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RE: Functional dependencies question
by Simon Peyton-Jones 19 May '03

19 May '03

Think of it like this. Should this be acceptable? f :: a -> a -> a f x y = x && y No, because the type (forall a. a->a->a) is plainly more general than the actual function. The type tempts you to think that you could write f 3 4 but you can't. But I suppose you could argue that the function definition should be allowed, but if you ever called (f 3 4) you should get an error "Int found where Bool expected". Your signature is bar :: (Foo Char t) => t and we know (from the instance decl) that t must be Bool. Any other call to bar would be illegal, just as any cal to 'f' above is illegal, except at type Bool. In general, it seems to make sense to reject definitions as early as possible, hence GHC and Hugs reject it. It is undoubtedly odd that adding an instance declaration makes a legal program illegal, and that might be a reason for relaxing the rule... but it's not entirely easy to implement such a relaxation. Simon | -----Original Message----- | From: haskell-cafe-admin(a)haskell.org [mailto:haskell-cafe-admin@haskell.org] On Behalf Of Andrew | J Bromage | Sent: 12 May 2003 06:10 | To: haskell-cafe(a)haskell.org | Subject: Functional dependencies question | | G'day everyone. | | Suppose I have code like this: | | class Foo a b | a -> b where | foo :: a -> b | | bar :: (Foo Char t) => t | bar = foo 'a' | | All well and good so far. Let's add an instance of Foo... | | instance Foo Char Bool where | foo c = isAlpha c | | Now neither GHC nor Hugs will allow this to compile, as the declared | type of bar is "too general". | | This seems intuitively the wrong behaviour. Either the original | program was incorrect (as there was no visible instance of Foo for | type Char) or the amended program is correct (since the type of bar | is no more general than it was in the first version). | | Thoughts? | | Cheers, | Andrew Bromage | _______________________________________________ | Haskell-Cafe mailing list | Haskell-Cafe(a)haskell.org | http://www.haskell.org/mailman/listinfo/haskell-cafe

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n-term recurrences
by William Lee Irwin III 16 May '03

16 May '03

-- coefs starts with constant term, then coef of a(n-k) in k-term -- recurrence and goes up by indices -- vals starts with x0 and goes up by indices -- solves linear recurrences -- where xs !! (n+1) = head coefs + -- sum (zipWith (*) (tail coefs) [xs !! k | k <- [n - length coefs - 1..n]]) -- and the first few terms are given by xs !! n = vals !! n recurrence coefs vals | length coefs /= length vals + 1 = error "mismatched lengths of coefficients and values" | otherwise = vals ++ map coefSum ys where xs = recurrence coefs vals ys = transpose [drop k xs | k <- [0..length coefs - 1]] coefSum xs = head coefs + sum (zipWith (*) (tail coefs) xs) transpose xs = map head xs : transpose (map tail xs) -- This could be used, say, to generate the list of orthonormal -- polynomials generated by a three-term recurrence and then -- expand some function in a function series via Gaussian quadrature. -- e^(-x^2/2)H_n(x) or e^(-x^c/2)L^c_n(x) look especially attractive. -- OTOH that tactic may not be numerically useful, or useful only -- when the function series expansion is known a priori. -- It could also be a fancier way to generate Fibonacci numbers or -- Apery numbers or whatever. -- Not terribly deep, but it amused me for a few minutes to think of it. -- wli

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ELearnChina Conference 2003
by Edwin Jones 16 May '03

16 May '03

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Need help on Ptr
by Christian Buschmann 15 May '03

15 May '03

Hi! Can someone tell me, how i can check, if a Ptr is a nullPtr. Now I am using this function, but I think its not a very nice one: import Ptr isNullPtr :: (Ptr a) -> Bool isNullPtr ptr = (minusPtr ptr nullPtr) == 0 Maybe someone has got a better idea... christian

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Re: on termination
by oleg＠pobox.com 15 May '03

15 May '03

I'd like to clarify the termination decision procedure. I believe it can take the body of a function in our language and tell if the function terminates for all inputs. This implies that we can decide on termination without evaluating the function on all inputs (whose set is countably infinite). A small clarification: the 'branch-on-zero' statement should have the form "if-nonnegative exp then exp1 else exp2". Because the domain includes 'undefined' (aka NaN), the only functions that terminate for all inputs are those that leave their arguments alone and either propagate them or return constants. For example, the following functions terminate on all inputs f1(x) = x f2(x,y) = y f3(x) = 0 f4(x) = if-nonnegative power(5,125) then x else NaN The following functions will fail to terminate on at least one input: g1(x) = x + x g2(x) = if-non-negative x then x else NaN g3(x) = if-non-negative x*x then 0 else 1 Although it seems that g3 is equivalent to f3, that is not true: g3(NaN) diverges whereas f3(NaN) returns 0. These examples clarify a decision procedure, which can be stated as follows: if a partial evaluation of f(x...z) -- that is, unfolding plus constant propagation -- reduces f() to f1, f2, or f3 types, the function f() terminates on all inputs. Otherwise, there is one value (most likely NaN), that, when passed to f(), will cause it to diverge. Our partial evaluation procedure first unfolds all function calls. That is possible because the call graph does not have direct cycles (NPFIX must be treated as a special node in the call graph). Any operation (multiplication, addition, comparison, NPFIX) on a dynamic value can lead to a non-termination (if that value is NaN). Thus the only operations that can still guarantee termination are "if exp1 then exp2 else exp3" and "NPFIX exp1 exp2 exp3" where exp1 is a constant expression. We can evaluate it. Because it contains only primitive recursive functions, we provably can do that in some finite time. Therefore, we can decide which branch to take and continue to analyze exp2 or exp3. It seems the decision procedure is total and correct.

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RE: on termination
by Simon Peyton-Jones 15 May '03

15 May '03

| If you guys think haskell-cafe is improper for this conversation we can | continue on theory-edge on yahoo groups. But I don't think it's improper, | after all there is all kind of theoretical weirdness talked of in the main | haskell discussion list. Why should this be any less relevant??? Anyway this is Haskell Café, where the rules are pretty loose! (If it were on the main Haskell list I'd certainly suggest moving the thread to Haskell-café, but it seem ok there, to me.) S

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re: re: 你好
by 张丽 14 May '03

14 May '03

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Re: on termination
by Markus.Schnell＠infineon.com 14 May '03

14 May '03

Somebody asked for coffee? / / / \ \ \ / / / +++++++++++ + +--\ + +___/ \+++++/ ============= ========= -- Markus Schnell Infineon Technologies AG, CPR ET Tel +49 (89) 234-20875

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DUKAT KABILA(Jnr)
by Dukat Kabila 13 May '03

13 May '03

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How to create an unboxed array?
by Ron de Bruijn 13 May '03

13 May '03

I finally succeeded in creating the unsafeWrite and unsafeRead functions, but the newArray_ function will not work. I get the following error: Compiling Try ( try.hs, interpreted ) try.hs:37: Could not deduce (Num i) from the context (MArray IOUArray Foo IO, HasBounds IOUArray, Monad IO, Ix i) Probable fix: Add (Num i) to the class or instance method `newArray_' arising from use of `*' at try.hs:37 In the first argument of `newArray_', namely `(fst (bound), (snd (bound)) * 2)' In a 'do' expression pattern binding: newArray_ (fst (bound), (snd (bound)) * 2) Failed, modules loaded: none. Prelude> newArray_ bound= do newArray_ (fst(bound),snd(bound)*2) I don't really understand why I have to do something with the Numclass. I think it is to be sure of the fact that the the arguments given in the tuple are indextypes of type Num, so the values can always be multiplied using the * operator. How can I make this work and why will it work the way you would say it would work? And why can't I just put the types of the functions just above the definition, just like I do with any other function, is it just because of the fact this is a method of a class? newArray_::(MArray a e m, Ix i) => (i, i) -> m (a i e) newArray_ bound= do newArray_ (fst(bound),snd bound)) And then yet another question: if I had another datatype, like Foo String Int, then I just had to create two different arrays, one of Int and one of String, right? Greets Ron __________________________________ Do you Yahoo!? The New Yahoo! Search - Faster. Easier. Bingo. http://search.yahoo.com

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