Re: [Haskell-cafe] pair (f,g) x = (f x, g x)?
Wenduan, you get a Int and Char out of the two composed functions, namely square.fst, Char.toUpper.snd.But in the type declaration of pair, which appeared to me,it meant its arguments must be two functions which are of the same type namely a,whereas Int and Char passed to as arguments are of different types here, and that's the reason I thought it wouldn't work. Well, actually the two argument functions are not required to be of exactly the same type. The only restriction is that the types of their parameters match: pair :: (a -> b) -> (a -> c) -> a -> (b, c) So, in pair (square . fst) (toUpper . snd) a matches the type of the parameters of (square . fst) and (toUpper . snd), i.e., (Int, Char), b matches the result type of (square . fst), i.e., Int, and c matches the result type of (toUpper . snd), i.e., Char; so the type of pair get instantiated with ((Int, Char) -> Int) -> ((Int, Char) -> Char) -> (Int, Char) -> (Int, Char) You might also want to use (***) :: (a -> c) -> (b -> d) -> (a, b) -> (c, d) (f *** g) (a, b) = (f a, g b) HTH, Stefan ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] pair (f,g) x = (f x, g x)?
> you are correct,but as in the following, > > > (square . fst) :: (Int,b) -> Int > > > > > > (Char.toUpper . snd) :: (a,Char) -> Char > > you get a Int and Char out of the two composed functions, namely square.fst, > Char.toUpper.snd.But in the type declaration of > pair, which appeared to me,it meant its arguments must be two functions which > are of the same type namely a,whereas Int and > Char passed to as arguments are of different types here, and that's the > reason I thought it wouldn't work. The signature says it takes two functions, which take the same type to *different* types ((a->b), (a->c)). In your case, 'a' is guaranteed the same type because you're applying it to the same value (in this case its type is (Int, Char)). So you are not passing Int or Char but (Int, Char) to 'fst' and 'snd'. The Int -> Int and Char -> Char functions never see the type they don't understand because the selectors 'fst' and 'snd' have stripped those values off. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] pair (f,g) x = (f x, g x)?
Marc A. Ziegert wrote: '.' is not always a namespace-separator like '::','.','->' in c++ or '.' in java. it is used as an operator, too. (.) :: (b->c) -> (a->b) -> (a->c) (f . g) x = f (g x) remember the types of fst and snd: fst :: (a,b)->a snd :: (a,b)->b so the function (.) combines square :: Int -> Int with fst to (square . fst) :: (Int,b) -> Int the same with toUpper: (Char.toUpper . snd) :: (a,Char) -> Char so you have with 'pair (f,g) x = (f x,g x)': pair (square . fst,Char.toUpper . snd) (2,'a') ==> ((square . fst) (2,'a'), (Char.toUpper . snd) (2,'a')) ==> ( square (fst(2,'a')), Char.toUpper (snd(2,'a')) ) ==> ( square 2 , Char.toUpper 'a' ) ==> (4,'A') - marc Am Samstag, 2. Juli 2005 08:32 schrieb wenduan: I came across a haskell function on a book defined as following: pair :: (a -> b,a -> c) -> a -> (b,c) pair (f,g) x = (f x,g x) I thought x would only math a single argument like 'a', 1, etc,but it turned out that it would match something else, for example, a pair as below: square x = x*x pair (square.fst,Char.toUpper.snd) (2,'a') (4,'A') The type declaration of pair is what confused me, pair :: (a -> b,a -> c) -> a -> (b,c),it says this function will take a pair of functions which have types of a->b,a->c,which I would take as these two functions must have argument of the same type, which is a,and I didn't think it would work on pairs as in the above instance,but surprisingly it did,can anybody enlighten me? -- X.W.D ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe you are correct,but as in the following, (square . fst) :: (Int,b) -> Int (Char.toUpper . snd) :: (a,Char) -> Char you get a Int and Char out of the two composed functions, namely square.fst, Char.toUpper.snd.But in the type declaration of pair, which appeared to me,it meant its arguments must be two functions which are of the same type namely a,whereas Int and Char passed to as arguments are of different types here, and that's the reason I thought it wouldn't work. Thank you, regards. -- X.W.D ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] pair (f,g) x = (f x, g x)?
'.' is not always a namespace-separator like '::','.','->' in c++ or '.' in java. it is used as an operator, too. (.) :: (b->c) -> (a->b) -> (a->c) (f . g) x = f (g x) remember the types of fst and snd: fst :: (a,b)->a snd :: (a,b)->b so the function (.) combines square :: Int -> Int with fst to (square . fst) :: (Int,b) -> Int the same with toUpper: (Char.toUpper . snd) :: (a,Char) -> Char so you have with 'pair (f,g) x = (f x,g x)': pair (square . fst,Char.toUpper . snd) (2,'a') ==> ((square . fst) (2,'a'), (Char.toUpper . snd) (2,'a')) ==> ( square (fst(2,'a')), Char.toUpper (snd(2,'a')) ) ==> ( square 2 , Char.toUpper 'a' ) ==> (4,'A') - marc Am Samstag, 2. Juli 2005 08:32 schrieb wenduan: > I came across a haskell function on a book defined as following: > > pair :: (a -> b,a -> c) -> a -> (b,c) > pair (f,g) x = (f x,g x) > > I thought x would only math a single argument like 'a', 1, etc,but > it turned out that it would match something else, for example, a pair as > below: > > square x = x*x > > pair (square.fst,Char.toUpper.snd) (2,'a') > (4,'A') > > The type declaration of pair is what confused me, > pair :: (a -> b,a -> c) -> a -> (b,c),it says this function will take a > pair of functions which have types of a->b,a->c,which I would take as > these two functions must have argument of the same type, which is a,and > I didn't think it would work on pairs as in the above instance,but > surprisingly it did,can anybody enlighten me? > > -- > X.W.D > > ___ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > > ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] pair (f,g) x = (f x, g x)?
I came across a haskell function on a book defined as following: pair :: (a -> b,a -> c) -> a -> (b,c) pair (f,g) x = (f x,g x) I thought x would only math a single argument like 'a', 1, etc,but it turned out that it would match something else, for example, a pair as below: square x = x*x pair (square.fst,Char.toUpper.snd) (2,'a') (4,'A') The type declaration of pair is what confused me, pair :: (a -> b,a -> c) -> a -> (b,c),it says this function will take a pair of functions which have types of a->b,a->c,which I would take as these two functions must have argument of the same type, which is a,and I didn't think it would work on pairs as in the above instance,but surprisingly it did,can anybody enlighten me? -- X.W.D ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] New-bee question about seq in Parsec.Pos
Hi, I try to understand, how Parsec works. In Pos.hs I found these definition of SourcePos: data SourcePos = SourcePos SourceName !Line !Column deriving (Eq,Ord) My poor haskell knowledge says me, that Line and Column is always strict. So, what does the following function? forcePos :: SourcePos -> SourcePos forcePos pos@(SourcePos name line column) = seq line (seq column (pos)) thanks for an enlightenment, Christoph Bauer -- let () = let rec f a w i j = Printf.printf "%.20f\r" a; let a1 = a *. i /. j in if w then f a1 false (i +. 2.0) j else f a1 true i (j +. 2.0) in f 2.0 false 2.0 1.0 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
