Scala：奇怪的类型不匹配错误

Question

Let's say I have a function like this (it's only an example, so do not offer me better ways to create 0,1,2,... style array):

def createArray(size: Int): Array[Int] = {
  for (i <- 0 until size) yield i
}

But the compiler get upset with some mysterious type mismatch error:

(fragment of compare-images.scala):39: error: type mismatch;
 found   : RandomAccessSeq.Projection[Int]
 required: Array[Int]
      for (i <- 0 until size) yield i
            ^
one error found
!!!
discarding <script preamble>

I'm sure, the reason has something to do with the fact that until method's return type is Range , not Array . Yet, why the compiler can't just cast the types? Interestingly the problem goes away when I remove the return type of my function:

def createArray(size: Int) = {
  for (i <- 0 until size) yield i
}

But I want my function to return Array ...

I also have another function, which goes like this:

def screateArray2: Array[Int] = {
  val a = Array(1,2,3,4,5)
  for (i <- a) yield i
}

It compiles without any problems. It yields values very similarly to the first example, but doesn't use until method...

Am I missing something about Scala's type system?

I'm quite new to Scala.

EDIT: I sort of solved my problem like this:

def crop(data: Array[Int]): Array[Int] = (
  for (i <- 0 until data.size) yield i
).toArray

But in my view it's anything but readable...

Answer 1

You can't cast the types just because an Array is not a Range, nor a superclass of it. I think their most common supertype would be IndexedSeq . So your example would be similar to a method which you declare to return an Int, while the implementation would return a String, for instance. Range does have a toArray method, however, so if you want to use the until style and return an Array, you can do it like this:

scala> (for (i <- 0 until 5) yield i).toArray
res0: Array[Int] = Array(0, 1, 2, 3, 4)

or, if do don't like this, there's another way, but that doesn't use the until style:

scala> for (i <- Array.range(0,5)) yield i          
res1: Array[Int] = Array(0, 1, 2, 3, 4)

Also, check this related question for returning custom types from a for comprehension

Answer 2

So, How about this one:

scala> import scala.collection.breakOut
import scala.collection.breakOut

scala> def createSet(size: Int): Set[Int] = (
     |     for(i <- 0 until size) yield i
     | )(breakOut)
createSet: (size: Int)Set[Int]

scala> def createList(size: Int): List[Int] = (
     |     for(i <- 0 until size) yield i
     | )(breakOut)
createList: (size: Int)List[Int]

scala> def createArray(size: Int): Array[Int] = (
     |     for(i <- 0 until size) yield i
     | )(breakOut)
createArray: (size: Int)Array[Int]

scala> createArray(10)
res3: Array[Int] = Array(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)

scala> createList(5)
res4: List[Int] = List(0, 1, 2, 3, 4)

scala> createSet(4)
res5: Set[Int] = Set(0, 1, 2, 3)

scala>

Answer 3

Arjan's solution can be shortened to:

 def createArray(size: Int) = (0 until size).toArray

[Edit] Of course you can remove values you don't need before you create the array

 def createArray(size: Int) = (0 until size).filter(LOTSOFINTERESTINGSTUFF).toArray