The problem here is even slightly deeper than you might realize. For
example, what if you have a list of functions. How do you compare two
functions to each other to see if they're equal? There is no good way really
to do it! So, not only is == not completely polymorphic, but it CAN'T be.
There is a nice solution for this, however, and it's very simple:
contain :: Eq a -> [a] -> Bool
contain x [] = False
contain x (y:ys) = if x == y then True else contain x ys
The "Eq a" in the type signature says that 'a' must be a member of the 'Eq'
typeclass. That says, in turn, that 'a' must have == defined for it.
Fortunately, most types have, or can easily derive that definition. Here is
the definition of the typeclass:
class Eqhttp://haskell.org/ghc/docs/latest/html/libraries/base/Data-Eq.html#t%3AEqa
where(==)http://haskell.org/ghc/docs/latest/html/libraries/base/Data-Eq.html#v%3A%3D%...::
a -> a ->
Boolhttp://haskell.org/ghc/docs/latest/html/libraries/ghc-prim/GHC-Bool.html#t%3...
(/=)http://haskell.org/ghc/docs/latest/html/libraries/base/Data-Eq.html#v%3A%2F%...::
a -> a ->
Boolhttp://haskell.org/ghc/docs/latest/html/libraries/ghc-prim/GHC-Bool.html#t%3...
That is, for 'a' to be a member of 'Eq', it must have a == operator which
can take 2 values of that type and return a Boolean, saying whether or not
they're equal, and it must also have a definition for the /= operator, which
is "not equal". These two are also defined in terms of each other, so if you
define ==, you get /= for free, and vice versa.
That's probably more information than you needed to know, but I hope it
helps.
2008/12/22 Raeck Zhao
contain :: a -> [a] -> Bool contain x [] = False contain x (y:ys) = if x == y then True else contain x ys it seems that
I am trying to define a containing function to see if a value is one of the elements within a list which is polymorphic, but failed with the following codes: the problem is the 'operator' == does not support a polymorphic check? Any way can solve the problem? or any alternative solution to achieve the purpose? Thanks! Raeck
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