是否可以在Haskell中编写任何没有显式参数的函数？

Question

For example I have a table tab = [(1,11), (2,22), (3,33)] and two functions. The first one will take as parameter the table and an Integer and will replace the Integer with the value from the table. Let's assume the Integer will be on the table.

replace :: [(Int, Int)] -> Int -> Int
replace t i = snd $ head [x | x <- t, fst x == i]

The second function will replace all the occurrences of 1, 2, 3 from a list with the values from the table.

replaceAll :: [(Int, Int)] -> [Int] -> [Int]

First I got:

replaceAll = \tab lst -> foldl(\acc x -> acc ++ [replace tab x]) [] lst

then,

replaceAll = \tab -> foldl(\acc x -> acc ++ [replace tab x]) []

Is there a way to write this without the lambda function? EDIT: Not necessary using the fold function.

Answer 1

This works for me

replaceAll = map . replace

Although I also like removing params, I think this is a bit easier to understand if you fill them the params for this definition....

replaceAll index vals = map (replace index) vals