On Fri, Jul 25, 2008 at 12:45 PM, Anatoly Vorobey
Hello Haskell-beginners,
I'm trying to work through the "Write Yourself a Scheme in 48 Hours" tutorial. One of the first exercises calls for writing code to sum up arguments on command line and display the sum.
After a few tries, this worked for me:
module Main where import System.Environment
main :: IO () main = do args <- getArgs putStrLn ("Hello, " ++ show (sumIt args)) where sumIt x = sum $ map read x
However, I have two basic questions:
1. I initially tried putStrLn("Hello, " ++ show $ sumIt args), but that didn't compile. Why not? If I wrap (show $ sumIt args) in parens, it works, but should I have to, if I didn't have to in the original code above?
2. I initially tried
where sumIt = sum $ map read
(the "point-free" style, I believe it's called?) but that didn't compile. Why not? A friend suggested
where sumIt = sum . map read
and that does work; I guess my real problem, then, is that I don't really understand the difference between the two and why the former doesn't work.
Thanks in advance, Anatoly.
-- Anatoly Vorobey, avorobey@gmail.com http://avva.livejournal.com (Russian) http://www.lovestwell.org (English)
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Hello, The difference between (.) and ($) is that (.) is composition and ($) is application. What this means is that (.) takes two /functions/ and generates a /function/ that is like applying the first function after the second function: (f . g) (x) = f (g x) Where f and g are functions and x is a value. ($), on the other hand, takes a /function/ and a /value/ and applies the function to the value: f $ x = f x So the reason you can't do sum $ map read is because map read is a function and sum is a function, so you need to compose the two functions. If you wanted to use application, you could make a lambda expression like \x -> sum $ map read x, because now map is fully applied and is a value for sum. (Note that I'm oversimplifying a bit here; Haskell doesn't actually make a distinction between functions and values, but it's easier to think about like this for simple cases.) Alex