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I tried to use readInt() to read two integers from the same line but that is not how it works.

val x = readInt() val y = readInt()

With an input of 1 727 I get the following exception at runtime:

Exception in thread "main" java.lang.NumberFormatException: For input string: "1 727" at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) at java.lang.Integer.parseInt(Integer.java:492) at java.lang.Integer.parseInt(Integer.java:527) at scala.collection.immutable.StringLike$class.toInt(StringLike.scala:231) at scala.collection.immutable.StringOps.toInt(StringOps.scala:31) at scala.Console$.readInt(Console.scala:356) at scala.Predef$.readInt(Predef.scala:201) at Main$$anonfun$main$1.apply$mcVI$sp(Main.scala:11) at scala.collection.immutable.Range.foreach$mVc$sp(Range.scala:75) at Main$.main(Main.scala:10) at Main.main(Main.scala)

I got the program to work by using readf but it seems pretty awkward and ugly to me:

 val (x,y) = readf2("{0,number} {1,number}") val a = x.asInstanceOf[Int] val b = y.asInstanceOf[Int] println(function(a,b))

Someone suggested that I just use Java's Scanner class, (Scanner.nextInt()) but is there a nice idiomatic way to do it in Scala?

Edit: My solution following paradigmatic's example:

val Array(a,b) = readLine().split(" ").map(_.toInt)

Followup Question: If there were a mix of types in the String how would you extract it? (Say a word, an int and a percentage as a Double)

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5 Answers 5

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6

If you mean how would you convert val s = "Hello 69 13.5%" into a (String, Int, Double) then the most obvious way is

val tokens = s.split(" ") (tokens(0).toString, tokens(1).toInt, tokens(2).init.toDouble / 100) // (java.lang.String, Int, Double) = (Hello,69,0.135)

Or as mentioned you could match using a regex:

val R = """(.*) (\d+) (\d*\.?\d*)%""".r s match { case R(str, int, dbl) => (str, int.toInt, dbl.toDouble / 100) }

If you don't actually know what data is going to be in the String, then there probably isn't much reason to convert it from a String to the type it represents, since how can you use something that might be a String and might be in Int? Still, you could do something like this:

val int = """(\d+)""".r val pct = """(\d*\.?\d*)%""".r val res = s.split(" ").map { case int(x) => x.toInt case pct(x) => x.toDouble / 100 case str => str } // Array[Any] = Array(Hello, 69, 0.135)

now to do anything useful you'll need to match on your values by type:

res.map { case x: Int => println("It's an Int!") case x: Double => println("It's a Double!") case x: String => println("It's a String!") case _ => println("It's a Fail!") }

Or if you wanted to take things a bit further, you could define some extractors which will do the conversion for you:

abstract class StringExtractor[A] { def conversion(s: String): A def unapply(s: String): Option[A] = try { Some(conversion(s)) } catch { case _ => None } } val intEx = new StringExtractor[Int] { def conversion(s: String) = s.toInt } val pctEx = new StringExtractor[Double] { val pct = """(\d*\.?\d*)%""".r def conversion(s: String) = s match { case pct(x) => x.toDouble / 100 } }

and use:

"Hello 69 13.5%".split(" ").map { case intEx(x) => println(x + " is Int: " + x.isInstanceOf[Int]) case pctEx(x) => println(x + " is Double: " + x.isInstanceOf[Double]) case str => println(str) }

prints

Hello 69 is Int: true 0.135 is Double: true

Of course, you can make the extrators match on anything you want (currency mnemonic, name begging with 'J', URL) and return whatever type you want. You're not limited to matching Strings either, if instead of StringExtractor[A] you make it Extractor[A, B].

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You came up with the extractor idea before I completed writing mine! I pushed the idea a bit further though - hope that illustrate better how extractors really help.
5

You can read the line as a whole, split it using spaces and then convert each element (or the one you want) to ints:

scala> "1 727".split(" ").map( _.toInt ) res1: Array[Int] = Array(1, 727)

For most complex inputs, you can have a look at parser combinators.

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And for moderately complex inputs you can use regex matchers.
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The input you are describing is not two Ints but a String which just happens to be two Ints. Hence you need to read the String, split by the space and convert the individual Strings into Ints as suggested by @paradigmatic.

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6 Comments

In Java you can do Scanner s = new Scanner(); and then scan for the next integer from the input by doing int a = s.nextInt(); int b = s.nextInt();. I assume there is nothing similar in the scala stdlib and that In Scala you're supposed to think of readInt as readNextLineAsInt?
Yes most methods in Console object read a whole line. But if you don't like this approach (and then the one I posted), you can stick to Java scanners: val s = new Scanner; val a = s.nextInt; val b = s.nextInt should also work.
Sure I could use the Java way, but how hard should I try to avoid using Java stuff when writing Scala? While I'm learning? On production code?
It seems to me that there is no need for the Scala team to waste time writing functionality that is already available in the JDK, unless it can be justified by the benefits it brings. In this case, Scanner looks perfectly good for the job
@oxbow_lakes: I agree, but I often feel more comfortable by using Scala stdlib as much as possible, to make my code portable outside the JVM world. (ie. Scala>/LLVM project or Scala to javascript translation).
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One way would be splitting and mapping:

// Assuming whatever is being read is assigned to "input" val input = "1 727" val Array(x, y) = input split " " map (_.toInt)

Or, if you have things a bit more complicated than that, a regular expression is usually good enough.

val twoInts = """^\s*(\d+)\s*(\d+)""".r val Some((x, y)) = for (twoInts(a, b) <- twoInts findFirstIn input) yield (a, b)

There are other ways to use regex. See the Scala API docs about them.

Anyway, if regex patterns are becoming too complicated, then you should appeal to Scala Parser Combinators. Since you can combine both, you don't loose any of regex's power.

import scala.util.parsing.combinator._ object MyParser extends JavaTokenParsers { def twoInts = wholeNumber ~ wholeNumber ^^ { case a ~ b => (a.toInt, b.toInt) } } val MyParser.Success((x, y), _) = MyParser.parse(MyParser.twoInts, input)

The first example was more simple, but harder to adapt to more complex patterns, and more vulnerable to invalid input.

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2

I find that extractors provide some machinery that makes this type of processing nicer. And I think it works up to a certain point nicely.

object Tokens { def unapplySeq(line: String): Option[Seq[String]] = Some(line.split("\\s+").toSeq) } class RegexToken[T](pattern: String, convert: (String) => T) { val pat = pattern.r def unapply(token: String): Option[T] = token match { case pat(s) => Some(convert(s)) case _ => None } } object IntInput extends RegexToken[Int]("^([0-9]+)$", _.toInt) object Word extends RegexToken[String]("^([A-Za-z]+)$", identity) object Percent extends RegexToken[Double]( """^([0-9]+\.?[0-9]*)%$""", _.toDouble / 100)

Now how to use:

List("1 727", "uptime 365 99.999%") collect { case Tokens(IntInput(x), IntInput(y)) => "sum " + (x + y) case Tokens(Word(w), IntInput(i), Percent(p)) => w + " " + (i * p) } // List[java.lang.String] = List(sum 728, uptime 364.99634999999995)

To use for reading lines at the console:

Iterator.continually(readLine("prompt> ")).collect{ case Tokens(IntInput(x), IntInput(y)) => "sum " + (x + y) case Tokens(Word(w), IntInput(i), Percent(p)) => w + " " + (i * p) case Tokens(Word("done")) => "done" }.takeWhile(_ != "done").foreach(println) // type any input and enter, type "done" and enter to finish

The nice thing about extractors and pattern matching is that you can add case clauses as necessary, you can use Tokens(a, b, _*) to ignore some tokens. I think they combine together nicely (for instance with literals as I did with done).

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