On Sun, Jan 3, 2010 at 2:22 AM, Ivan Lazar Miljenovic <ivan.miljeno...@gmail.com> wrote: > jberryman <brandon.m.simm...@gmail.com> writes: > >> This may be a dumb question, but why can we not declare a Monad >> instance of a type synonym? This question came to me while working >> with the State monad recently and feeling that the requirement that we >> wrap our functions in the State constructor is a bit... kludgy. >> > > Because type defines an _alias_. If you define "type Foo = Maybe Int", > then everywhere you have a "Foo" the compiler should be able to replace > it with "Maybe Int". > > As such, if you have a custom instance on your type synonym (say a > custom Show instance for Foo), then which instance will the compiler > use?
Thanks. I guess what I'm really asking is if there is any way to redefine the monad instance for (->) such that we can have a State monad without the data constructor wrapper. It sounds like there probably isn't, but it seems like that would be a pretty useful thing to be able to do generally. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
