![]() Declaring Byte in Kotlin does compile-time error 'The integer literal does not conform to the expected type Byte'.How to find out the latest version of Kotlin in Android Studio.How to omit the constructor parameter with a default value when calling Kotlin in Java?.Kotlin data class: how to read the value of property if I don't know its name at compile time?.How Kotlin converts Long to Int? Getting different outputs for out of range values.How do I step through a fixed range in a loop in Kotlin where the step value changes on every iteration.Kotlin long literal: The value is out of range.Kotlin's hexadecimal literals don't represent negative integers, so 0xaaaaaaaaaaaaaaaaL is out of range, because it represents 12297829382473034410, rather than -6148914691236517206. this is the same hex you get when you translate to Kotlin! You can see that this is the case in Java by printing the number in hex: long a = 0xaaaaaaaaaaaaaaaaL So in Java, 0xaaaaaaaaaaaaaaaaL doesn't represent the positive number 12297829382473034410Īfter all, Long in both languages only has the max value of 9223372036854775807 (Note that decimal literals in Java OTOH can only represent non-negative integers). It is a compile-time error if a hexadecimal, octal, or binary long literal does not fit in 64 bits. From the JLS 3.10.1:Ī hexadecimal numeral consists of the leading ASCII characters 0x or 0X followed by one or more ASCII hexadecimal digits interspersed with underscores, and can represent a positive, zero, or negative integer. Val java_decode10 = reason why that literal works in Java is because Java's hexadecimal integer literals can represent negative numbers, and 0xaaaaaaaaaaaaaaaaL just so happens to fit 64 bits. Val java_parseLong10Radix = ("10", 8) // 8 as "octal" parsing is forced Val java_parseLong010Radix = ("010", 8) // 8 as "octal" parsing is forced Val kotlin_toLong10Radix = "10".toLong(8) // 8 as "octal" parsing is forced Val kotlin_toLong010Radix = "010".toLong(8) // 8 as "octal" parsing is forced Val java_parseLong10 = ("10") // 10 as parsed as decimal Val java_parseLong010 = ("010") // 10 as parsed as decimal Val kotlin_toLong10 = "10".toLong() // 10 as parsed as decimal ![]() val kotlin_toLong010 = "010".toLong() // 10 as parsed as decimal That means that decode can parse Strings like "0x412", where other methods will result in a NumberFormatException. (Sign) DecimalNumeral | (Sign) 0x HexDigits | (Sign) 0X HexDigits | (Sign) # HexDigits | (Sign) 0 OctalDigits Numbers given by the following grammar: DecodableString: ![]() The characters in the string must all be digits of the specified radix …Īnd here comes (String) into the picture:ĭecodes a String into a Long. Parses the string argument as a signed long in the radix specified by Kotlin’s String.toLong(radix: Int) is equivalent to Java’s e Long.parseLong(String, int): Radix 10 were given as arguments to the parseLong(, int) method. Resulting long value is returned, exactly as if the argument and the Parses the string argument as a signed decimal long. ![]() Kotlin’s String.toLong() is equivalent to Java’s Long.parseLong(String): One good old Java possibility what’s not mentioned in the answers is (String). ![]()
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