How to convert any number to long

Question

Suppose I have:

val number:AnyVal

and I know that x can be any number (for our purposes, Float, Double, Int, Long).

What is the easiest way to convert that number to Long:

val l = number.toLong   //fails for AnyVal

Answer 1

What about:

scala> import scala.util.Try
import scala.util.Try

scala> val i1: Int = 23
i1: Int = 23

scala> val l1: Long = 42
l1: Long = 42

scala> val f1: Float = 14.9f
f1: Float = 14.9

scala> val d1: Double = 14.96
d1: Double = 14.96

scala> val b1: Boolean = true
b1: Boolean = true

scala> List(i1, l1, f1, d1, b1) map (x => Try(x.asInstanceOf[Number].longValue)) foreach (println(_))
Success(23)
Success(42)
Success(14)
Success(14)
Failure(java.lang.ClassCastException: java.lang.Boolean cannot be cast to java.lang.Number)

scala> List(i1, l1, f1, d1, b1) map (x => Try(x.asInstanceOf[Number].longValue)) foreach (n => println(n.get))
23
42
14
14
java.lang.ClassCastException: java.lang.Boolean cannot be cast to java.lang.Number
    at $anonfun$1$$anonfun$apply$1.apply$mcJ$sp(<console>:14)
    at $anonfun$1$$anonfun$apply$1.apply(<console>:14)
    at $anonfun$1$$anonfun$apply$1.apply(<console>:14)
    at scala.util.Try$.apply(Try.scala:161)
    at $anonfun$1.apply(<console>:14)
    at $anonfun$1.apply(<console>:14)
    at scala.collection.TraversableLike$$anonfun$map$1.apply(TraversableLike.scala:244)
    at scala.collection.TraversableLike$$anonfun$map$1.apply(TraversableLike.scala:244)
    at scala.collection.immutable.List.foreach(List.scala:318)
    at scala.collection.TraversableLike$class.map(TraversableLike.scala:244)
    at scala.collection.AbstractTraversable.map(Traversable.scala:105)
    at .<init>(<console>:14)

Answer 2

You can direct it to the number if you know that it will definitely be Float, Double, Int or Long. Then you can call longValue:

val number:AnyVal = 10
val l:Long = number.asInstanceOf[Number].longValue

Answer 3

, .

Answer 4

Scala 2.11

Scala, RichLong .

Number, Scala 2.11:

object LongNumber {
  def cast(number: Any): Long = number match {
    case n: Number => n.longValue()
    case x         => throw new IllegalArgumentException(s"$x is not a number.")
  }

  // Test cases
  def main(args: Array[String]): Unit = {
    val twelveByte:     Byte   = 0x0c
    val twelveString:   String = "12"

    println(s"Converting a long:   ${cast(12L)}")
    println(s"Converting an int:   ${cast(12)}")
    println(s"Converting a double: ${cast(12.0)}")
    println(s"Converting a byte:   ${cast(twelveByte)}")
    println(s"Converting a string: $twelveString")
  }
}

- , .

Original answer for older versions of Scala

Trying to implicitly convert to RichLong in a block of matches seems very nice:

import scala.runtime.RichLong

...

  def cast(number: Any): Long = number match {
    case n: RichLong => n.toLong
    case x => throw new IllegalArgumentException(s"$x is not a number.")
  }

It is also possible to add a case for matching strings in numerical format if you want to satisfy this possibility.