Question or issue of Kotlin Programming:
So, due to lack of methods like Long.valueOf(String s) I am stuck.
How to convert String to Long in Kotlin?
How to solve this issue?
Solution no. 1:
1. string.toLong()
Parses the string as a [Long] number and returns the result.
@throws NumberFormatException if the string is not a valid
representation of a number.
2. string.toLongOrNull()
Parses the string as a [Long] number and returns the result or null
if the string is not a valid representation of a number.
3. str.toLong(10)
Parses the string as a [Long] number and returns the result.
@throws NumberFormatException if the string is not a valid
representation of a number.
@throws IllegalArgumentException when
[radix] is not a valid radix for string to number conversion.
public inline fun String.toLong(radix: Int): Long = java.lang.Long.parseLong(this, checkRadix(radix))
4. string.toLongOrNull(10)
Parses the string as a [Long] number and returns the result or null
if the string is not a valid representation of a number.
@throws IllegalArgumentException when [radix] is not a valid radix for string
to number conversion.
public fun String.toLongOrNull(radix: Int): Long? {...}
5. java.lang.Long.valueOf(string)
public static Long valueOf(String s) throws NumberFormatException
Solution no. 2:
String has a corresponding extension method:
"10".toLong()
Solution no. 3:
Extension methods are available for Strings to parse them into other primitive types. Examples below:
"true".toBoolean()"10.0".toFloat()"10.0".toDouble()"10".toByte()"10".toShort()"10".toInt()"10".toLong()
Solution no. 4:
Note: Answers mentioning jet.String are outdated. Here is current Kotlin (1.0):
Any String in Kotlin already has an extension function you can call toLong(). Nothing special is needed, just use it.
All extension functions for String are documented. You can find others for standard lib in the api reference
Solution no. 5:
Actually, 90% of the time you also need to check the ‘long’ is valid, so you need:
"10".toLongOrNull()
There is an ‘orNull’ equivalent for each ‘toLong’ of the basic types, and these allow for managing invalid cases with keeping with the Kotlin? idiom.
Solution no. 6:
It’s interesting. Code like this:
val num = java.lang.Long.valueOf("2");
println(num);
println(num is kotlin.Long);
makes this output:
2 true
I guess, Kotlin makes conversion from java.lang.Long and long primitive to kotlin.Long automatically in this case. So, it’s solution, but I would be happy to see tool without java.lang package usage.
Solution no. 7:
string.toLong()
where string is your variable.
Solution no. 8:
One good old Java possibility what’s not mentioned in the answers is java.lang.Long.decode(String).
Decimal Strings:
Kotlin’s String.toLong() is equivalent to Java’s Long.parseLong(String):
Parses the string argument as a signed decimal long. … The
resulting long value is returned, exactly as if the argument and the
radix 10 were given as arguments to the parseLong(java.lang.String, int) method.
Non-decimal Strings:
Kotlin’s String.toLong(radix: Int) is equivalent to Java’s eLong.parseLong(String, int):
Parses the string argument as a signed long in the radix specified by
the second argument. The characters in the string must all be digits of the specified radix …
And here comes java.lang.Long.decode(String) into the picture:
Decodes a String into a Long. Accepts decimal, hexadecimal, and octal
numbers given by the following grammar: DecodableString:
(Sign) DecimalNumeral | (Sign) 0x HexDigits | (Sign) 0X HexDigits | (Sign) # HexDigits | (Sign) 0 OctalDigits
Sign: – | +
That means that decode can parse Strings like "0x412", where other methods will result in a NumberFormatException.
val kotlin_toLong010 = "010".toLong() // 10 as parsed as decimal val kotlin_toLong10 = "10".toLong() // 10 as parsed as decimal val java_parseLong010 = java.lang.Long.parseLong("010") // 10 as parsed as decimal val java_parseLong10 = java.lang.Long.parseLong("10") // 10 as parsed as decimal val kotlin_toLong010Radix = "010".toLong(8) // 8 as "octal" parsing is forced val kotlin_toLong10Radix = "10".toLong(8) // 8 as "octal" parsing is forced val java_parseLong010Radix = java.lang.Long.parseLong("010", 8) // 8 as "octal" parsing is forced val java_parseLong10Radix = java.lang.Long.parseLong("10", 8) // 8 as "octal" parsing is forced val java_decode010 = java.lang.Long.decode("010") // 8 as 0 means "octal" val java_decode10 = java.lang.Long.decode("10") // 10 as parsed as decimal
Solution no. 9:
To convert a String to Long (that represents a 64-bit signed integer) in Kotlin 1.3 is quite simple.
You can use any of the following three methods:
val number1: Long = "789".toLong() println(number1) // 789 val number2: Long? = "404".toLongOrNull() println("number = $number2") // number = 404 val number3: Long? = "Error404".toLongOrNull() println("number = $number3") // number = null val number4: Long? = "111".toLongOrNull(2) println("numberWithRadix(2) = $number4") // numberWithRadix(2) = 7
Solution no. 10:
If you don’t want to handle NumberFormatException while parsing
var someLongValue=string.toLongOrNull() ?: 0
Solution no. 11:
Actually, there are several ways:
Given:
var numberString : String = "numberString" // number is the Long value of numberString (if any) var defaultValue : Long = defaultValue
Then we have:
+—————————————————————————————————————————————+——————————+———————————————————————+ | numberString is a valid number ? | true | false | +—————————————————————————————————————————————+——————————+———————————————————————+ | numberString.toLong() | number | NumberFormatException | +—————————————————————————————————————————————+——————————+———————————————————————+ | numberString.toLongOrNull() | number | null | +—————————————————————————————————————————————+——————————+———————————————————————+ | numberString.toLongOrNull() ?: defaultValue | number | defaultValue | +—————————————————————————————————————————————+——————————+———————————————————————+
