Why can String not be used in a constraint?

Question

How come I can't set a String constraint, but a Num works just fine?

This works:

ltest' :: (Num a) => a -> a
ltest' a = (\b -> b * 2) a

This does not:

test' :: (String a) => a -> a
test' a = a ++ " good job"

Error:

• Expecting one fewer argument to ‘String’
  Expected kind ‘* -> Constraint’, but ‘String’ has kind ‘*’
• In the type signature:
    test' :: (String a) => a -> a

Chris Martin · Accepted Answer · 2017-09-12T18:05:41.877

13

String is a type:

type String = [Char]

Num is a class:

class Num a where
    (+), (-), (*)       :: a -> a -> a
    negate              :: a -> a
    abs                 :: a -> a
    signum              :: a -> a
    fromInteger         :: Integer -> a
    negate x            = 0 - x

Constraints only involve classes, not types.

Your ltest' has a type parameter, a.¹ This can be expressed explicitly:

ltest' :: forall a. (Num a) => a -> a

This means it works with any type a that has an instance of the Num class.

Your test' doesn't need type parameters, because it only works with one type² (the argument can only be of type String). So it also doesn't need any constraints, because there are no type parameters to constrain.

test' :: String -> String
test' a = a ++ " good job"

¹ In other words, ltest' has a polymorphic type.

² In other words, test' has a monomorphic type.

edited Sep 12 '17 at 18:05

answered Sep 20 '16 at 18:15

Chris Martin

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Thanks! I wanted to make sure you pass only String values as function param like in C for example function x(char a){} – v0d1ch Sep 20 '16 at 18:22
That is achieved by declaring the type as `String -> String`. – Chris Martin Sep 20 '16 at 18:24
So there is no way of having something like this (String a) => a -> a ? – v0d1ch Sep 20 '16 at 18:25
I'm confused - What do you want that to do? – Chris Martin Sep 20 '16 at 18:26
1

@vodich No, since `String` is a type, not a constraint on some type. – Ingo Sep 20 '16 at 18:26
Ok thanks for the help, cheers! – v0d1ch Sep 20 '16 at 18:28
@vodich What do you want `String a => a -> a` to mean that is different from what you believe `String -> String` means? – Daniel Wagner Sep 20 '16 at 19:36
I guess I thought you can have String x like in Typeclass Num x and then x is used instead of String -> String x -> x. Nevermind I just read a section about types in "Haskell Programming from first principles" so all is clear now – v0d1ch Sep 20 '16 at 19:42

Why can String not be used in a constraint?

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