scala - Initializing an array so that each element of it would be equal to its index -

I want to initialize an array so that each element is equal to its index:

  Def method1 (obj: AnyRef) = {// .... if (obj.isInstanceOf [Array [int]] {val arr1 = obj.asInstanceOf [Array [int]] val arr2: Array [Int] = New Array [int] (arr1.length) // initial2 arr2. how? Arr2 / arr2 [0] = 0, arr2 [1] = 1, etc ....}    
 
 
 You can use  range  to  array .        P> 
  var arr2 = arr1.indices.toArray  method  index  / code>  
 if you have an array and fill it out You can use the  copyToArray  method:  
  (up to 0 arr2.length) .copyToArray (arr2)   
 There is no performance difference between this method:  toArray  method  copyToArray . Only <3> code  field:  start ,  expiration  and  phase  
  range  , then there is almost no memory overhead.  
 There is also  category  in the object:  
  val arr2 = Array.range (0, Arr1.length) // contains 0, Arr1.length is not included   
 There are some other methods to fill the array even before. This method is not very useful in this case.  
 method  fill :  
  var i = -1 val arr2 = array.phil (arr1 .length) ({i + = 1; I})   
 method  apply :  
  val arr2 = Array.apply (0, (up to 1 arr1.length ): _ *) // Array (0) on empty arr1   
 If you have not stored the  array  and you want to change your elements and You can use the  collection.breakOut  method to get the  array :  
  // It is not like that case arr arr2: Array [int] = (up to 0 arr1.length) .map (intro) (breakout)