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is there something special about the Num instance?

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Re: [Haskell-cafe] is there...

Anatoly Yakovenko

3 Dec 2008 3 Dec '08

6:05 p.m.

module Test where --why does this work: data Test = Test class Foo t where foo :: Num v => t -> v -> IO () instance Foo Test where foo _ 1 = print $ "one" foo _ _ = print $ "not one" --but this doesn't? class Bar t where bar :: Foo v => t -> v -> IO () instance Bar Test where bar _ Test = print $ "test" bar _ _ = print $ "not test"

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Daniel Fischer

3 Dec 3 Dec

6:34 p.m.

Am Donnerstag, 4. Dezember 2008 00:05 schrieb Anatoly Yakovenko:

module Test where --why does this work: data Test = Test

class Foo t where foo :: Num v => t -> v -> IO ()

instance Foo Test where foo _ 1 = print $ "one" foo _ _ = print $ "not one"

--but this doesn't?

class Bar t where bar :: Foo v => t -> v -> IO ()

instance Bar Test where bar _ Test = print $ "test" bar _ _ = print $ "not test"

Because bar has to work for all types which belong to class Foo, but actually uses the type Test. This is what the error message Test.hs:18:10: Couldn't match expected type `v' against inferred type `Test' `v' is a rigid type variable bound by the type signature for `bar' at Test.hs:15:15 In the pattern: Test In the definition of `bar': bar _ Test = print $ "test" In the definition for method `bar' tells you. In the signature of bar, you've said that bar works for all types v which are members of Foo. Test is a monomorphic value of type Test, so it can't have type v for all v which belong to Foo. It doesn't matter that there is so far only the one instance of Foo, there could be others defined in other modules. The first works because the type of 1 in the definition of foo is defaulted to Integer (or whatever you specified in the default declaration).

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Ryan Ingram

6:47 p.m.

Yes; I had a similar question, and it turns out Num is special, or rather, pattern matching on integer literals is special. See the thread http://www.nabble.com/Pattern-matching-on-numbers--td20571034.html The summary is that pattern matching on a literal integer is different than a regular pattern match; in particular:

foo 1 = print "one" foo _ = print "not one"

turns into

foo x = if x == fromInteger 1 then "one" else "not one"

whereas

bar Test = print "Test" bar _ = print "Not Test"

turns into

bar x = case x of { Test -> print "Test" ; _ -> print "Not Test" }

In the former case, the use of (y == fromInteger 1) means that "foo" works on any argument within the class Num (which requires Eq), whereas in the latter case, the use of the constructor Test directly turns into a requirement for a particular type for "bar". There's no way to get special pattern matching behavior for other types; this overloading is specific to integer literals. -- ryan On Wed, Dec 3, 2008 at 3:05 PM, Anatoly Yakovenko wrote:

module Test where --why does this work: data Test = Test

class Foo t where foo :: Num v => t -> v -> IO ()

instance Foo Test where foo _ 1 = print $ "one" foo _ _ = print $ "not one"

--but this doesn't?

class Bar t where bar :: Foo v => t -> v -> IO ()

instance Bar Test where bar _ Test = print $ "test" bar _ _ = print $ "not test" _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

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Anatoly Yakovenko

7:08 p.m.

Thanks for your help. On Wed, Dec 3, 2008 at 3:47 PM, Ryan Ingram wrote:

Yes; I had a similar question, and it turns out Num is special, or rather, pattern matching on integer literals is special. See the thread

http://www.nabble.com/Pattern-matching-on-numbers--td20571034.html

The summary is that pattern matching on a literal integer is different than a regular pattern match; in particular:

...
foo 1 = print "one" foo _ = print "not one"

turns into

...
foo x = if x == fromInteger 1 then "one" else "not one"

whereas

...
bar Test = print "Test" bar _ = print "Not Test"

turns into

...
bar x = case x of { Test -> print "Test" ; _ -> print "Not Test" }

In the former case, the use of (y == fromInteger 1) means that "foo" works on any argument within the class Num (which requires Eq), whereas in the latter case, the use of the constructor Test directly turns into a requirement for a particular type for "bar".

There's no way to get special pattern matching behavior for other types; this overloading is specific to integer literals.

-- ryan

On Wed, Dec 3, 2008 at 3:05 PM, Anatoly Yakovenko wrote:

...
module Test where --why does this work: data Test = Test

class Foo t where foo :: Num v => t -> v -> IO ()

instance Foo Test where foo _ 1 = print $ "one" foo _ _ = print $ "not one"

--but this doesn't?

class Bar t where bar :: Foo v => t -> v -> IO ()

instance Bar Test where bar _ Test = print $ "test" bar _ _ = print $ "not test" _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

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Anatoly Yakovenko
Daniel Fischer
Ryan Ingram