Re: [Haskell-cafe] Existential question
On Wed, Aug 17, 2011 at 4:49 PM, Tom Schouten t...@zwizwa.be wrote:
{-# LANGUAGE ExistentialQuantification #-}
-- Dear Cafe, this one works.
data Kl' s i o = Kl' (i - s - (s, o))
iso' :: (i - Kl' s () o) - Kl' s i o
iso' f = Kl' $ \i s - (\(Kl' kl') - kl' () s) (f i)
-- Is there a way to make this one work also?
data Kl i o = forall s. Kl (i - s - (s, o))
iso :: (i - Kl () o) - Kl i o
iso f = Kl $ \i s - (\(Kl kl) - kl () s) (f i)
Not without moving the quantifier, as Oleg says. Here is why:
Existential types are equivalent to packing up a type with the constructor;
imagine KI as
data KI i o = KI @s (i - s - (s,o)) -- not legal haskell
where @s represents hold a type s which can be used in the other elements
of the structure. An example element of this type:
KI @[Int] (\i s - (i:s, sum s)) :: KI Int Int
Trying to implement iso as you suggest, we end up with
iso f = KI ?? (\i s - case f i of ...)
What do we put in the ??. In fact, it's not possible to find a solution;
consider this:
ki1 :: KI () Int
ki1 = KI @Int (\() s - (s+1, s))
ki2 :: KI () Int
ki2 = KI @() (\() () - ((), 0))
f :: Bool - KI () Int
f x = if x then ki1 else ki2
iso f = KI ?? ??
The problem is that we have multiple possible internal state types!
-- ryan
{-
Couldn't match type `s0' with `s'
because type variable `s' would escape its scope
This (rigid, skolem) type variable is bound by
a pattern with constructor
Kl :: forall i o s. (i - s - (s, o)) - Kl i o,
in a lambda abstraction
The following variables have types that mention s0
s :: s0 (bound at /home/tom/meta/haskell/iso.hs:**11:17)
In the pattern: Kl kl
In the expression: \ (Kl kl) - kl () s
In the expression: (\ (Kl kl) - kl () s) (f i)
-}
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[Haskell-cafe] Simon PJ in Australia
This is a message for Australian FPers -- sorry to spam the rest of you. I'm crossing the planet to Australia for YOW (Melbourne Dec 1-2, Brisbane Dec 5-6). http://www.yowconference.com.au/YOW2011/ I'm committed for the conference dates of course, but I could spend a few days before or afterwards, hanging out with some of you guys. Hack on GHC, give a talk, eat pizza, university departments, Haskell users groups, surfing (on water, that is) Do drop me (not the list) a line if any of that sounds fun. Simon PS: John Hughes is coming to Yow too, so you could try luring him into your lair too. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Simon PJ in Australia
On 08/19/2011 05:56 PM, Simon Peyton-Jones wrote: This is a message for *Australian FPers* -- sorry to spam the rest of you. I'm crossing the planet to Australia for YOW (Melbourne Dec 1-2, Brisbane Dec 5-6). http://www.yowconference.com.au/YOW2011/ I'm committed for the conference dates of course, but I could spend a few days before or afterwards, hanging out with some of you guys. Hack on GHC, give a talk, eat pizza, university departments, Haskell users groups, surfing (on water, that is) Do drop me (not the list) a line if any of that sounds fun. Simon PS: John Hughes is coming to Yow too, so you could try luring him into your lair too. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ...and I will be running a Haskell/FP workshop. I'm as keen as Simon to meet others from further afield with similar interests. http://www.yowconference.com.au/YOW2011/general/workshopDetails.html?eventId=3552 PS: we have a significant FP community here in Brisbane with nearly 250 members http://bfpg.org/ -- Tony Morris http://tmorris.net/ ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] custom SQL-to-Haskell type conversion in HDBC
On Fri, Aug 19, 2011 at 07:23, Henry House hajho...@hajhouse.org wrote: Does there exist any sample code or other resources on writing a custom SQL-to-Haskell datatype converter instance for use with HDBC that would be accessible to someone just starting with Haskell? The reason I need this is because of this problem (using PostgreSQL): Prelude Database.HDBC.PostgreSQL Database.HDBC res - (quickQuery db select 1::numeric(5,4); []) Prelude Database.HDBC.PostgreSQL Database.HDBC res [[SqlRational (1 % 1)]] Prelude Database.HDBC.PostgreSQL Database.HDBC res - (quickQuery db select 1::numeric(5,0); []) [[SqlRational (1 % 1)]] where db is a database connection. The SQL values 1::numeric(5,4) and 1::numeric(5,0) are supposed to be fixed-precision numbers having 4 and zero significant decimal figures after the decimal point, respectively. Both are offered by HDBC as the same SqlValue, SqlRational (1 % 1) but they are not really the same at all. The precision information has been lost. The native outputs of PostgreSQL, before HDBC's type conversion, are 1. and 1 for 'select 1::numeric(5,4);' and 'select 1::numeric(5,0);', respectively. Do you really need the precision info about the column, or do you just need the values at the right precision? Because you get the last thing already: Prelude Database.HDBC.PostgreSQL Database.HDBC Data.Ratio Control.Monad (fromSql . head . head) `liftM` (quickQuery db select 1.231 ::numeric(5,0); []) :: IO Rational 1 % 1 Prelude Database.HDBC.PostgreSQL Database.HDBC Data.Ratio Control.Monad (fromSql . head . head) `liftM` (quickQuery db select 1.231 ::numeric(5,4); []) :: IO Rational 1231 % 1000 If you need the precision information, perhaps 'describeResult' will give you what you need. I've never used it, but it looks like it might. Erik ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Simon PJ in Australia
You heard it here first; Simon's down here in Aus to eat pizza, university departments, Haskell users groups. Will he leave nothing left!? ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Problem with types
Hi all, Suppose I have a compound data type - data M o = M (String,o) Now, I can define a function that works for ALL M irrespective of o. For example - f :: M o - M o f (M (s,o)) = M (s++!, o) I can also use this function in an expression, applying it to different types without problem - p = (m1',m2') where m1 = M (1, ()) m2 = M (2, True) m1' = f m1 m2' = f m2 Main* p (M (1!,()),M (2!,True)) However, if I try to parameterise over the function 'f' it does not work! - p f = (m1',m2') where m1 = M (1, ()) m2 = M (2, True) m1' = f m1 m2' = f m2 It doesn't even typecheck, producing the error - Couldn't match expected type 'Bool' with actual type '()' Is there a particular reason for this? How can I define a function like 'p' within Haskell? Thanks, Anupam Jain ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Problem with types
p :: (forall o. M o - M o) - ... Отправлено с iPad 19.08.2011, в 16:06, Anupam Jain ajn...@gmail.com написал(а): Hi all, Suppose I have a compound data type - data M o = M (String,o) Now, I can define a function that works for ALL M irrespective of o. For example - f :: M o - M o f (M (s,o)) = M (s++!, o) I can also use this function in an expression, applying it to different types without problem - p = (m1',m2') where m1 = M (1, ()) m2 = M (2, True) m1' = f m1 m2' = f m2 Main* p (M (1!,()),M (2!,True)) However, if I try to parameterise over the function 'f' it does not work! - p f = (m1',m2') where m1 = M (1, ()) m2 = M (2, True) m1' = f m1 m2' = f m2 It doesn't even typecheck, producing the error - Couldn't match expected type 'Bool' with actual type '()' Is there a particular reason for this? How can I define a function like 'p' within Haskell? Thanks, Anupam Jain ___ 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
Re: [Haskell-cafe] Problem with types
On Fri, Aug 19, 2011 at 14:06, Anupam Jain ajn...@gmail.com wrote: Hi all, Suppose I have a compound data type - data M o = M (String,o) Now, I can define a function that works for ALL M irrespective of o. For example - f :: M o - M o f (M (s,o)) = M (s++!, o) I can also use this function in an expression, applying it to different types without problem - p = (m1',m2') where m1 = M (1, ()) m2 = M (2, True) m1' = f m1 m2' = f m2 Main* p (M (1!,()),M (2!,True)) However, if I try to parameterise over the function 'f' it does not work! - p f = (m1',m2') where m1 = M (1, ()) m2 = M (2, True) m1' = f m1 m2' = f m2 It doesn't even typecheck, producing the error - Couldn't match expected type 'Bool' with actual type '()' Is there a particular reason for this? How can I define a function like 'p' within Haskell? If you write down the type for 'p', you get something like this: p :: (forall a. M a - M a) - (M b, M b) That is, the type variable 'a' isn't top level, but is forall'ed in the function you pass. This is called a rank 2 type, and it cannot be inferred automatically by GHC. You have to specify it yourself, and turn on the Rank2Types or RankNTypes extension. Erik ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Problem with types
Anupam Jain ajn...@gmail.com wrote: However, if I try to parameterise over the function 'f' it does not work! - p f = (m1',m2') where m1 = M (1, ()) m2 = M (2, True) m1' = f m1 m2' = f m2 It doesn't even typecheck, producing the error - Couldn't match expected type 'Bool' with actual type '()' Is there a particular reason for this? How can I define a function like 'p' within Haskell? Try to write the type signature for 'p'. And as a general advice: Always write type signatures. Greets, Ertugrul -- nightmare = unsafePerformIO (getWrongWife = sex) http://ertes.de/ ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Failed link to mixed-language shared object.
On Thu, 2011-08-18 at 20:32 -0700, David Banas wrote:
Hi all,
Does this trigger recollection in anyone:
dbanas@dbanas-eeepc:~/prj/haskell/AMIParse/trunk$ make
ghc -dynamic -o ami_test -L. -lami ami_test.o
./libami.so: undefined reference to `__stginit_haskell98_MarshalArray_'
./libami.so: undefined reference to `__stginit_haskell98_MarshalError_'
collect2: ld returned 1 exit status
make: *** [ami_test] Error 1
dbanas@dbanas-eeepc:~/prj/haskell/AMIParse/trunk$
?
Know what I need to do?
Thanks,
-db
So, I was able to make my link errors go away, by adding `-shared' to my
command line:
dbanas@dbanas-eeepc:~/prj/haskell/AMIParse/trunk$ make
ghc -no-hs-main -shared -dynamic -o ami_test -L. -lami ami_test.o
However, when I try to execute the resultant program, I get a
segmentation fault:
dbanas@dbanas-eeepc:~/prj/haskell/AMIParse/trunk$ ./ami_test test.ami
Segmentation fault
Tracing this shows that the `main' pointer of the program appears to be
NULL:
dbanas@dbanas-eeepc:~/prj/haskell/AMIParse/trunk$ gdb --args ./ami_test
test.ami
GNU gdb (GDB) 7.1-ubuntu
Copyright (C) 2010 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later
http://gnu.org/licenses/gpl.html
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law. Type show
copying
and show warranty for details.
This GDB was configured as i486-linux-gnu.
For bug reporting instructions, please see:
http://www.gnu.org/software/gdb/bugs/...
Reading symbols
from /home/dbanas/prj/haskell/AMIParse/trunk/ami_test...done.
(gdb) l
2 #include stdlib.h
3 #include string.h
4 //#include HsFFI.h
5 #include ami_model.h
6
7 #define DEF_AMI_FILE test.ami
8 #define VEC_SIZE 8
9 #define MAX_LINE_LEN 256
10
11 int main(int argc, char *argv[]) {
(gdb) b 11
Breakpoint 1 at 0x87d: file ami_test.c, line 11.
(gdb) run
Starting program: /home/dbanas/prj/haskell/AMIParse/trunk/ami_test
test.ami
Program received signal SIGSEGV, Segmentation fault.
0x0002 in ?? ()
I assume that's because of the `-shared' flag.
Any thoughts?
Thanks,
-db
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Re: [Haskell-cafe] custom SQL-to-Haskell type conversion in HDBC
On Friday, 19 August 2011, Erik Hesselink wrote: Do you really need the precision info about the column, or do you just need the values at the right precision? Because you get the last thing already: Prelude Database.HDBC.PostgreSQL Database.HDBC Data.Ratio Control.Monad (fromSql . head . head) `liftM` (quickQuery db select 1.231 ::numeric(5,0); []) :: IO Rational 1 % 1 Prelude Database.HDBC.PostgreSQL Database.HDBC Data.Ratio Control.Monad (fromSql . head . head) `liftM` (quickQuery db select 1.231 ::numeric(5,4); []) :: IO Rational 1231 % 1000 I'm not sure I understand the distinction --- to my way of thinking, getting the value at the right precision means getting the correct number of significant decimal digits, which both your example and mine fail to provide. Prelude Database.HDBC.PostgreSQL Database.HDBC Data.Ratio Control.Monad (fromSql . head . head) `liftM` (quickQuery db select 1.231 ::numeric(10,4); []) :: IO Rational -- gives 1231 % 1000 == 1.231 in decimal notation Prelude Database.HDBC.PostgreSQL Database.HDBC Data.Ratio Control.Monad (fromSql . head . head) `liftM` (quickQuery db select 1.231 ::numeric(10,8); []) :: IO Rational -- still gives 1231 % 1000 but should be 1.2131 in decimal notation -- or 1231000 % 100 in rational notation If you need the precision information, perhaps 'describeResult' will give you what you need. I've never used it, but it looks like it might. Thanks for the suggestion. That will work, but ideally i'd like to have the normal usage of quickQuery, etc return correct data. Unless I am missing something here, the method of getting precision information using describeResult would be to fetch the result set, then find any columns having a numeric(x,y) SQL type and transform them. Even just getting the raw string response from the RDBMS would be better than that since the raw response is actually already a correctly formatted string representation of the data so I could use it for output without any further transformation. Is there a way to get raw response data from the RDBMS through HDBC? -- Henry House +1 530 848-1238 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] custom SQL-to-Haskell type conversion in HDBC
On Fri, Aug 19, 2011 at 16:09, Henry House hajho...@hajhouse.org wrote: On Friday, 19 August 2011, Erik Hesselink wrote: Do you really need the precision info about the column, or do you just need the values at the right precision? Because you get the last thing already: Prelude Database.HDBC.PostgreSQL Database.HDBC Data.Ratio Control.Monad (fromSql . head . head) `liftM` (quickQuery db select 1.231 ::numeric(5,0); []) :: IO Rational 1 % 1 Prelude Database.HDBC.PostgreSQL Database.HDBC Data.Ratio Control.Monad (fromSql . head . head) `liftM` (quickQuery db select 1.231 ::numeric(5,4); []) :: IO Rational 1231 % 1000 I'm not sure I understand the distinction --- to my way of thinking, getting the value at the right precision means getting the correct number of significant decimal digits, which both your example and mine fail to provide. Prelude Database.HDBC.PostgreSQL Database.HDBC Data.Ratio Control.Monad (fromSql . head . head) `liftM` (quickQuery db select 1.231 ::numeric(10,4); []) :: IO Rational -- gives 1231 % 1000 == 1.231 in decimal notation Prelude Database.HDBC.PostgreSQL Database.HDBC Data.Ratio Control.Monad (fromSql . head . head) `liftM` (quickQuery db select 1.231 ::numeric(10,8); []) :: IO Rational -- still gives 1231 % 1000 but should be 1.2131 in decimal notation -- or 1231000 % 100 in rational notation The % notation is a rational, so 'infinite' precision. So '1 % 1' and '1000 % 1000' are exactly the same, semantically. It's like fractions instead of decimal digits. Why exactly do you need the precision information? Erik ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] custom SQL-to-Haskell type conversion in HDBC
On Fri, Aug 19, 2011 at 10:09, Henry House hajho...@hajhouse.org wrote: I'm not sure I understand the distinction --- to my way of thinking, getting the value at the right precision means getting the correct number of significant decimal digits, which both your example and mine fail to provide. So you want display format data instead of something that you can do computations with? HDBC is giving you the latter; the precision is correct for computation, although it could be argued that a Fixed type would be better if you intend to propagate exact precision through operations. (That said, I believe Fixed obeys exact mathematical precision instead of replicating your SQL backend's limitations, so you'd be unhappy again in case of multiplication or division.) I would suggest that if you want SQL formatted string output, you describe that in the query instead of expecting HDBC to convert numeric data to string while duplicating your SQL backend's formatting. Otherwise, perhaps you'd be better suited to a language with a looser type system, so you can pretend string and numeric values are the same thing and usually get something resembling the right result. -- brandon s allbery allber...@gmail.com wandering unix systems administrator (available) (412) 475-9364 vm/sms ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] custom SQL-to-Haskell type conversion in HDBC
On Friday, 19 August 2011, Erik Hesselink wrote: On Fri, Aug 19, 2011 at 16:09, Henry House hajho...@hajhouse.org wrote: On Friday, 19 August 2011, Erik Hesselink wrote: Do you really need the precision info about the column, or do you just need the values at the right precision? Because you get the last thing already: Prelude Database.HDBC.PostgreSQL Database.HDBC Data.Ratio Control.Monad (fromSql . head . head) `liftM` (quickQuery db select 1.231 ::numeric(5,0); []) :: IO Rational 1 % 1 Prelude Database.HDBC.PostgreSQL Database.HDBC Data.Ratio Control.Monad (fromSql . head . head) `liftM` (quickQuery db select 1.231 ::numeric(5,4); []) :: IO Rational 1231 % 1000 I'm not sure I understand the distinction --- to my way of thinking, getting the value at the right precision means getting the correct number of significant decimal digits, which both your example and mine fail to provide. Prelude Database.HDBC.PostgreSQL Database.HDBC Data.Ratio Control.Monad (fromSql . head . head) `liftM` (quickQuery db select 1.231 ::numeric(10,4); []) :: IO Rational -- gives 1231 % 1000 == 1.231 in decimal notation Prelude Database.HDBC.PostgreSQL Database.HDBC Data.Ratio Control.Monad (fromSql . head . head) `liftM` (quickQuery db select 1.231 ::numeric(10,8); []) :: IO Rational -- still gives 1231 % 1000 but should be 1.2131 in decimal notation -- or 1231000 % 100 in rational notation The % notation is a rational, so 'infinite' precision. So '1 % 1' and '1000 % 1000' are exactly the same, semantically. It's like fractions instead of decimal digits. You're right, of course. My example was something of an abuse of notation to include a notion of precision where it is actually an incompatible concept. Why exactly do you need the precision information? Empirical measurements (e.g., sizes of some fields in hectares) are precise only to a certain level of measurement error. Thus, the area measurements 1 ha and 1.000 ha are not equivalent or interchangeable. Database engines recognize this fact by providing different data types for rational numbers and fixed-precision decimal numbers. The bottom line for me is that the conversion of a fixed-precision decimal number as a rational is both throwing away information (the precision) as well as introducing bogus information (the notion that the result value has greater --- i.e., infinite --- precision that was in fact intended when that value was stored). -- Henry House +1 530 848-1238 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] custom SQL-to-Haskell type conversion in HDBC
On Fri, Aug 19, 2011 at 16:53, Henry House hajho...@hajhouse.org wrote: On Friday, 19 August 2011, Erik Hesselink wrote: Why exactly do you need the precision information? Empirical measurements (e.g., sizes of some fields in hectares) are precise only to a certain level of measurement error. Thus, the area measurements 1 ha and 1.000 ha are not equivalent or interchangeable. Database engines recognize this fact by providing different data types for rational numbers and fixed-precision decimal numbers. The bottom line for me is that the conversion of a fixed-precision decimal number as a rational is both throwing away information (the precision) as well as introducing bogus information (the notion that the result value has greater --- i.e., infinite --- precision that was in fact intended when that value was stored). Note that PostgreSQL also doesn't work with decimals as precision: postgres=# select 1::decimal(4,2) * 1::decimal(4,2); ?column? -- 1. (1 row) That should be 1.00 instead if you want the precision correctly represented. Perhaps a solution would be to not treat the database precision as your primary source of information, but represent that in Haskell using some data type that correctly propagates precision information, and marshall your database data to and from that. This means some duplication of information (precision in both database and Haskell) but you do the same with NULL and Maybe, etc. I guess that's inherent to (the way HDBC does) database access. Erik ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Can I have a typeclass for topological spaces?
You may want to, instead, check out the methods of point-free topology (also known as pointless topology) where you, basically, study topological spaces in terms of their open-set lattices. This allows you to use a more algebraic language which probably fits better into Haskell (I think this is used, for example, when studying topology within type theories). A search for stone-duality should give you somewhere to start (I know there are a sequence of posts on http://topologicalmusings.wordpress.com/ on the subject). Somewhat less known is the theory of formal topology (see for example n-lab http://ncatlab.org/nlab/show/formal+topology). Which might also be an approach to doing some topology in Haskell. Hope that's of some help :) On 11/08/2011, Grigory Sarnitskiy sargrig...@ya.ru wrote: Hello! I just wonder whether it is possible to have a typeclass for topological spaces? ___ 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
Re: [Haskell-cafe] why is Random in System?
Yep, but don't conflate determinism with splitting. In the imperative world, you normally know how many CPUs you have, so you initialize one PRNG per CPU, and simply go from there; there's no need for splitting. In the parallel community, people are going out of their way to *avoid* splitting. I'm having trouble thinking of scenarios where per-CPU does the trick. Do you mean one per pthread rather than one per CPU? In the Cilk case, you've got to deal with work stealing of course. So you want rand() to generate a result that is determined by the current stack-frame's position in the binary-tree of spawns. In the work I was referring to: http://groups.csail.mit.edu/sct/wiki/index.php?title=Other_Projects#Deterministic_Parallel_Random-Number_Generation ... they try a bunch of different methods and I can't remember if any of them split the RNG eagerly as they go down the spawn tree or if they just record the tree-index on the way down and then read it out when they generated randoms. (I think the latter.) Cheers, -Ryan ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] custom SQL-to-Haskell type conversion in HDBC
On Fri, Aug 19, 2011 at 11:45, Erik Hesselink hessel...@gmail.com wrote: Note that PostgreSQL also doesn't work with decimals as precision: postgres=# select 1::decimal(4,2) * 1::decimal(4,2); ?column? -- 1. (1 row) That should be 1.00 instead if you want the precision correctly represented. Er? Last I checked, that was exactly how precision worked over multiplication; otherwise you are incorrectly discarding precision present in the original values. Unless you're assuming the OP actually wants an incorrect flat precision model -- brandon s allbery allber...@gmail.com wandering unix systems administrator (available) (412) 475-9364 vm/sms ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Question about data
I've created a simple type declaration: data MathExpression = Float | Add MathExpression MathExpression | Subtract MathExpression MathExpression | Multiply MathExpression MathExpression | Divide MathExpression MathExpression deriving (Show) Now how do I create an instance of MathExpression which is just a Float? This doesn't work: *Main let pi = 3.14 :: MathExpression interactive:1:10: No instance for (Fractional MathExpression) arising from the literal `3.14' Possible fix: add an instance declaration for (Fractional MathExpression) In the expression: 3.14 :: MathExpression In an equation for `pi': pi = 3.14 :: MathExpression ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Question about data
This is not a valid data declaration. You can't have a Float field without any constructor name and have it still of type MathExpression. I suggest you do something like: data MathExpr = MathFloat Float So you may declare pi: let mathPi = MathFloat pi -- note pi is defined in the prelude alread Cheers, Thomas On Fri, Aug 19, 2011 at 1:40 PM, Paul Reiners paul.rein...@gmail.com wrote: I've created a simple type declaration: data MathExpression = Float | Add MathExpression MathExpression | Subtract MathExpression MathExpression | Multiply MathExpression MathExpression | Divide MathExpression MathExpression deriving (Show) Now how do I create an instance of MathExpression which is just a Float? This doesn't work: *Main let pi = 3.14 :: MathExpression interactive:1:10: No instance for (Fractional MathExpression) arising from the literal `3.14' Possible fix: add an instance declaration for (Fractional MathExpression) In the expression: 3.14 :: MathExpression In an equation for `pi': pi = 3.14 :: MathExpression ___ 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
Re: [Haskell-cafe] Question about data
On Fri, Aug 19, 2011 at 1:45 PM, Thomas DuBuisson thomas.dubuis...@gmail.com wrote: This is not a valid data declaration. You can't have a Float field without any constructor name and have it still of type And the reason why it accepts 'data MathExpr = Float', is because data constructors and types live in separate namespaces. There is already a type called Float in the Prelude, but there is no 'Float' data constructor. (Just pointing this out in case it's not obvious!) Jason ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Question about data
Others have pointed out your first bug. To get floating-point numeric literals you will need to define instances of Num and then Fractional for your symbolic type. On 19 August 2011 21:40, Paul Reiners paul.rein...@gmail.com wrote: *Main let pi = 3.14 :: MathExpression ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Question about data
Your MathExpression data type has nothing to do with numbers of any kind. Your Float data constructor doesn't mean that float numbers are a part of your type; instead it means that you have a SINGLE value of type MathExpression, and this value is named Float. You should modify your data declaration as data MathExpression = Float Float | …, and after that you can write something like a = Float 4.2, which would automatically make a a value of type MathExpession. Отправлено с iPad 20.08.2011, в 0:40, Paul Reiners paul.rein...@gmail.com написал(а): I've created a simple type declaration: data MathExpression = Float | Add MathExpression MathExpression | Subtract MathExpression MathExpression | Multiply MathExpression MathExpression | Divide MathExpression MathExpression deriving (Show) Now how do I create an instance of MathExpression which is just a Float? This doesn't work: *Main let pi = 3.14 :: MathExpression interactive:1:10: No instance for (Fractional MathExpression) arising from the literal `3.14' Possible fix: add an instance declaration for (Fractional MathExpression) In the expression: 3.14 :: MathExpression In an equation for `pi': pi = 3.14 :: MathExpression ___ 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
Re: [Haskell-cafe] Simon PJ in Australia
On 11-08-19 06:54 AM, Arlen Cuss wrote: You heard it here first; Simon's down here in Aus to eat pizza, university departments, Haskell users groups. Will he leave nothing left!? He is not eating newbies. But I am. I love newbies. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Simon PJ in Australia
On Fri, Aug 19, 2011 at 20:23, Albert Y. C. Lai tre...@vex.net wrote: On 11-08-19 06:54 AM, Arlen Cuss wrote: You heard it here first; Simon's down here in Aus to eat pizza, university departments, Haskell users groups. Will he leave nothing left!? He is not eating newbies. But I am. I love newbies. Hasn't he (or his brainchild) already eaten Haskell users groups? -- brandon s allbery allber...@gmail.com wandering unix systems administrator (available) (412) 475-9364 vm/sms ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
