练习多重组合，列表中的函数

Question

这是一个练习：

-- 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

作为初学者，我无法解决这个问题。我试了很多方法，这个不行:

multiCompose [] = (1*) $
multiCompose fs = (multiCompose (init fs)) (last fs) $

根据我目前的理解，它应该可以工作，因为它可以开发如下：

multicompose [(+1), (2^), (3*)] = multiCompose [(+1), (2^)]           (3*) $ 
                                = multiCompose [(+1)]         (2^) $  (3*) $ 
                                = multiCompose []     (+1) $  (2^) $  (3*) $ 
                                =              (1*) $ (+1) $  (2^) $  (3*) $

我的问题

你能帮我给出这个练习的有效答案吗？
你能帮我理解为什么我的解决方案不起作用吗？

非常感谢

Answer 1

你忘记了参数，还有一个$:

     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

而不是 [a,b,c,d] = [a,b,c] ++ [d]，使用身份 [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

你可以从这里拿走。特别是，multiCompose [] x = x 必须成立，并且是 [] 参数案例的有效定义。

练习多重组合，列表中的函数

Exercise Multicomposition , functions in list

list

haskell

function-composition