This is an exercise:
-- Ex 12: recall the binary function composition operation
-- (f . g) x = f (g x). In this exercise, your task is to define a function
-- that takes any number of functions given as a list and composes them in the
-- same order than they appear in the list.
--
-- Examples:
-- multiCompose [] "foo" ==> "foo"
-- multiCompose [] 1 ==> 1
-- multiCompose [( "bar")] "foo" ==> "foobar"
-- multiCompose [reverse, tail, ( "bar")] "foo" ==> "raboo"
-- multiCompose [(3*), (2^), ( 1)] 0 ==> 6
-- multiCompose [( 1), (2^), (3*)] 0 ==> 2
As beginner, I am not able to solve that. I tried many approach, this one does not work :
multiCompose [] = (1*) $
multiCompose fs = (multiCompose (init fs)) (last fs) $
Following my current understanding, it should work since it can be developed as follow :
multicompose [( 1), (2^), (3*)] = multiCompose [( 1), (2^)] (3*) $
= multiCompose [( 1)] (2^) $ (3*) $
= multiCompose [] ( 1) $ (2^) $ (3*) $
= (1*) $ ( 1) $ (2^) $ (3*) $
My questions
- Could you help me with a valid answer to this exercise ?
- Could you help me to understand why my solution does not work ?
Thank you so much
CodePudding user response：
Your forgot the argument, and one more $:
multicompose [( 1), (2^), (3*)] $ x
-- = multiCompose [( 1), (2^)] (3*) $ x
-- here |
= multiCompose [( 1), (2^)] $ (3*) $ x
= multiCompose [( 1)] $ (2^) $ (3*) $ x
= multiCompose [] $ ( 1) $ (2^) $ (3*) $ x
= ( 1) $ (2^) $ (3*) $ x
Instead of [a,b,c,d] = [a,b,c] [d], use the identity [a,b,c,d] = [a] [b,c,d] = a : [b,c,d]:
= ( 1) $ (2^) $ (3*) $ x
= ( 1) $ (2^) $ (3*) $ multiCompose [ ] $ x
= ( 1) $ (2^) $ multiCompose [ (3*)] $ x
= ( 1) $ multiCompose [ (2^), (3*)] $ x
You can take it from here. In particular, multiCompose [] x = x must hold, and is a valid definition for the [] argument case.