[Haskell-cafe] Question of a manual computation on Reader Monad.

[Kim-Ee Yeoh]

一首诗 wrote:
> 
> runReader (do { b <- Reader $ show; return b } )  -- This is the initial
> expression, it should equals "show"
> 
> runReader (Reader $ show >>= \b -> return b)   -- remove do notion
> 

I'm not sure that's the right un-do-ization. It so happens that the 
exponent monad ((->) r) and the Reader monads are semantically 
identically. Is this a homework question?


一首诗 wrote:
> 
> runReader (Reader $ \e -> return( show e ) e)  -- apply the definition of
> ">>="
> 
> runReader (Reader $ \e -> (Reader $ \e1 -> show(e)) e)  -- apply the
> definition of "return"
> 
> But the last expression is incorrect, and I don't know how to go on.
> 

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