 — 1, use Kotlin’s function, otherwise use Java’s. A third option would be to use ULong.toString if you want to avoid using Java methods.

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How to convert Int to Hex String in Kotlin?

Solution 1: You can still use the Java conversion by calling the static function on : And, starting with Kotlin 1.1, there is a function in the Kotlin standard library that does the conversion, too: Note, however, that this will still be different from , because the latter performs the unsigned conversion: But with experimental Kotlin unsigned types, it is now possible to get the same result from negative number unsigned conversion as with : Solution 2: You can simply do it like this: Solution 3: If you need to add zero before bytes which less than 10(hex), for example you need string — «0E» then use: Question: i’m still new to kotlin so take my question with a grain of salt so i’ve made an example that hopefully will help people hope this helped someone Question: I am working on a console application in Kotlin where I accept multiple arguments in function I want to check whether the is a valid integer and convert the same or else I have to throw some exception.

How to convert Int to Hex String in Kotlin?

I’m looking for a similar function to Java’s Integer.toHexString() in Kotlin. Is there something built-in, or we have to manually write a function to convert Int to String ?

You can still use the Java conversion by calling the static function on java.lang.Integer :

val hexString = java.lang.Integer.toHexString(i) 

And, starting with Kotlin 1.1, there is a function in the Kotlin standard library that does the conversion, too:

fun Int.toString(radix: Int): String

Returns a string representation of this Int value in the specified radix .

Note, however, that this will still be different from Integer.toHexString() , because the latter performs the unsigned conversion:

println((-50).toString(16)) // -32 println(Integer.toHexString(-50)) // ffffffce 

But with experimental Kotlin unsigned types, it is now possible to get the same result from negative number unsigned conversion as with Integer.toHexString(-50) :

println((-50).toUInt().toString(16)) // ffffffce 

You can simply do it like this: «%x».format(1234)

If you need to add zero before bytes which less than 10(hex), for example you need string — «0E» then use: «%02x».format(14)

Best Way to Convert ArrayList to String in Kotlin, Kotlin as well has method for that, its called joinToString.. You can simply call it like this: list.joinToString()); Because by default it uses comma as separator but you can also pass your own separator as parameter, this method takes quite a few parameters aside from separator, which allow to do a lot of …

Kotlin: why would you want to convert Int.toString?

i’m still new to kotlin so take my question with a grain of salt

so i’ve been learning about kotlin and in one of the articles i was reading about had this code as an example of how to use .toString

val sum1 = < a: Int, b: Int ->val num = a + b num.toString() //convert Integer to String > fun main(args: Array)

The sum of two numbers is: 5 

as i understand it, it’s a «5» but of type string.. but why??

we can just simplify the code and get the same outcome:

 val sum1 = < a: Int, b: Int ->a + b > val result1 = sum1(2,3) println("The sum of two numbers is: $result1") 

The sum of two numbers is: 5 

so.. i’ve seen examples of how to use it, but i still didn’t find anything to explain why you’d want to use it

.toString() returns a string representation of the object. It is not just restricted to converting Int to String . You can use it to convert other data type like Boolean to String .

So whenever a a developer need to store Int as a String , he/she can cast Int to String with the help of .toString() .

The above code snippet shared by you is way generic, hence you got confused.

To summarize, whenever you want to store/map any data-type into String, then use .toString() .

everyone talking about «how you can use it» but still no one saying «why you’d want to use it»

so i’ve made an example that hopefully will help people

fun main() < inference() >fun inference() < val sum1 = < a: Int, b: Int ->val num = a + b // convert Integer to String num.toString() > val result1 = sum1(2, 3) println("The sum of the two numbers is: $result1") // compiler error: can't use "result1" because it's a String and not an Int //val result2 = sum1(result1, 3) // let's say your code only takes strings but you want to insert an Int val sum2 = val str = a + b str > var int: Int? = 10 //int = null // uncomment this line to get the other outcome if(int != null) < val result3 = sum2("the number we got is: ", int.toString()) println(result3) >else < val result4 = sum2("we did not get any number, ", " check if int is null") println(result4) >> 

Convert String obtained from edittext to Integer in Kotlin, You can use .toInt (): val myNumber: Int = «25».toInt () Note that it throws a NumberFormatException if the content of the String is not a valid integer. If you don’t like this behavior, you can use .toIntOrNull () instead (since Kotlin 1.1):

How to convert String to Int in Kotlin?

I am working on a console application in Kotlin where I accept multiple arguments in main() function

I want to check whether the String is a valid integer and convert the same or else I have to throw some exception.

You could call toInt() on your String instances:

fun main(args: Array) < for (str in args) < try < val parsedInt = str.toInt() println("The parsed int is $parsedInt") >catch (nfe: NumberFormatException) < // not a valid int >> > 

Or toIntOrNull() as an alternative:

If you don’t care about the invalid values, then you could combine toIntOrNull() with the safe call operator and a scope function, for example:

Actually, there are several ways:

var numberString : String = "numberString" // number is the int value of numberString (if any) var defaultValue : Int = defaultValue 

Operation	Numeric value	Non-numeric value
numberString.toInt()	number	NumberFormatException
numberString.toIntOrNull()	number	null
numberString.toIntOrNull() ?: defaultValue	number	defaultValue

If numberString is a valid integer, we get number , else see a result in column Non-numeric value .

Keep in mind that the result is nullable as the name suggests.

As suggested above, use toIntOrNull() .

Parses the string as an [Int] number and returns the result or null if the string is not a valid representation of a number.

val a = "11".toIntOrNull() // 11 val b = "-11".toIntOrNull() // -11 val c = "11.7".toIntOrNull() // null val d = "11.0".toIntOrNull() // null val e = "abc".toIntOrNull() // null val f = null?.toIntOrNull() // null 

Convert Long to String in Kotlin, Format in Kotlin string templates. 219. How to convert List to Map in Kotlin? 497. Converting int values in Java to human readable strings more hot questions Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader.

Kotlin String to Int or zero (default value)

How can I covert String to Int in Kotlin and if it can’t be then return 0 (default value).

I think the best solution is to tell value is Int and use Elvis operator to assign value 0 if it can’t be converted.

val a:String="22" val b:Int = a.toIntOrNull()?:0//22 val c:String="a" val d:Int = c.toIntOrNull()?:0//0 

To make code more concise you can create Extension Function

fun String?.toIntOrDefault(default: Int = 0): Int

Java — Extract numbers from a string kotlin, For some cases like if the text contains formatted amount, then the above answers might fail. For example, if the text is «Rs. 1,00,000.00» and if we only filter . and digit, then the output we get is .100000.00 this will be a wrong amount.So better first extract the string between digits and then filter for digit …

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android, kotlin преобразовать длинную строку в шестнадцатеричную

Преобразование java-кода в kotlin, кажется, что kotlin имеет Long.toString(16) , но не уверен, что лучше продолжать звонить в java, какие предложения лучше?

java.lang.Long.toHexString(theLong) 

which: public static String toHexString(long i) < return toUnsignedString0(i, 4); >static String toUnsignedString0(long val, int shift) < // assert shift >0 && shift static int formatUnsignedLong(long val, int shift, char[] buf, int offset, int len) < int charPos = len; int radix = 1 >>= shift; > while (val != 0 && charPos > 0); return charPos; > 

public actual inline fun Long.toString(radix: Int): String = java.lang.Long.toString(this, checkRadix(radix)) which: public static String toString(long i, int radix) < if (radix < Character.MIN_RADIX || radix >Character.MAX_RADIX) radix = 10; if (radix == 10) return toString(i); char[] buf = new char[65]; int charPos = 64; boolean negative = (i < 0); if (!negative) < i = -i; >while (i buf[charPos] = Integer.digits[(int)(-i)]; if (negative) < buf[--charPos] = '-'; >return new String(buf, charPos, (65 - charPos)); > 

1 ответ

• kotlin.Long.toString(radix: Int) подписан, что означает, что он вернет строку с отрицательным знаком для отрицательных значений (или значений более 2^63 — 1, если мы думаем в терминах чисел без знака).
• Long.toHexString(long i) без знака, вы никогда не увидите префикс со знаком минус. Этот метод также оптимизирован для степени двойки с использованием операторов сдвига вместо деления, поэтому производительность может быть немного лучше. Однако это должно быть буквально последней из ваших проблем.

Поэтому, если вы не планируете форматировать значения более 2^63 — 1, используйте функцию Kotlin, в противном случае используйте Java. Третий вариант — использовать ULong.toString если вы не хотите использовать методы Java.

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