[Haskell-cafe] A law for MonadReader

[Benjamin Fox]

No, newIORef does not itself change any state. It returns an IO value
because each invocation returns a different IORef, but there isn't any way
to tell how many or which IORefs were created. So newIORef 0 >> newIORef 0
doesn't differ in any way from newIORef 0, except in that an additional
object being created and then immediately garbage collected. This is
exactly the same as how the expression Just 0 >> Just 0 is the equivalent
to Just 0, except again with an additional object being created and then
garbage collected behind the scenes.

If you'd agree that Just 0 >> Just 0 ≡ Just 0, then I think you have to
also agree that newIORef 0 >> newIORef 0 ≡ newIORef 0. In both cases, each
side of the equivalence has the exact same semantics (save perhaps for
additional allocations, which we generally ignore when doing this kind of
analysis). No program which you could write could differentiate between the
two sides (again, save by doing some trickery such as benchmarking the
program and seeing which allocates more.)


On Sun, Jun 3, 2018 at 6:56 PM Ivan Perez <ivanperezdominguez at gmail.com>
wrote:

> > This obeys law (1): (newIORef 0 >> newIORef 0) == newIORef 0.
>
> Doesn't this change the state in the IO monad (which is why (2) does not
> hold for this instance)? If so, would it still be true?
>
> Ivan
>
> On 3 June 2018 at 07:55, Benjamin Fox <foxbenjaminfox at gmail.com> wrote:
>
>> Here, as in general in the definitions of laws, the relevent question is
>> referential transparency, not Eq instances.
>>
>> (You'll note that generally in the definitions of laws the symbol "=" is
>> used, not "==". Sometimes that's written as "≡", to be even clearer about
>> what it represents, as for instance the Monad Laws page
>> <https://wiki.haskell.org/Monad_laws> on the Haskell wiki does.)
>>
>> For some laws, like the "fmap id = id" Functor law, this is obviously the
>> only possible interpretation, as both sides of that equation are
>> necessarily functions, and functions don't have an Eq instance.
>>
>> So in this case, what the first law is asking for is that "ask >> ask" is
>> the same as "ask", in that any instance of "ask" in a program can be
>> replaced with "ask >> ask", or vice versa, without that changing the
>> program's semantics.
>>
>>
>> On Sun, Jun 3, 2018 at 9:47 AM Viktor Dukhovni <ietf-dane at dukhovni.org>
>> wrote:
>>
>>>
>>>
>>> > On Jun 3, 2018, at 3:32 AM, Benjamin Fox <foxbenjaminfox at gmail.com>
>>> wrote:
>>> >
>>> > Here is the counterexample:
>>> >
>>> > instance MonadReader (IORef Int) IO where
>>> >     ask = newIORef 0
>>> >     local _ = id
>>> >
>>> > This obeys law (1): (newIORef 0 >> newIORef 0) == newIORef 0.
>>>
>>> Can you explain what you mean?
>>>
>>> Prelude> :m + Data.IORef
>>> Prelude Data.IORef> let z = 0 :: Int
>>> Prelude Data.IORef> a <- newIORef z
>>> Prelude Data.IORef> b <- newIORef z
>>> Prelude Data.IORef> let c = newIORef z
>>> Prelude Data.IORef> let d = newIORef z
>>> Prelude Data.IORef> a == b
>>> False
>>> Prelude Data.IORef> c == d
>>>
>>> <interactive>:8:1: error:
>>>     • No instance for (Eq (IO (IORef Int))) arising from a use of ‘==’
>>>     • In the expression: c == d
>>>       In an equation for ‘it’: it = c == d
>>>
>>> --
>>>         Viktor.
>>>
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