[Haskell-cafe] Why can't we make an instance declaration on a type synonym?

Brandon Simmons brandon.m.simmons at gmail.com
Mon Jan 4 11:50:57 EST 2010

On Sun, Jan 3, 2010 at 2:22 AM, Ivan Lazar Miljenovic
<ivan.miljenovic at gmail.com> wrote:
> jberryman <brandon.m.simmons at 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.

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