[Haskell-cafe] Inferred type is not general enough
I start some ip networks related work in haskell and wrote two basic classes Location and Packet. Before writing IpLocation and IpPacket instances i have written simple TestLocation and TestPacket instances just to compile this and check for errors in class definitions. But looks like i misunderstand some haskell principles... == class Location a where point :: a - String class Packet a where source, destination :: Location b = a - b size :: Num b = a - b -- data TestLocation = TestSource | TestDestination data TestPacket = TestPacket -- instance Location TestLocation where point a = location instance Packet TestPacket where source p = TestSource destination p = TestDestination size p = 99 $ hugs Test.hs ERROR Test.hs:20 - Inferred type is not general enough *** Expression: source *** Expected type : (Packet TestPacket, Location a) = TestPacket - a *** Inferred type : (Packet TestPacket, Location TestLocation) = TestPacket - TestLocation == But if i remove source and destination from class and instance definitions alone size compiles well. How write this in a haskell way? ___ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] Inferred type is not general enough
Ivan, It's the type of source and destination that is playing you parts. Let's write them explicitly quantified, source :: forall a b . (Packet a, Location b) = a - b destination :: forall a b . (Packet a, Location b) = a - b and reflect on this for a while. This basicly says that whenever I have a value of a type a that is known to be an instance of Packet, then for each type b that is known to be an instance of Location, source will give me an instance of that type b. (The same holds for destination, of course.) So now, I can define a new data type data Foo = Foo , declare it an instance of Location instance Location Foo where point foo = foo , and write the following function, knowing that TestPacket is an instance of Packet, getFoo:: Packet - Foo getFoo packet = source packet . But, clearly, your implementation of source for TestPacket is not prepared to produce a value of Foo. Instead, it always produce a value of TestLocation. To be well-typed, however, it shoud be capable of producing a value of any type that is declared an instance of Location. A possible solution would be to fix the type of the returned location for each instance of Packet, class (Location b) = Packet a b where source, destination :: a - b but this requires multi-parameter type classes, which are not part of Haskell 98. Another approach would be to enrich the interface of the Location class, so that it provides support for the construction of locations. HTH, Stefan ___ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Inferred type is not general enough
Ivan, I don't yet know how to explain this formally, but I know how to fix your problem.. You need to add a parameter to the Packet class so the compiler knows what type to use for the Location. .. The following code works for me.. You'll need to use a compiler/interpreter which supports multi-parameter type classes.. like ghci with the glasgow extensions turned on - class Location a where point :: a - String class Packet a loc where source :: Location loc = a - loc destination :: Location loc = a - loc size :: Num b = a - b -- data TestLocation = TestSource | TestDestination data TestPacket = TestPacket -- instance Location TestLocation where point a = location instance Packet TestPacket TestLocation where source p= TestSource destination p = TestDestination size p = 99 ___ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] Inferred type is not general enough
Ivan, Oops ... SH and write the following function, knowing that SH TestPacket is an instance of Packet, SH SH getFoo:: Packet - Foo SH getFoo packet = source packet . Clearly, I meant to write getFoo :: TestPacket - Foo here. Regards, Stefan ___ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Inferred type is not general enough
Stefan Holdermans wrote: Huh. I'm really get stuck. Can someone write me working implementation of my crap? ___ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Inferred type is not general enough
Ivan.. Uh.. by 'works' I meant 'compiles' :) Here is a fixed version.. As I understand it, the 2 parameter class (Location loc = Packet p loc) means loc is a Location, and types p and loc are related by class Packet with just that information, if you try (as you did) something like $ source TestPacket Then you've given the compiler the type for 'p' (TestPacket), but it won't go the distance and decide that type for 'loc' is TestLocation.. even though that's the only instance you've given it. You can tell the compiler that The type of 'p' sets the type of 'loc' with a functional dependency.. which is the | p - loc annotation at the end of the class definition in the new version below. There is a paper which covers exactly this problem, in detail. I suggest you read it (I did). Type classes with functional dependencies, Mark P. Jones, 2000 You can get it from his homepage at, http://www.cse.ogi.edu/~mpj/ Ben. Ivan Tihonov wrote: It compiles well, but doesn't work for me :( BTW: (email answers are not always real-time :) ) new version: -- class Location a where point:: a - String class Location loc = Packet p loc | p - loc where source :: p - loc destination :: p - loc size :: Num b = p - b -- data TestLocation = TestSource | TestDestination deriving (Show) data TestPacket = TestPacket -- instance Location TestLocation where point a= location instance Packet TestPacket TestLocation where source p = TestSource destination p = TestDestination size p = 99 ___ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] Inferred type is not general enough
Ivan,
IT Stefan Holdermans wrote:
IT
IT Huh. I'm really get stuck. Can someone write me working
IT implementation of my crap?
Huh? Did *I* wrote that ... ? ;)
\begin{code}
{-# -OPTIONS -fglasgow-exts #-}
class Location a where
point :: a - String
class (Location b) = Packet a b where
source :: a - b
destination :: a - b
size:: forall c . (Num c) = a - c
data TestLocation = TestSource | TestDestination
deriving (Show)
instance Location TestLocation where
point = show
data TestPacket = TestPacket
instance Packet TestPacket TestLocation where
source = const TestSource
destination = const TestDestination
size= const 99
main :: IO ()
main = print $ (source TestPacket :: TestLocation)
\end{code}
As I mentioned, depending on your needs, there are other options: enriching
the Location class, parameterizing the packet types, ...
HTH,
Stefan
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Re: [Haskell-cafe] Inferred type is not general enough
Ivan Tihonov wrote: I start some ip networks related work in haskell and wrote two basic classes Location and Packet. Before writing IpLocation and IpPacket instances i have written simple TestLocation and TestPacket instances just to compile this and check for errors in class definitions. But looks like i misunderstand some haskell principles... == class Location a where point :: a - String class Packet a where source, destination :: Location b = a - b This says that source can return any type which is an instance of Location. IOW, any instance of Packet can have any instance of Location as its source and destination (and can even use different instances of Location for the two functions). data TestLocation = TestSource | TestDestination data TestPacket = TestPacket instance Packet TestPacket where source p = TestSource However, this definition of source has type: TestPacket - TestLocation when, according to the class definition, it should have type: (Location b) = TestPacket - b ERROR Test.hs:20 - Inferred type is not general enough *** Expression: source *** Expected type : (Packet TestPacket, Location a) = TestPacket - a *** Inferred type : (Packet TestPacket, Location TestLocation) = TestPacket - TestLocation Hence this error message; the function returns a specific instance of Location, when it should return an arbitry instance. You can use functional dependencies, which aren't in the Haskell98 standard, but are supported by Hugs (use the -98 switch to enable extensions) and GHC (use the -fglasgow-exts switch). I.e.: class (Location b) = Packet a b | a - b where source, destination :: a - b size :: Num c = a - c ... instance Packet TestPacket TestLocation where The class definition says that each instance of Packet uses a specific instance of Location. The instance declaration says that, for TestPacket, source and destination will always have type TestLocation. -- Glynn Clements [EMAIL PROTECTED] ___ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Inferred type is not general enough
thanks a lot, functional dependencies is my choice. ___ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe
