Haskell function and let statement

Question

In GHCi, I can do following:

ghci> let ddd n = [0..n]
ghci> ddd 10
[0,1,2,3,4,5,6,7,8,9,10]

However, if I define similar function in a file as :

ddd :: Int -> [a]
ddd n = [0..n]

After loading the file in GHCi, I got following error

ghci> :l dsp.hs
[1 of 1] Compiling Main             ( dsp.hs, interpreted )

dsp.hs:43:13:
    Couldn't match expected type `a' with actual type `Int'
      `a' is a rigid type variable bound by
          the type signature for ddd :: Int -> [a] at dsp.hs:42:8
    In the expression: n
    In the expression: [0 .. n]
    In an equation for `ddd': ddd n = [0 .. n]
Failed, modules loaded: none.

Any reason??

Thanks

Answer 1

That's because you are returning a list of [a] and not [Int] , this should work:

ddd :: Int -> [Int]
ddd n = [0..n]

Or if you want a more generic solution:

ddd :: (Num a, Enum a) => a -> [a]
ddd n = [0..n]

Answer 2

If we enable the ExplicitForAll extension, we can see that the type signature you gave is the same as

ddd :: forall a. Int -> [a]

The forall is implied if it isn't written out, but the ExplicitForAll extension makes it, well, explicit.

But, this isn't the case! Your definition doesn't work for all types a , just for one specific type: Int . So the correct signature is Int -> [Int] .