在Haskell中无法将类型“ a”与“ Int”匹配

Question

place nx is meant to find the place of integer n in the list x , for example place 2 [1,2,3] will return 1 :

place :: Int -> [a] -> Int
place n x = length $ takeWhile (/=n) x

But it gets the error Couldn't match type 'a' with 'Int'

Why? takeWhile should return a list and its length is an integer, hence place should output an Int eventually.

Answer 1

The correct type signature for place is:

place :: Int -> [Int] -> Int

When you use [a] in place of [Int] you are saying that place will work on a list of any type. For instance, I could call it like this:

place 3 [True, False, False, True]

or call it with a list of Char:

place 4 "this is a test"

But neither of these can work because takeWhile will compare each element of the list against the Int n . This forces the list to be a list of Ints.

Answer 2

You are using (/=) :: Eq ty => ty -> ty -> Bool where that second ty ~ Int due to the type of n . If ty is Int then the type of the other argument of (/=) ( x ) must be Int too.

Answer 3

find the place of integer n in the list x

ah, but there are no integers in the list at all. The type is [a] . Only if the type were [Int] would it be possible to find an n in it.

Now, you might say “ [Int] is a possible instantiation of [a] , so it could still work in that case”. Well, it could – but only in that case, and therefore it would be stupid not to enforce a ~ Int in the type signature.

More reasonably, you might want to find any type of number which happens to be equal to the integer n . Well, you can do that:

place' :: (Num a, Eq a) => Int -> [a] -> Int
place' n = length . takeWhile (/= fromIntegral n)

Here, I first convert n to the type of number that's contained in the list, and then compare it with those numbers.