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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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Web Server
by Shawn P. Garbett 13 May '03

13 May '03

I got into a discussion with folks in my local circle of computer heads. I was talking about what I liked about Haskell. Most of the local circle are sold on the idea that Perl is the best language ever written. They eventually challenged me: show us a web site that is being served by Haskell. Does anyone out there run a web site using Haskell? How much effort would be involved in creating a haskell_mod to plug into apache? Shawn

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Re: CC: Functional dependencies question
by oleg＠pobox.com 13 May '03

13 May '03

> > then the user can write > > hbar = 6.62606876e-34 `scalarm` (joule `mult` second) > That works, but IMO it's a bit unwieldy. It also doesn't help with > actually defining a reasonable type signature for hbar which is the > point of the exercise. Well, how about this? hbar = (PhysicalUnit 6.62606876e-34)::t where x::t = (undefined::Energy) `mult` (undefined::Time) the use of bottom might be a bit bizarre -- OTH, a physicist of a kind that cares about Planck constant should get used to the infinities -- take QED for example. A physicist might even appreciate that equation. It's interesting that in every thread I have participated over the last couple of months I mentioned 'undefined'. It must be my favorite Haskell value.

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CC: Functional dependencies question
by oleg＠pobox.com 13 May '03

13 May '03

Hello! Andrew J Bromage wrote: > 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". The type of bar is indeed too general. It should be bar:: forall t. (Foo Char t | Char -> t) => t or bar:: exists t. (Foo Char t) => t because the functional dependency implies that there can be only one such type that relates to Char. Yet such types are not expressible. Perhaps because of that, GHC and Hugs delay reporting an error. For example, if we omit the instance declaration, both GHC and Hugs seem happy. But they only seem. In Hugs, if we try to actually evaluate 'bar', we get Main> bar ERROR - Unresolved overloading *** Type : Foo Char a => a *** Expression : bar Indeed, an instance declaration is required. Incidentally, GHC can report the error even before running the code: if we just want to see the type of bar: *Main> :t bar No instance for (Foo Char t) arising from use of `bar' at <No locn> If we add an instance of Foo, such as "instance Foo Char Bool", then the signature of bar is clearly too polymorphic: there can be only one type t (namely, Bool) for which the constraint Foo Char t is true. Incidentally, local type variable again seem to help -- they can help express types that are normally inexpressible. If we define bar as bar :: (Foo Char t) => t bar ::tbar = (foo::(Char->tbar)) 'a' we get an error: /tmp/f.hs:10: Inferred type is less polymorphic than expected Quantified type variable `t' escapes It is mentioned in the environment: tbar = t is bound by the pattern type signature at /tmp/f.hs:1 When trying to generalize the type inferred for `bar' Signature type: forall t. (Foo Char t) => t Type to generalise: t When checking the type signature for `bar' When generalising the type(s) for `bar' which alerts us that the given type of bar is indeed too general. Now, we get the error right away. If we drop the explicit type declaration, bar ::tbar = (foo::(Char->tbar)) 'a' we get another error, /tmp/f.hs:12: No instance for (Foo Char tbar) arising from use of `foo' at /tmp/f.hs:12 When checking the type signature of the expression: foo :: Char -> tbar In the definition of `bar': (foo :: Char -> tbar) 'a' The compiler demands to see an instance of Foo Char t. If we add the instance now, the compiler is happy. bar ::tbar = (foo::(Char->tbar)) 'a' instance Foo Char Bool where foo c = c == 'a' *Main> bar True and the code even runs.

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