如何将函数应用于Haskell列表中的每个元素？

Question

I want to apply resolve_on to every element in list c1, like

for(Char c:c1){
resolve_on c c1 c2;}

So how can I do this using map function?

resolvents :: [Char] -> [Char] -> [[Char]]
resolvents c1 c2 =  map (//what should I do) c1

resolve_on :: Char -> [Char] -> [Char] -> [Char]
resolve_on c c1 c2

Answer 1

When in doubt, start with list comprehensions:

-- for (Char c) in c1:                     do { c <- c1
--   yield (resolve_on c c1 c2)               ; return (resolve_on c c1 c2) }

resolvents c1 c2 = [ resolve_on c c1 c2 | c <- c1 ]

    -- read it:    a list of
    --              (resolve_on c c1 c2)
    --                                  for        -- or: for every
    --                                    c
    --                                      in c1

This is indeed a map ,

                 = map (\c -> resolve_on c c1 c2) c1

This uses a lambda (ie, unnamed) function , receiving one argument named c .

The do code in the comments would also work. It is in do notation, and is equivalent to the list comprehension.