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type declarations in 'where' -- what's going on?

While reading the QuickCheck Manual, I came across the following example:

prop_RevRev xs = reverse (reverse xs) == xs
  where types = xs::[Int]

The manual goes on to say:

Properties must have monomorphic types. `Polymorphic' properties, such as the one above, must be restricted to a particular type to be used for testing. It is convenient to do so by stating the types of one or more arguments in a

where types = (x1 :: t1, x2 :: t2, ...)

clause. Note that types is not a keyword; this is just a local declaration which provides a convenient place to restrict the types of x1, x2 etc.

I have never seen such a trick in Haskell before. Here's what I'm really having problems with:

  1. Why does this syntax for type declarations even exist? What can it do for me that the following couldn't?

    prop_RevRev :: [Int] -> Bool
    prop_RevRev xs = reverse (reverse xs) == xs
    
  2. Does this use of where constitute 'special' syntax for type declarations? Or is it consistent and logical (and if so, how?)?

  3. Is this usage standard or conventional Haskell?

like image 205
Matt Fenwick Avatar asked Oct 03 '11 13:10

Matt Fenwick


1 Answers

where is not special syntax for type declarations. For example, this works:

prop_RevRev :: [Int] -> Bool
prop_RevRev xs = ys == xs
  where ys = reverse (reverse xs)

and so does this:

prop_RevRev xs = ys == xs
  where ys = reverse (reverse xs)
        ys :: [Int]

The advantage of where type = (xs :: [Int]) over prop_RevRev :: [Int] -> Bool is that in the latter case you have to specify the return type, while in the former case the compiler can infer it for you. This would matter if you had a non-Boolean property, for example:

prop_positiveSum xs = 0 < length xs && all (0 <) xs ==> 0 < sum xs
  where types = (xs :: [Int])
like image 150
dave4420 Avatar answered Oct 08 '22 10:10

dave4420