应用内联函数进行转换

Question

Starting from a simple case of "fold" (I used (+) but can be anything else):

Prelude.foldl (+) 0 [10,20,30]

is it possible apply an inline transformation similar to (that doesn't work):

Prelude.foldl ((+) . (\x -> read x :: Int)) 0 ["10","20","30"]

In case not, is there an alternative to fold, to apply a generic function and inline transformation (apart from using specific functions like 'sum', 'max' etc)?

Answer 1

The read lambda applies to the first argument and the first argument to the function given to foldl is the accumulator. Those two arguments are the opposite for foldr . So, expanded, it looks like this:

foldl (\acc element -> (read acc :: Int) + element) 0 ["10", "20", "30"]

Since acc is an Int , this doesn't work.

So, with this information in hand, you can do this with foldr since it has the opposite argument order:

foldr ((+) . (read :: String -> Int)) 0 ["10","20","30"]

If you want an inline foldl version, you can use flip to achieve this.

You could also use map first (everything else being equal, this would be the solution I'd prefer):

foldl (+) 0 $ map (read :: String -> Int) ["10","20","30"]

Also, you probably want foldl' instead of foldl .