CODEWITHSUNDEEP
haskell
Elliot Gorokhovsky
Elliot Gorokhovsky

Reputation: 3762

Haskell chain of functions

I have a composition of functions, many of them operators, to raise a constant to the power of a constant minus the size of the ith set in a list. Here's my code:

(k^).(n-).size.(phi !!) i

When I tried it on a test case I got 

<interactive>:64:16-25:
    Couldn't match expected type ‘a -> Set a0’
                with actual type ‘Set a1’
    Relevant bindings include
      it :: a -> c (bound at <interactive>:64:1)
    Possible cause: ‘phi !!’ is applied to too many arguments
    In the second argument of ‘(.)’, namely ‘(phi !!) 3’
    In the second argument of ‘(.)’, namely ‘size . (phi !!) 3’

However, 

(k^)$(n-)$size$(phi !!)$i

works. What's wrong? Why does composition not work but application work? Also, is composing the operators in parentheses the most idiomatic way to do it? It feels weird.

Upvotes: 1

Views: 1977

Answers (2)

karakfa
karakfa

Reputation: 67507

Note that you can use parenthesis to group function compositions, ((k^).(n-).size.(phi !!)) i should work.

Upvotes: 1

amalloy
amalloy

Reputation: 92067

Rather than f . g . h x, you want f . g . h $ x: the former calls h x immediately and then composes that, as a function, with f and g The latter composes together f, g, and h into a new function and then calls that on i.

Upvotes: 3

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