map with multiple parameters

Question

The type of map is:

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

So if I would like to map a function with multiple parameters onto an array, something like:

myObviouslyFakeFunction :: Int -> Char -> String -> Char -> Integer -> String
myObviouslyFakeFunction -pattern- = -Very complex transform-

and do something like:

map (myObviouslyFakeFunction 1 'a' "abc" 'b' 2) ["abc", "def", "ghi"]

How could I do this? Would a in the type signature represent the first parameter? A tuple with all of them? A list?

(myObviouslyFakeFunction 1 'a' "abc" 'b' 2) is a String, not a function. — Benesh
– Benesh, Commented Mar 26, 2014 at 9:47
any chance you could clarify? I feel like I'm missing something incredibly basic... Why isn't it a function? Doesn't the above define it as a function (that returns a string?) To paraphrase a language I am more familiar with is that not equivalent to public string myObviouslyFakeFunction(int i, char c, string s, char c2, int i2){}? How does it differ? — Abraham P
– Abraham P, Commented Mar 26, 2014 at 9:50
@AbrahamP - Its a string because you've provided all the arguments to myObvisoulyFakeFunction. You could partially apply a function a1 -> a2 .. -> aN -> String -> b to create a function String -> b and pass that to map, but myObviouslyFakeFunction does not follow that pattern. — Lee
– Lee, Commented Mar 26, 2014 at 9:52
Let's say you've got a function f :: Int -> Int. Then f 1 is an Int. — Benesh
– Benesh, Commented Mar 26, 2014 at 9:52
@AbrahamP In Java (or whatever that is) it would be equivalent to myObviouslyFakeFunction(1, 'a', "abc", 'b', 2) which would be a string, not a function. — molbdnilo
– molbdnilo, Commented Mar 26, 2014 at 9:57

Lee Duhem · Accepted Answer · 2014-03-26 10:24:24Z

4

If you want to use myObviouslyFakeFunction like

map (myObviouslyFakeFunction 1 'a' "abc" 'b' 2) ["abc", "def", "ghi"]

then the type of it should be

myObviouslyFakeFunction :: Int -> Char -> String -> Char -> Integer -> String -> ResultType

Because map needs a one-parameter function, in this case with type String -> b (b is a type variable, could be any valid type), as its first argument, and if myObviouslyFakeFunction has the above type,

(myObviouslyFakeFunction 1 'a' "abc" 'b' 2)

will be a String -> ResultType function.

answered Mar 26, 2014 at 10:24

Lee Duhem

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Comments

Kl232 · Accepted Answer · 2014-03-26 12:12:17Z

2

I think, if i got you right, what you want to do is just

map (\x->myObviouslyFakeFunction 1 'a' x 'b' 2) ["abc", "def", "ghi"]

answered Mar 26, 2014 at 12:12

Kl232

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Comments

Emil · Accepted Answer · 2014-03-26 16:11:01Z

This might clarify a little bit:

myObviouslyFakeFunction :: Int -> Char -> String -> Char -> Integer -> String 
myObviouslyFakeFunction 1 :: Char -> String -> Char -> Integer -> String 
myObviouslyFakeFunction 1 'a' :: String -> Char -> Integer -> String 
myObviouslyFakeFunction 1 'a' "abc" :: Char -> Integer -> String 
myObviouslyFakeFunction 1 'a' "abc" 'b' :: Integer -> String
myObviouslyFakeFunction 1 'a' "abc" 'b' 2 :: String

What you want to give to map is something of type a -> b, but this is of type String. If you had redefined your function so that it would have the following type (for some type b)

myObviouslyFakeFunction' :: Int -> Char -> String -> Char -> Integer -> String -> b

then we would have that

myObviouslyFakeFunction' 1 'a' "abc" 'b' 2 :: String -> b

Then you can apply it to the list of Strings.

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