CODEWITHSUNDEEP
functionhaskell
Dieter Brow
Dieter Brow

Reputation: 17

multiple functions on one argument with map

Is it possible to code a simple map function which allows to have multiple functions on one argument? for example mapmultiple [square, pred] 4 --> [16, 3]

map1 :: (a -> b) -> [a] -> [b]
map1 f [] = []
map1 f (x:xs) = foldr (\y ys -> (f y):ys) [] xs

map1 allows it with one function but how will it work out with two or three functions?

Upvotes: 1

Views: 3063

Answers (4)

Silvio Mayolo
Silvio Mayolo

Reputation: 70367

This is just map with the arguments flipped.

map :: (a -> b) -> [a] -> [b]
mapmultiple :: [a -> b] -> a -> [b]

The "function" of map is going to be a higher-order function applying an argument.

mapmultiple fs x = map ($ x) fs

If you're not comfortable with the $ section, this is equivalent[1] to

mapmultiple fs x = map (\f -> f x) fs

[1] For a time, there was a GHC bug where sections would compile incorrectly to similar (but not quite equivalent) code. I'm uncertain as to whether it's been fixed. It shouldn't be relevant here, but I figure I should mention it.

Upvotes: 5

Franky
Franky

Reputation: 2421

Consider a list comprehension:

map' fs x = [ f x | f <- fs ]

Upvotes: 4

chepner
chepner

Reputation: 532153

If you aren't committed to those exact arguments or order of arguments, two options are 

>>> [(^ 2), pred] <*> pure 4
[16, 3]

>>> fmap ($ 4) [(^ 2), pred]
[16, 3]

Upvotes: 1

karakfa
karakfa

Reputation: 67547

Here is an alternative with sequence

> mmap = map . sequence
> mmap [(^2), pred] [1..3]
[[1,0],[4,1],[9,2]]

Upvotes: 4

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