CODEWITHSUNDEEP
functionhaskellmap-functionfunction-call
zebraeleven
zebraeleven

Reputation: 45

Functions in Haskell - Understanding

I have the following code which gives out [2,4,6] :

j :: [Int]
j = ((\f x -> map x) (\y -> y + 3) (\z -> 2*z)) [1,2,3]

Why? It seems that just the "z-function" is being used, what happens to the "y-function"? And how does map work in this particular case?

Upvotes: 3

Views: 78

Answers (1)

chi
chi

Reputation: 116139

Let's compute:

((\f x -> map x) (\y -> y + 3) (\z -> 2*z)) [1,2,3]
                 ^^^ f ^^^^^^^ ^^^ x ^^^^^
=
(map x) [1,2,3]
   where f = \y -> y +3
         x = \z -> 2*z
=
[x 1,x 2,x 3]
   where f = \y -> y +3
         x = \z -> 2*z
=
[2*1,2*2,2*3]
   where f = \y -> y +3
         x = \z -> 2*z
=
[2,4,6]
   where f = \y -> y +3
         x = \z -> 2*z

As we can see, f was taken as an argument, but was never used after that. Consequently \y -> y+3 never affected the final result.

The function map x is the function that applies function x to each element of a list. Note that, above, (map x) [1,2,3] is the same as map x [1,2,3]. Indeed, each function application g x1 x2 x3 x4 can be equivalently written as (((g x1) x2) x3) x4 by left-associating the applications.

Upvotes: 6

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