I have a pretty general question about Haskell's type system. I'm trying to become familiar with it, and I have the following function:
getN :: Num a => a
getN = 5.0 :: Double
When I try to compile this, I get the following error:
Couldn't match expected type `a' against inferred type `Double'
`a' is a rigid type variable bound by
the type signature for `getN' at Perlin.hs:15:12
In the expression: 5.0 :: Double
In the definition of `getN': getN = 5.0 :: Double
As I understand this, the function is set up to "return" a type in the class Num. Double is in this class (http://www.zvon.org/other/haskell/Outputprelude/Num_c.html), so I would have expected that it would be okay to "return" a Double in this case.
Can someone explain this please?
A function with signature Num a => a
is expected to work for any type in the class Num
. The implementation 5.0 :: Double
just works for one type, not for all types of the class, so the compiler complains.
An example of a generic function would be:
square :: (Num a) => a -> a
square x = x * x
This works for any type that is a Num
. It works for doubles, integers and whatever other numbers you want to use. Because of that it can have a generic type signature that just requires the parameter to be in class Num
. (Type class Num
is necessary because the function uses multiplication with *
, which is defined there)
To add to sth's answer: Haskell is not object-oriented. It's not true that Double
is a subclass of Num
, so you cannot return a Double
if you promise to return a polymorphic Num
value, like you can in, say, Java.
When you write getN :: Num a => a
you promise to return a value that is fully polymorphic within the Num
constraint. Effectively this means that you can only use functions from the Num
type class, such as +
, *
, -
and fromInteger
.
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With