Haskell无限循环

Question

I am implementing a calculator in Haskell to brush up on the language but I am hitting a snag in my main when I want it to enter an infinite loop until the user inputs q . Heres my main let me know if you see what I'm doing wrong and ill also post my error

error: No instances for (Floating (IO a0), Read (IO a0))
      arising from a use of `compute'
    Possible fix:
      add instance declarations for (Floating (IO a0), Read (IO a0))
    In a stmt of a 'do' block: compute e
    In the expression:
      do { compute e;
           evaluate_input }
    In an equation for `evaluate_expression':
        evaluate_expression e
          = do { compute e;
                 evaluate_input }

Answer 1

In your do-statement

 compute e 
 evaluate_input

both function need to be of the same monadic type, in here IO (declared by evaluate_input :: IO () ). So GHC now can expect that compute is a function that takes the String e and returns an IO a0 ( == a ). Yet, it could not find any a0 so that IO a0 is an instance of Floating or Read , which a must be.

I'd assume that you want to output the result of the computation (and a is an instance of Show rather than Read ), so use

do
 putStrLn . show $ compute e
 evaluate_input