Converting from newtype to Int and from Int to newtype










2















How do I convert newtype to Int and vice versa?



I tried:



newtype NT1 = NT1 Integer

fromNT1toInt :: NT1 -> Int
fromNT1toInt x = let y = x :: Int
in y


but I get could't match expected type error



I tried making NT1 instance of Enum class
but I don't understand well how does toEnum work



newtype NT1 = NT1 Integer

instance Enum NT1 where
toEnum x = let x = x :: Integer
in if x >= 0
then x :: NT1
else x :: NT1


when I call toEnum 5 :: NT1 this should return 5 NT1 but I get StackOverflow error.
Where am I making mistake?



Edited: newtype name










share|improve this question




























    2















    How do I convert newtype to Int and vice versa?



    I tried:



    newtype NT1 = NT1 Integer

    fromNT1toInt :: NT1 -> Int
    fromNT1toInt x = let y = x :: Int
    in y


    but I get could't match expected type error



    I tried making NT1 instance of Enum class
    but I don't understand well how does toEnum work



    newtype NT1 = NT1 Integer

    instance Enum NT1 where
    toEnum x = let x = x :: Integer
    in if x >= 0
    then x :: NT1
    else x :: NT1


    when I call toEnum 5 :: NT1 this should return 5 NT1 but I get StackOverflow error.
    Where am I making mistake?



    Edited: newtype name










    share|improve this question


























      2












      2








      2








      How do I convert newtype to Int and vice versa?



      I tried:



      newtype NT1 = NT1 Integer

      fromNT1toInt :: NT1 -> Int
      fromNT1toInt x = let y = x :: Int
      in y


      but I get could't match expected type error



      I tried making NT1 instance of Enum class
      but I don't understand well how does toEnum work



      newtype NT1 = NT1 Integer

      instance Enum NT1 where
      toEnum x = let x = x :: Integer
      in if x >= 0
      then x :: NT1
      else x :: NT1


      when I call toEnum 5 :: NT1 this should return 5 NT1 but I get StackOverflow error.
      Where am I making mistake?



      Edited: newtype name










      share|improve this question
















      How do I convert newtype to Int and vice versa?



      I tried:



      newtype NT1 = NT1 Integer

      fromNT1toInt :: NT1 -> Int
      fromNT1toInt x = let y = x :: Int
      in y


      but I get could't match expected type error



      I tried making NT1 instance of Enum class
      but I don't understand well how does toEnum work



      newtype NT1 = NT1 Integer

      instance Enum NT1 where
      toEnum x = let x = x :: Integer
      in if x >= 0
      then x :: NT1
      else x :: NT1


      when I call toEnum 5 :: NT1 this should return 5 NT1 but I get StackOverflow error.
      Where am I making mistake?



      Edited: newtype name







      haskell






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 13 '18 at 13:26









      Micha Wiedenmann

      10.3k1364103




      10.3k1364103










      asked Nov 13 '18 at 4:03









      cheshirecheshire

      1898




      1898






















          2 Answers
          2






          active

          oldest

          votes


















          4














          Use fromIntegral to convert between integral types, such as Int and Integer. Use NT1 to convert from Integer to NT1, and pattern-matching from NT1 to Integer. Finally, you can compose functions to connect things.



          toNT1 :: Int -> NT1
          toNT1 = NT1 . fromIntegral

          fromNT1 :: NT1 -> Int
          fromNT1 (NT1 x) = fromIntegral x





          share|improve this answer
































            6














            e :: t doesn’t mean “convert expression e to type t”, it’s just an annotation that says “e has type t (already)”. So this:



            let y = x :: Int in y


            Means: assert that x has type Int, set y equal to x, and return y. That’s why you got a type mismatch: x does not have type Int as you are claiming to the compiler. And this:



            let x = x :: Integer
            in if x >= 0 then x :: NT1 else x :: NT1


            Means: declare a new variable x, set it equal to itself (an infinite loop), asserting that it has type Integer, then test whether that infinite loop returns a nonnegative value; either way, return x, asserting that it has type NT1 (this contradicts the Integer from before).



            To convert between Integer and Int, you can use fromIntegral :: (Integral a, Num b) => a -> b, which converts any integral type (such as Int or Integer) into any numeric type (such as Int, Integer, Float, Double, or Ratio).



            For converting from newtypes, you can use pattern-matching:



            fromNT1ToInt :: NT1 -> Int
            fromNT1ToInt (NT1 x) = fromIntegral x


            Or add a record accessor function to the newtype and use that:



            newtype NT1 = NT1 nt1Val :: Integer 
            -- Note: nt1Val :: NT1 -> Integer

            fromNT1ToInt :: NT1 -> Int
            fromNT1ToInt nt = fromIntegral (nt1Val nt)
            -- Or, with function composition (.):
            fromNT1ToInt = fromIntegral . nt1Val


            Or, finally, use coerce from Data.Coerce:



            import Data.Coerce (coerce)

            fromNT1ToInt :: NT1 -> Int
            fromNT1ToInt nt = fromIntegral (coerce nt :: Integer)


            And of course, to construct a newtype you just use its constructor—in this case, NT1 :: Integer -> NT1, e.g. NT1 5.






            share|improve this answer






















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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









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              active

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              4














              Use fromIntegral to convert between integral types, such as Int and Integer. Use NT1 to convert from Integer to NT1, and pattern-matching from NT1 to Integer. Finally, you can compose functions to connect things.



              toNT1 :: Int -> NT1
              toNT1 = NT1 . fromIntegral

              fromNT1 :: NT1 -> Int
              fromNT1 (NT1 x) = fromIntegral x





              share|improve this answer





























                4














                Use fromIntegral to convert between integral types, such as Int and Integer. Use NT1 to convert from Integer to NT1, and pattern-matching from NT1 to Integer. Finally, you can compose functions to connect things.



                toNT1 :: Int -> NT1
                toNT1 = NT1 . fromIntegral

                fromNT1 :: NT1 -> Int
                fromNT1 (NT1 x) = fromIntegral x





                share|improve this answer



























                  4












                  4








                  4







                  Use fromIntegral to convert between integral types, such as Int and Integer. Use NT1 to convert from Integer to NT1, and pattern-matching from NT1 to Integer. Finally, you can compose functions to connect things.



                  toNT1 :: Int -> NT1
                  toNT1 = NT1 . fromIntegral

                  fromNT1 :: NT1 -> Int
                  fromNT1 (NT1 x) = fromIntegral x





                  share|improve this answer















                  Use fromIntegral to convert between integral types, such as Int and Integer. Use NT1 to convert from Integer to NT1, and pattern-matching from NT1 to Integer. Finally, you can compose functions to connect things.



                  toNT1 :: Int -> NT1
                  toNT1 = NT1 . fromIntegral

                  fromNT1 :: NT1 -> Int
                  fromNT1 (NT1 x) = fromIntegral x






                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Nov 13 '18 at 4:23

























                  answered Nov 13 '18 at 4:07









                  Li-yao XiaLi-yao Xia

                  12k1327




                  12k1327























                      6














                      e :: t doesn’t mean “convert expression e to type t”, it’s just an annotation that says “e has type t (already)”. So this:



                      let y = x :: Int in y


                      Means: assert that x has type Int, set y equal to x, and return y. That’s why you got a type mismatch: x does not have type Int as you are claiming to the compiler. And this:



                      let x = x :: Integer
                      in if x >= 0 then x :: NT1 else x :: NT1


                      Means: declare a new variable x, set it equal to itself (an infinite loop), asserting that it has type Integer, then test whether that infinite loop returns a nonnegative value; either way, return x, asserting that it has type NT1 (this contradicts the Integer from before).



                      To convert between Integer and Int, you can use fromIntegral :: (Integral a, Num b) => a -> b, which converts any integral type (such as Int or Integer) into any numeric type (such as Int, Integer, Float, Double, or Ratio).



                      For converting from newtypes, you can use pattern-matching:



                      fromNT1ToInt :: NT1 -> Int
                      fromNT1ToInt (NT1 x) = fromIntegral x


                      Or add a record accessor function to the newtype and use that:



                      newtype NT1 = NT1 nt1Val :: Integer 
                      -- Note: nt1Val :: NT1 -> Integer

                      fromNT1ToInt :: NT1 -> Int
                      fromNT1ToInt nt = fromIntegral (nt1Val nt)
                      -- Or, with function composition (.):
                      fromNT1ToInt = fromIntegral . nt1Val


                      Or, finally, use coerce from Data.Coerce:



                      import Data.Coerce (coerce)

                      fromNT1ToInt :: NT1 -> Int
                      fromNT1ToInt nt = fromIntegral (coerce nt :: Integer)


                      And of course, to construct a newtype you just use its constructor—in this case, NT1 :: Integer -> NT1, e.g. NT1 5.






                      share|improve this answer



























                        6














                        e :: t doesn’t mean “convert expression e to type t”, it’s just an annotation that says “e has type t (already)”. So this:



                        let y = x :: Int in y


                        Means: assert that x has type Int, set y equal to x, and return y. That’s why you got a type mismatch: x does not have type Int as you are claiming to the compiler. And this:



                        let x = x :: Integer
                        in if x >= 0 then x :: NT1 else x :: NT1


                        Means: declare a new variable x, set it equal to itself (an infinite loop), asserting that it has type Integer, then test whether that infinite loop returns a nonnegative value; either way, return x, asserting that it has type NT1 (this contradicts the Integer from before).



                        To convert between Integer and Int, you can use fromIntegral :: (Integral a, Num b) => a -> b, which converts any integral type (such as Int or Integer) into any numeric type (such as Int, Integer, Float, Double, or Ratio).



                        For converting from newtypes, you can use pattern-matching:



                        fromNT1ToInt :: NT1 -> Int
                        fromNT1ToInt (NT1 x) = fromIntegral x


                        Or add a record accessor function to the newtype and use that:



                        newtype NT1 = NT1 nt1Val :: Integer 
                        -- Note: nt1Val :: NT1 -> Integer

                        fromNT1ToInt :: NT1 -> Int
                        fromNT1ToInt nt = fromIntegral (nt1Val nt)
                        -- Or, with function composition (.):
                        fromNT1ToInt = fromIntegral . nt1Val


                        Or, finally, use coerce from Data.Coerce:



                        import Data.Coerce (coerce)

                        fromNT1ToInt :: NT1 -> Int
                        fromNT1ToInt nt = fromIntegral (coerce nt :: Integer)


                        And of course, to construct a newtype you just use its constructor—in this case, NT1 :: Integer -> NT1, e.g. NT1 5.






                        share|improve this answer

























                          6












                          6








                          6







                          e :: t doesn’t mean “convert expression e to type t”, it’s just an annotation that says “e has type t (already)”. So this:



                          let y = x :: Int in y


                          Means: assert that x has type Int, set y equal to x, and return y. That’s why you got a type mismatch: x does not have type Int as you are claiming to the compiler. And this:



                          let x = x :: Integer
                          in if x >= 0 then x :: NT1 else x :: NT1


                          Means: declare a new variable x, set it equal to itself (an infinite loop), asserting that it has type Integer, then test whether that infinite loop returns a nonnegative value; either way, return x, asserting that it has type NT1 (this contradicts the Integer from before).



                          To convert between Integer and Int, you can use fromIntegral :: (Integral a, Num b) => a -> b, which converts any integral type (such as Int or Integer) into any numeric type (such as Int, Integer, Float, Double, or Ratio).



                          For converting from newtypes, you can use pattern-matching:



                          fromNT1ToInt :: NT1 -> Int
                          fromNT1ToInt (NT1 x) = fromIntegral x


                          Or add a record accessor function to the newtype and use that:



                          newtype NT1 = NT1 nt1Val :: Integer 
                          -- Note: nt1Val :: NT1 -> Integer

                          fromNT1ToInt :: NT1 -> Int
                          fromNT1ToInt nt = fromIntegral (nt1Val nt)
                          -- Or, with function composition (.):
                          fromNT1ToInt = fromIntegral . nt1Val


                          Or, finally, use coerce from Data.Coerce:



                          import Data.Coerce (coerce)

                          fromNT1ToInt :: NT1 -> Int
                          fromNT1ToInt nt = fromIntegral (coerce nt :: Integer)


                          And of course, to construct a newtype you just use its constructor—in this case, NT1 :: Integer -> NT1, e.g. NT1 5.






                          share|improve this answer













                          e :: t doesn’t mean “convert expression e to type t”, it’s just an annotation that says “e has type t (already)”. So this:



                          let y = x :: Int in y


                          Means: assert that x has type Int, set y equal to x, and return y. That’s why you got a type mismatch: x does not have type Int as you are claiming to the compiler. And this:



                          let x = x :: Integer
                          in if x >= 0 then x :: NT1 else x :: NT1


                          Means: declare a new variable x, set it equal to itself (an infinite loop), asserting that it has type Integer, then test whether that infinite loop returns a nonnegative value; either way, return x, asserting that it has type NT1 (this contradicts the Integer from before).



                          To convert between Integer and Int, you can use fromIntegral :: (Integral a, Num b) => a -> b, which converts any integral type (such as Int or Integer) into any numeric type (such as Int, Integer, Float, Double, or Ratio).



                          For converting from newtypes, you can use pattern-matching:



                          fromNT1ToInt :: NT1 -> Int
                          fromNT1ToInt (NT1 x) = fromIntegral x


                          Or add a record accessor function to the newtype and use that:



                          newtype NT1 = NT1 nt1Val :: Integer 
                          -- Note: nt1Val :: NT1 -> Integer

                          fromNT1ToInt :: NT1 -> Int
                          fromNT1ToInt nt = fromIntegral (nt1Val nt)
                          -- Or, with function composition (.):
                          fromNT1ToInt = fromIntegral . nt1Val


                          Or, finally, use coerce from Data.Coerce:



                          import Data.Coerce (coerce)

                          fromNT1ToInt :: NT1 -> Int
                          fromNT1ToInt nt = fromIntegral (coerce nt :: Integer)


                          And of course, to construct a newtype you just use its constructor—in this case, NT1 :: Integer -> NT1, e.g. NT1 5.







                          share|improve this answer












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                          share|improve this answer










                          answered Nov 13 '18 at 5:32









                          Jon PurdyJon Purdy

                          37.9k668125




                          37.9k668125



























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