Problems when trying to solve yaht exercise 3.10

Hi, all I'm new to haskell and currently reading yaht. I find some problems when trying to solve exercise 3.10. The exersices asks to read a list of numbers terminated by a zero, and figure out the sum and product of the list. My program is as follows: ex3_10 = do hSetBuffering stdin LineBuffering numbers <- getNumber let sum = foldr (+) 0 numbers product = foldr (*) 1 numbers putStrLn "The sum is " ++ show(sum) putStrLn "The product is " ++ show(product) getNumber = do putStrLn "Give me a number (or 0 to stop):" num <- getLine if read num == 0 then return [] else do rest <- getNumber return (read num : rest) But when i load the program, ghci reports error: Couldn't match expected type `[a]' against inferred type `IO ()' In the first argument of `(++)', namely `putStrLn "The sum is "' In a stmt of a 'do' expression: putStrLn "The sum is " ++ show (sum) And i just don't understand the first sentence. Could you tell what does it mean? Thanks for your reply

Ah, you came very close to getting this right. The problem is on your two
putStrLn lines. For example, putStrLn "The sum is " ++ show(sum)
This parses as (putStrLn "The sum is ") ++ show(sum). Thus, it's trying to
use ++ on an IO () and a string. This obviously won't work. You need one of
the following:
putStrLn ("The sum is " ++ show(sum))
putStrLn $ "The sum is " ++ show(sum)
Ditto for the product line. Also, if I can make 2 other quick comments:
product and sum are already defined in the prelude, so there's no need to
define them here. You should also get used to putting type signatures on
your functions, because it's a good practice. Otherwise, good job!
2009/3/20 ZelluX
Hi, all
I'm new to haskell and currently reading yaht. I find some problems when trying to solve exercise 3.10.
The exersices asks to read a list of numbers terminated by a zero, and figure out the sum and product of the list. My program is as follows:
ex3_10 = do hSetBuffering stdin LineBuffering numbers <- getNumber let sum = foldr (+) 0 numbers product = foldr (*) 1 numbers putStrLn "The sum is " ++ show(sum) putStrLn "The product is " ++ show(product)
getNumber = do putStrLn "Give me a number (or 0 to stop):" num <- getLine if read num == 0 then return [] else do rest <- getNumber return (read num : rest)
But when i load the program, ghci reports error: Couldn't match expected type `[a]' against inferred type `IO ()' In the first argument of `(++)', namely `putStrLn "The sum is "' In a stmt of a 'do' expression: putStrLn "The sum is " ++ show (sum)
And i just don't understand the first sentence. Could you tell what does it mean?
Thanks for your reply _______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

On 20 Mar 2009, at 13:45, ZelluX wrote:
Hi, all
I'm new to haskell and currently reading yaht. I find some problems when trying to solve exercise 3.10.
The exersices asks to read a list of numbers terminated by a zero, and figure out the sum and product of the list. My program is as follows:
ex3_10 = do hSetBuffering stdin LineBuffering numbers <- getNumber let sum = foldr (+) 0 numbers product = foldr (*) 1 numbers putStrLn "The sum is " ++ show(sum) putStrLn "The product is " ++ show(product)
getNumber = do putStrLn "Give me a number (or 0 to stop):" num <- getLine if read num == 0 then return [] else do rest <- getNumber return (read num : rest)
But when i load the program, ghci reports error: Couldn't match expected type `[a]' against inferred type `IO ()' In the first argument of `(++)', namely `putStrLn "The sum is "' In a stmt of a 'do' expression: putStrLn "The sum is " ++ show (sum)
And i just don't understand the first sentence. Could you tell what does it mean?
It means that you can't use ++ on an IO action, ++'s type reveals why: (++) :: [a] -> [a] -> [a] It accepts two lists, not a list and an IO action. So the question then is, which IO action are you using ++ on? The answer is (putStrLn "The sum is"). Your last two statements are being parsed as: (putStrLn "The sum is ") ++ (show sum) (putStrLn "The product is ") ++ (show product) As a random aside, it's usually a good plan in Haskell to get out of IO based computations as fast as possible, and use pure functions instead. You might want to consider something like this: ex3_10 = do hSetLineBuffering stdin LineBuffering interact sumAndProduct sumAndProduct :: String -> String sumAndProduct = (\ns -> sumText ns ++ productText ns) . map read . lines sumText :: (Num a, Show a) => [a] -> String sumText = ("The sum is " ++) . show . sum productText :: (Num a, Show a) => [a] -> String productText = ("The product is " ++) . show . product Why might you want to do this? Well, firstly, the code becomes more readable, secondly, a computation that was not sequential is no longer described as being sequential, and finally, the results are more composible. We can now use sumText and productText, safe in the knowledge that they will never have side effects etc. Bob

Got it. Many thanks to Thomas Davie's and Andrew Wagner's replies. ^_^
On Fri, Mar 20, 2009 at 9:00 PM, Thomas Davie
On 20 Mar 2009, at 13:45, ZelluX wrote:
Hi, all
I'm new to haskell and currently reading yaht. I find some problems when trying to solve exercise 3.10.
The exersices asks to read a list of numbers terminated by a zero, and figure out the sum and product of the list. My program is as follows:
ex3_10 = do hSetBuffering stdin LineBuffering numbers <- getNumber let sum = foldr (+) 0 numbers product = foldr (*) 1 numbers putStrLn "The sum is " ++ show(sum) putStrLn "The product is " ++ show(product)
getNumber = do putStrLn "Give me a number (or 0 to stop):" num <- getLine if read num == 0 then return [] else do rest <- getNumber return (read num : rest)
But when i load the program, ghci reports error: Couldn't match expected type `[a]' against inferred type `IO ()' In the first argument of `(++)', namely `putStrLn "The sum is "' In a stmt of a 'do' expression: putStrLn "The sum is " ++ show (sum)
And i just don't understand the first sentence. Could you tell what does it mean?
It means that you can't use ++ on an IO action, ++'s type reveals why:
(++) :: [a] -> [a] -> [a]
It accepts two lists, not a list and an IO action. So the question then is, which IO action are you using ++ on? The answer is (putStrLn "The sum is"). Your last two statements are being parsed as:
(putStrLn "The sum is ") ++ (show sum) (putStrLn "The product is ") ++ (show product)
As a random aside, it's usually a good plan in Haskell to get out of IO based computations as fast as possible, and use pure functions instead. You might want to consider something like this:
ex3_10 = do hSetLineBuffering stdin LineBuffering interact sumAndProduct
sumAndProduct :: String -> String sumAndProduct = (\ns -> sumText ns ++ productText ns) . map read . lines
sumText :: (Num a, Show a) => [a] -> String sumText = ("The sum is " ++) . show . sum
productText :: (Num a, Show a) => [a] -> String productText = ("The product is " ++) . show . product
Why might you want to do this? Well, firstly, the code becomes more readable, secondly, a computation that was not sequential is no longer described as being sequential, and finally, the results are more composible. We can now use sumText and productText, safe in the knowledge that they will never have side effects etc.
Bob
participants (3)
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Andrew Wagner
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Thomas Davie
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ZelluX