Am Donnerstag, 8. Januar 2009 01:31 schrieb David Schonberger:

Hello all. This is my first post and I'm essentially a rank beginner with Haskell, having used it briefly some 7 years ago. Glad to be back but struggling with syntax and understanding error messages.

This post actually ties to a prior post, http://www.haskell.org/pipermail/beginners/2008-December/000583.html

My troubles are with exercise 3.10, p45 of from HDIII's "Yet Another Haskell Tutorial" (YAHT). The exercise states:

Write a program that will repeatedly ask the user for numbers until she types in zero, at which point it will tell her the sum of all the numbers, the product of all the numbers, and, for each number, its factorial. For instance, a session might look like:

Note that the sample session accompanying the exercise suggests a session such as:

Give me a number (or 0 to stop): 5 Give me a number (or 0 to stop): 8 Give me a number (or 0 to stop): 2 Give me a number (or 0 to stop): 0 The sum is 15 The product is 80 5 factorial is 120 8 factorial is 40320 2 factorial is 2

The following code handles the sum and products pieces--I built the module incrementally--but fails on the factorial part. In fact my current code, if it worked, would only output something like:

The sum is 15 The product is 80 120 40320 2

But I'm not even getting that much. Here's the code:

--begin code

module AskForNumbers whereimport IO askForNums = do putStrLn "Enter a pos int or 0 to end: " numStr <- getLine let num = read numStr if num == 0 then return [] else do rest <- askForNums return (num:rest)

listFactorial l = if length l == 0 then return 1 else do fact (head l) listFactorial (tail l)

fact n = if n == 0 then return 1 else return foldr (*) 1 [1..n] f = do nums <- askForNums putStr ("Sum is " ++ (show (foldr (+) 0 nums)) ++ "\n") putStr ("Product is " ++ (show (foldr (*) 1 nums)) ++ "\n") listFactorial nums

--end code

Here is the error msg I get when I load into WinHugs (Sept 2006 version):

ERROR file:.\AskForNumbers.hs:22 - Ambiguous type signature in inferred type *** ambiguous type : (Num a, Num [a], Monad ((->) [b]), Num c, Num (b -> [a] -> [a]), Enum a, Monad ((->) (c -> c -> c))) => a -> [b] -> [a] *** assigned to : fact

Ambiguous type assigned to fact. Ok. Not sure what to make of that or how to correct it. I though I was passing fact an Integer, since I think nums is a list of Integers, since the sum and product lines in f work ok.

Help? Thanks.

Okay, that's a nice one :) Don't be disappointed by the rather prosaic reason for it. First, if you don't already know what ambiguous type means: in the context "(Num a, ..., Monad ((->) (c -> c -> c)))" there appears a type variable, c, which doesn't appear on the right hand side a -> [b] -> [a]. I will not explain how that monstrous type arises, though it is a nice exercise in type inferring. module AskForNumbers where import IO askForNums = do putStrLn "Enter a pos int or 0 to end: " numStr <- getLine let num = read numStr if num == 0 then return [] else do rest <- askForNums return (num:rest) listFactorial l = if length l == 0 then return 1 else do fact (head l) listFactorial (tail l) ^^^^ There are several things that can be improved here. First, don't use "if length l == 0". That has to traverse the whole list and if that's long (or even infinite), you're doing a lot of superfluous work. To check for an empty list, use "null", so "if null l then ...", that's far more efficient because null is defined as null [] = True null _ = False Also, it would be better to define listFactorial by pattern matching: listFactorial [] = ... listFactorial (k:ks) = do fact k listFactorial ks Next, why do you return 1 for an empty list? Probably to match the type of the other branch, but that that isn't so well chosen either. And since fact only returns something and does nothing else, listFactorial actually is "do something without any effects and then return 1", that isn't what you want. What you want is that for each k in the list it prints k factorial is ... That suggests that you either let fact do the output, so that fact would get the type Int(eger) -> IO () or let fact be a pure function calculating the factorial and have listFactorial (k:ks) = do putStrLn (show k ++ " factorial is " ++ show (fact k)) listFactorials ks f = do nums <- askForNums putStr ("Sum is " ++ (show (foldr (+) 0 nums)) ++ "\n") putStr ("Product is " ++ (show (foldr (*) 1 nums)) ++ "\n") listFactorial nums There are library functions "sum" and "product", you could use them. Instead of putStr (some string ++ "\n") you could use putStrLn (some string) but all this is just a matter of personal taste. Now comes the culprit: fact n = if n == 0 then return 1 else return foldr (*) 1 [1 .. n] The else-branch should be return ( foldr (*) 1 [1 .. n] ) or return $ foldr (*) 1 [1 .. n] Without the parentheses or ($), it is parsed as ( ( (return foldr) (*) ) 1 ) [1 .. n], which leads to the interesting type of the error message. If you're interested, I could explain how hugs infers that.

David

HTH, Daniel