CODEWITHSUNDEEP
haskelllambdahaskell-platform
Sven Kuffer
Sven Kuffer

Reputation: 81

Haskell map functions and lambda expressions

I've got this expression in Haskell, now I don't understand exactly how this is applied. Typing in j returns [2,3,4] 

j :: [Int]
j = map (\a -> a 1) (map (\a x -> x + a) [1,2,3])

Upvotes: 5

Views: 5291

Answers (1)

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 476567

There are two maps here, let us first analyze the subexpression:

map (\a x -> x + a) [1,2,3]

We can write the lambda expression in a form that is probably better for the above case:

map (\a -> (\x -> x + a)) [1,2,3]

So this is a function that takes a parameter a and returns a function. So it will return a function that takes a parameter x that maps to x + a. Thus that means that the second map produces a list of functions. Indeed, so the above expression is equivalent to:

[(+1), (+2), (+3)]

or more verbose:

[\x -> x+1, \x -> x+2, \x -> x+3]

here the xs in the lambda expressions are different variables.

Now the first map takes these functions, and maps these on calling the function with value one, so, this expression:

map (\a -> a 1) [(+1), (+2), (+3)]

is equivalent to:

[(+1) 1, (+2) 1, (+3) 1]

and thus equivalent to:

[2,3,4]

as you found out.

We can syntactically simplify this function to:

j :: Num a => [a]
j = map ($ 1) (map (+) [1,2,3])

which is semantically equivalent to:

j :: Num a => [a]
j = map (+1) [1,2,3]

and hence:

j :: Num a => [a]
j = [2,3,4]

Upvotes: 5

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