Another option is to just add the instance:
> instance FooBar Char
instead of
> instance (Foo a, Bar a) => FooBar a
Now you don't need any extensions, the disadvantage is that you have to
add an instance for each type...
There has also been an proposal for type class synonyms:
http://repetae.net/john/recent/out/classalias.html
Maybe you like it,
Gerrit
Sean Seefried wrote:
On 24/03/2006, at 12:45 PM, Fritz Ruehr wrote:
What is the easiest way to name a combination of type classes, i.e.,
to abbreviate the fact that a certain type is an instance of several
classes simultaneously? I have a vague sense that this is do-able,
but that I am messing up by trying to use an empty class body as below.
So in the code below, I try to use FooBar to abbreviate the
conjunction of Foo and Bar. But while f (which uses a FooBar
constraint) has a valid definition, it can't be used. On the other
hand, g (which uses the long-winded constraint), is both a valid
defined and useable.
(In a real example, imagine that FooBar names a conjunction of a
half dozen things, so that the g-like form really is onerous,
whereas the f-like form would be sweet and tidy :) .)
Hi Fritz!
You only need to do a couple of things to get this working. Add an
instance declaration:
instance (Foo a, Bar a) => FooBar a
But for this to work you need to allow undecidable instances (and -
fglasgow-exts).
To have this type class synonym trick work you need both the class
and instance declaration:
class (Foo a, Bar a) => FooBar a
instance (Foo a, Bar a) => FooBar a
The first ensures that members of class FooBar will inherit the
methods of classes Foo and Bar. The second ensures that if there is a
Foo and a Bar instance then there will be a FooBar instance. You were
lacking this in your code hence the error message:
> f 'a'
No instance for (FooBar Char)
arising from use of `f' at <interactive>:1:0
Probable fix: add an instance declaration for (FooBar Char)
In the definition of `it': it = f 'a'
This is a neat trick. I've also used it to reduce onerous contexts.
Sean
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