CODEWITHSUNDEEP
haskellprefix
user2842807
user2842807

Reputation: 23

Inferred type is not general enough

I'm wondering if a list is a prefix of a second list, using the following code:

prefix :: [a] -> [b] -> Bool
prefix [] _ = True
prefix _ [] = False
prefix (x:xs) (y:ys) = if (x==y) then prefix xs ys else False

But it returns an error:

Inferred type is not general enough
*** Expression    : prefix
*** Expected type : [a] -> [b] -> Bool
*** Inferred type : [a] -> [a] -> Bool

Can someone help me get this to work?

Upvotes: 2

Views: 2051

Answers (2)

Yann Vernier
Yann Vernier

Reputation: 15877

That's because you're comparing the types you've named a and b using (x==y). Checking the type of ==, that means they are the same type, which has an equality test:

Prelude> :t (==)
(==) :: Eq a => a -> a -> Bool

So the type signature that is inferred is actually Eq a => [a] -> [a] -> Bool.

Upvotes: 7

augustss
augustss

Reputation: 23014

Your type signature claims the two lists can have different types, but they can't. So the compiler complains that it inferred a type that was less general than the one you asked for.

Upvotes: 10

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