Re: [Haskell-cafe] A law for MonadReader

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 wrote:

...

On Jun 3, 2018, at 3:32 AM, Benjamin Fox 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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