scala - restricted way of converting string to number

Question

I have a string variable that needs to be parsed to a double

Code:

val myString = "9d"
myString.toDouble // produces "9.0" which i don't want. this should throws an error

How do I make my string to be interpreted literally so that it produces an exception?

Interesting question! This applies to pure Java too. One solution: BigDecimal(myString).toDouble — ygor
– ygor, Commented Nov 22, 2018 at 11:01

ygor · Accepted Answer · 2018-11-22 11:36:15Z

3

So, apparently, Java's java.lang.Double#parseDouble is able to parse all kinds of strings, e.g. NaN, Infinity and other.

The answer to your problem seems to be BigDecimal(myString).toDouble

Out of curiosity, here is, what three different approaches return for various kinds of input strings:

  def main(args: Array[String]): Unit = {
    val strings = Seq("Infinity", "NaN", "9.0d", "9d", "9f", "9.0", "9.1", "1.4e14")
    val parsers = Seq(
      ("Double", (s: String) => s.toDouble),
      ("BigDecimal", (s: String) => BigDecimal(s).toDouble),
      ("NumberFormat", (s: String) => NumberFormat.getNumberInstance.parse(s).doubleValue()),
    )

    for (string <- strings) {
      println(s"\n------------- $string ------------")
      for ((name, parser) <- parsers) {
        val result = Try(parser(string)) match {
          case scala.util.Success(value) => value
          case scala.util.Failure(ex) => ex.toString
        }
        println(name.reverse.padTo(20, " ").reverse.mkString + " -> " + result)
      }

    }
  }

Result:

------------- Infinity ------------
              Double -> Infinity
          BigDecimal -> java.lang.NumberFormatException
        NumberFormat -> java.text.ParseException: Unparseable number: "Infinity"

------------- NaN ------------
              Double -> NaN
          BigDecimal -> java.lang.NumberFormatException
        NumberFormat -> java.text.ParseException: Unparseable number: "NaN"

------------- 9.0d ------------
              Double -> 9.0
          BigDecimal -> java.lang.NumberFormatException
        NumberFormat -> 9.0

------------- 9d ------------
              Double -> 9.0
          BigDecimal -> java.lang.NumberFormatException
        NumberFormat -> 9.0

------------- 9f ------------
              Double -> 9.0
          BigDecimal -> java.lang.NumberFormatException
        NumberFormat -> 9.0

------------- 9.0 ------------
              Double -> 9.0
          BigDecimal -> 9.0
        NumberFormat -> 9.0

------------- 9.1 ------------
              Double -> 9.1
          BigDecimal -> 9.1
        NumberFormat -> 9.1

------------- 1.4e14 ------------
              Double -> 1.4E14
          BigDecimal -> 1.4E14
        NumberFormat -> 1.4

edited Nov 22, 2018 at 11:36

answered Nov 22, 2018 at 11:12

ygor

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2 Comments

Kolmar Over a year ago

The second one doesn't work as desired: val n = NumberFormat.getNumberInstance.parse("9d") results in n: Number = 9

Kolmar Over a year ago

For the second one to work, you'd need something like:

val s = "9d"; val pos = new ParsePosition(0); val n = NumberFormat.getNumberInstance.parse(s, pos); if (pos.getIndex != s.length) throw new NumberFormatException

pme · Accepted Answer · 2018-11-22 11:08:24Z

0

I found a solution with scalaz:

import scalaz.Scalaz._

println("34b".parseDouble) // Failure(java.lang.NumberFormatException: For input string: "34b")

println("34.4".parseDouble) // Success(34.4)

answered Nov 22, 2018 at 11:08

pme

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1 Comment

Kolmar Over a year ago

No, that still doesn't work: "9d".parseDouble -> Success(9.0)

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