On Thu, Dec 02, 1999 at 11:11:12AM +0100, Michael Erik Florentin Nielsen wrote:
>
> > My problem:
> > ----------
> > One of the algorithms I have to implement is the
> > addition of symbolic expressions. It should have
> > two symbolic expressions as arguments and should
> > produce a symbolic expression as the result. But
> > how the result is produced is depending on series
> > of flags that control how the expressions is to
> > be manipulated. This set of flags should then be
> > passed as a third argument to addition function.
> > This is the correct way of doing it. But, being
> > a Mathematics application, my system should preserve
> > the tradicional Math notation (that is, infix
> > operators with suitable associations defined). So
> > my symbolic expression type should be an instance
> > of the Num class so that the (+) operator can
> > be overloaded for it. But, as the function has
> > now three arguments, it cannot be a binary operator
> > anymore.
>
> There are no problems in having infix operators with more than two
> parameters, eg. the operators `plus` and `times` below both take a third
> parameter:
>
> infixl 6 `plus`
> infixl 7 `times`
>
> int :: Int->(Int->Int)
> int n x = n
>
> x :: (Int->Int)
> x y = y
>
> plus :: (Int->Int)->(Int->Int)->(Int->Int)
> (a `plus` b) x = a x + b x
>
> times :: (Int->Int)->(Int->Int)->(Int->Int)
> (a `times` b) x = a x * b x
>
> This allows for expressions such as (x `times` (int 1 `plus` x)) 7 that
> evaluates to 56. Replacing + and * for `plus` and `times`, that is
> making (a suitable newtype over) Int->Int into an instance of Num is
> more or less trivial - except for some silly choices when making the Eq
> instance: Whether expressions of this type are equal should probably
> depend on the environment. But as the return type of == is Bool the
> environment will not be available and you would have to make == a dummy
> function.
This solution seems to fit my problem. I will just investigate it
a little more. Thanks.
> In fact you would probably be better of by hiding the prelude
> and overloading + and friends on your own.
Is there any directions on how to hide the prelude and still use the
definitions it exports?
Romildo.
--
Prof. José Romildo Malaquias <[EMAIL PROTECTED]>
Departamento de Computação
Universidade Federal de Ouro Preto
Brasil
