While reading the Haskell language report I noticed that function type is
not an instance of class Read.
I was told that one cannot define them as an instance of class Show
without breaking "referential transparency" or printing a constant.
f :: (a->b)->String
f x = "bla bla bla"
How can I define a function to do the inverse operation ?
g :: String -> ( a -> b )
This time I cannot see how referential transparency will deny it.
What's the excuse now ?
I'm at the begining of chapter 7, but I have the feeling I'll not find the
answer in there.
Thank you.
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