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POWhy is Scala's type inference not as powerful as Haskell's?
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copied!<p>The type inference engine of Haskell is much more powerful than Scala's. In Haskell I rarely have to explicitly write the types whereas in Scala the types can only be inferred in expressions but not in method definitions.</p> <p>For example, see following Haskell code snippet:</p> <pre><code>size xs = loop xs 0 where loop [] acc = acc loop (_ : xs) acc = loop xs (acc+1) </code></pre> <p>It returns the size of a List. The Haskell compiler can recognize what types are used and what the function definition is. The equivalent Scala code:</p> <pre><code>def size[A]: List[A] => Int = xs => { def loop: (List[A], Int) => Int = { case (Nil, acc) => acc case (_ :: xs, acc) => loop(xs, acc+1) } loop(xs, 0) } </code></pre> <p>Or with method definitions:</p> <pre><code>def size[A](xs: List[A]) = { def loop(xs: List[A], acc: Int): Int = xs match { case Nil => acc case _ :: xs => loop(xs, acc+1) } loop(xs, 0) } </code></pre> <p>My question is: Why can't I write them like the following?</p> <pre><code>def size = xs => { def loop = { case (Nil, acc) => acc case (_ :: xs, acc) => loop(xs, acc+1) } loop(xs, 0) } </code></pre> <p>Once again with method definitions:</p> <pre><code>def size(xs) = { def loop(xs, acc) = xs match { case Nil => acc case _ :: xs => loop(xs, acc+1) } loop(xs, 0) } </code></pre> <p>Is it because nobody has implemented it yet? Is the type system of Scala not as powerful as needed for this case? Or are there other reasons?</p>

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