вопрос хаскельщикам

[juan_gandhi]

true x y = x
false x y = y

lampair :: a -> b -> (a -> b -> c) -> c
lampair x y p = p x y

pa :: ((a -> b -> c) ->c) -> a
pa p = p true

And I see

   • Couldn't match type ‘a’ with ‘c’
      ‘a’ is a rigid type variable bound by
        the type signature for:
          pa :: forall a b c. ((a -> b -> c) -> c) -> a
        at sample1.hs:8:1-30
      ‘c’ is a rigid type variable bound by
        the type signature for:
          pa :: forall a b c. ((a -> b -> c) -> c) -> a
        at sample1.hs:8:1-30
      Expected type: a -> b -> c
        Actual type: c -> b -> c
    • In the first argument of ‘p’, namely ‘true’
      In the expression: p true
      In an equation for ‘pa’: pa p = p true
    • Relevant bindings include
        p :: (a -> b -> c) -> c (bound at sample1.hs:9:4)
        pa :: ((a -> b -> c) -> c) -> a (bound at sample1.hs:9:1)
  |

What have I done wrong? What fails here? Is Hindley-Milner "not smart enough"? What am I missing?

Comments

[juan_gandhi]

А, хитро. В смысле, я это и хотел нарисовать, но тут хитро, лямбды просто вставляем. А если эти лямбды вынуть?

[sassa_nf]

Prelude> first = \t -> t (\x y -> x)
Prelude> :t first
first :: ((t1 -> t2 -> t1) -> t) -> t

That is, if you don't use lambdas, you get your code, but with the corrected types :)

pa :: ((a->b->a)->c)->c