Haskell pass contents of list as arguments

Asked Sep 20 '12 at 13:25

Active Sep 20 '12 at 13:25

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Why is such a function definition not allowed in haskell?

Good day!

I'm trying to write/to find a function g that translate a list of arbitrary finite size to function arguments, such as

    g k [a,b,c,d] = k a b c d
    g f [a,b,c] = f a b c = ((((f) $ a) $ b) $ c)

Syntactically that seems like g = foldl ($) f, but the compiler reasonably fails to apply that function, answering:

     ghci> let (><) f = foldl ($) f
     <interactive>:1:19:
     Occurs check: cannot construct the infinite type: b = b1 -> b
     Probable cause: `$' is applied to too many arguments
     In the first argument of `foldl', namely `($)'
     In the expression: foldl ($) f

So I fail to produce some working implementation. Do you have any ideas? Thank you in advance!

edited May 23 '17 at 12:27

Community

asked Sep 20 '12 at 13:25

ownclo

2

Try starting by writing down a type signature for `g` first, then figuring out how to implement it. ;] – C. A. McCann Sep 20 '12 at 13:27
I tried that at first, but I cannot define it `g :: f->[a]->b :: (a->a->...->b)->[a]->b`. I don't know the `f` type in advance. – ownclo Sep 20 '12 at 13:35
See that as a hint. Haskell is statically typed, so if you could implement it, you could also write down its type. – Daniel Fischer Sep 20 '12 at 13:38
It is the same question indeed! Many thanks, I didn't find it by myself. Migrating there... – ownclo Sep 20 '12 at 13:40
Little wonder, since the other question has an *atrocious* title. I should probably fix that... – C. A. McCann Sep 20 '12 at 14:07

Haskell pass contents of list as arguments

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