Printing array in Scala

Question

I am having problem with most basic Scala operation and it is making me crazy.

val a = Array(1,2,3)

println(a)   and result is [I@1e76345

println(a.toString()) and result is [I@1e76345

println(a.toString) and result is [I@1e76345

Can anyone tell me how to print array without writing my own function for doing that because that is silly. Thanks!

Answer 1

mkString will convert collections (including Array ) element-by-element to string representations.

println(a.mkString(" "))

is probably what you want.

Answer 2

You can do the normal thing (see either Rex's or Jiri's answer), or you can:

scala> Array("bob","sue")
res0: Array[String] = Array(bob, sue)

Hey, no fair! The REPL printed it out real nice.

scala> res0.toString
res1: String = [Ljava.lang.String;@63c58252

No joy, until:

scala> runtime.ScalaRunTime.stringOf(res0)
res2: String = Array(bob, sue)

scala> runtime.ScalaRunTime.replStringOf(res0, res0.length)
res3: String = 
"Array(bob, sue)
"

scala> runtime.ScalaRunTime.replStringOf(res0, 1)
res4: String = 
"Array(bob)
"

I wonder if there's a width setting in the REPL. Update: there isn't. It's fixed at

val maxStringElements = 1000  // no need to mkString billions of elements

But I won't try billions:

scala> Array.tabulate(100)(identity)
res5: Array[Int] = Array(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99)

scala> import runtime.ScalaRunTime.replStringOf
import runtime.ScalaRunTime.replStringOf

scala> replStringOf(res5, 10)
res6: String = 
"Array(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
"

scala> res5.take(10).mkString(", ")
res7: String = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

Wait, let's make that:

scala> res5.take(10).mkString("Array(", ", ", ")")
res8: String = Array(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)

This might be obvious:

scala> var vs = List("1")
vs: List[String] = List(1)

scala> vs = null
vs: List[String] = null

scala> vs.mkString
java.lang.NullPointerException

So instead:

scala> import runtime.ScalaRunTime.stringOf
import runtime.ScalaRunTime.stringOf

scala> stringOf(vs)
res16: String = null

Also, an array doesn't need to be deep to benefit from its stringPrefix:

scala> println(res0.deep.toString)
Array(bob, sue)

Whichever method you prefer, you can wrap it up:

implicit class MkLines(val t: TraversableOnce[_]) extends AnyVal { 
  def mkLines: String = t.mkString("", EOL, EOL)
  def mkLines(header: String, indented: Boolean = false, embraced: Boolean = false): String = { 
    val space = "\u0020"
    val sep = if (indented) EOL + space * 2 else EOL
    val (lbrace, rbrace) = if (embraced) (space + "{", EOL + "}") else ("", "")
    t.mkString(header + lbrace + sep, sep, rbrace + EOL)
  } 
} 

But arrays will need a special conversion because you don't get the ArrayOps:

implicit class MkArrayLines(val a: Array[_]) extends AnyVal {
  def asTO: TraversableOnce[_] = a
  def mkLines: String = asTO.mkLines
  def mkLines(header: String = "Array", indented: Boolean = false, embraced: Boolean = false): String =
    asTO.mkLines(header, indented, embraced)
}

scala> Console println Array("bob","sue","zeke").mkLines(indented = true)
Array
  bob
  sue
  zeke

Answer 3

Here are two methods.

One is to use foreach :

val a = Array(1,2,3)
a.foreach(println)

The other is to use mkString :

val a = Array(1,2,3)
println(a.mkString(""))

Answer 4

If you use list instead, toString() method prints the actual elenents (not the hashCode)

var a = List(1,2,3)
println(a)

or

var a = Array(1,2,3)
println(a.toList)

Answer 5

Rather than manually specifying all the parameters for mkString yourself (which is a bit more verbose if you want to add start and end markers in addition to the delimiter) you can take advantage of the WrappedArray class, which uses mkString internally . Unlike converting the array to a List or some other data structure, the WrappedArray class just wraps an array reference, it's created in effectively constant time.

scala> val a = Array.range(1, 10)                
a: Array[Int] = Array(1, 2, 3, 4, 5, 6, 7, 8, 9)

scala> println(a)                               
[I@64a2e69d                                     

scala> println(x: Seq[_]) // implicit                      
WrappedArray(a, b, c, d)                        

scala> println(a.toSeq)   // explicit                        
WrappedArray(1, 2, 3, 4, 5, 6, 7, 8, 9)         

Answer 6

For a simple Array of Ints like this, we can convert to a Scala List ( scala.collection.immutable.List ) and then use List.toString() :

var xs = Array(3,5,9,10,2,1)
println(xs.toList.toString)
// => List(3, 5, 9, 10, 2, 1)
println(xs.toList)
// => List(3, 5, 9, 10, 2, 1)

If you can convert to a List earlier and do all your operations with Lists, then you'll probably end up writing more idiomatic Scala, written in a functional style.

Note that using List.fromArray is deprecated (and has been removed in 2.12.2) .

Answer 7

The method deep in ArrayLike recursively converts multidimensional arrays to WrappedArray , and overwrites a long prefix "WrappedArray" with "Array".

def deep: scala.collection.IndexedSeq[Any] = new scala.collection.AbstractSeq[Any] with scala.collection.IndexedSeq[Any] {
  def length = self.length
  def apply(idx: Int): Any = self.apply(idx) match {
    case x: AnyRef if x.getClass.isArray => WrappedArray.make(x).deep
    case x => x
  }
  override def stringPrefix = "Array"
}

Usage:

scala> val arr = Array(Array(1,2,3),Array(4,5,6))
arr: Array[Array[Int]] = Array(Array(1, 2, 3), Array(4, 5, 6))

scala> println(arr.deep)
Array(Array(1, 2, 3), Array(4, 5, 6))