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    <p>It seems Scala is doing something different from Java when it comes to varargs. At least, I can't get that array shuffled any way I try. Supposedly, the shuffle on the list would shuffle the array because the list is array-backed. Well, it seems that Scala will create a <em>new</em> array when passing vararg arguments to Java, therefore making the aforementioned example useless.</p> <pre><code>scala&gt; val b = java.util.Arrays.asList(a: _*) b: java.util.List[java.lang.String] = [a, b, c] scala&gt; java.util.Collections.shuffle(b); println(a.toString+" "+b.toString) Array(a, b, c) [a, b, c] scala&gt; java.util.Collections.shuffle(b); println(a.toString+" "+b.toString) Array(a, b, c) [c, b, a] scala&gt; java.util.Collections.shuffle(b); println(a.toString+" "+b.toString) Array(a, b, c) [a, c, b] scala&gt; java.util.Collections.shuffle(b); println(a.toString+" "+b.toString) Array(a, b, c) [b, a, c] scala&gt; java.util.Collections.shuffle(b); println(a.toString+" "+b.toString) Array(a, b, c) [a, b, c] scala&gt; java.util.Collections.shuffle(b); println(a.toString+" "+b.toString) Array(a, b, c) [c, a, b] </code></pre> <p>It does works with Ints, despite the claim otherwise, though:</p> <pre><code>scala&gt; val a = Array(1,2,3) a: Array[Int] = Array(1, 2, 3) scala&gt; val b = java.util.Arrays.asList(a: _*) b: java.util.List[Int] = [1, 2, 3] scala&gt; java.util.Collections.shuffle(b); println(a.toString+" "+b.toString) Array(1, 2, 3) [2, 3, 1] scala&gt; java.util.Collections.shuffle(b); println(a.toString+" "+b.toString) Array(1, 2, 3) [3, 2, 1] scala&gt; java.util.Collections.shuffle(b); println(a.toString+" "+b.toString) Array(1, 2, 3) [3, 2, 1] scala&gt; java.util.Collections.shuffle(b); println(a.toString+" "+b.toString) Array(1, 2, 3) [1, 2, 3] </code></pre> <p>On Scala 2.8, there's a simpler way:</p> <pre><code>scala&gt; scala.util.Random.shuffle(a) res32: Sequence[Int] = Array(1, 2, 3) scala&gt; scala.util.Random.shuffle(a) res33: Sequence[Int] = Array(2, 1, 3) scala&gt; scala.util.Random.shuffle(a) res34: Sequence[Int] = Array(3, 2, 1) </code></pre> <p>Note that, in a Scala fashion, it does not change the original array.</p>
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    1. POHow to use Java Collections.shuffle() on a Scala array?
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    1. USDaniel C. Sobral
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    1. CODaniel, if I do this: val a = Array("a", "b", "c") java.util.Collections.shuffle(java.util.Arrays.asList(a: _*)) then my Scala array does get shuffled. And if I make a an array of Ints instead of Strings, I get a ClassCastException.
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      1. USJesper
    2. COAnd thanks for the tip about Scala 2.8, so there will be a shuffle() method in the library in 2.8. Ofcourse it returns a new array, that's the typical functional way of programming.
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    3. CO@Daniel I think you are using 2.8.0 repl. Try your examples in 2.7.5 and you will see differences from your results. Apparently, 2.8.0 generates a new Seq when it sees ":_*" but 2.7.5 just wraps a Seq around the array.
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      1. USWalter Chang
 

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