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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