molleweide

molleweide

Effective Haskell: Left / Right types in Calculator?

@RebeccaSkinner,

page 141. calculator.

I suggest you explain a little more about why you introduced the Left and Right types in the Calculator parser or where they come from.

If I am correct you use them to distinguish between the first and second output of readEither but I am not sure.

You do pattern matching against Left / Right in the case statement for readEither in parser' but it is not clear to me where they come from exactly. When I look at :info for readEither it sais String -> Either String a and so I am thinking does readEither return Left/Right values implicitly?

Edit: Now, reading through the code again, I realise that the above is wrong, however, there is something about the source code to the calculator that I don’t really understand.
I see now that the readEither is used as the case expression in case expr of but I have never seen a function being used alone as the case expr. Does this mean that lit is being passed into the case expression and then is passed as argument to readEither?

parse' :: [String] -> Either String (Expr, [String])
parse' [] = Left "unexpected end of expression"
parse' (token:rest) =
  case token of
    "+" -> parseBinary Add rest
    "*" -> parseBinary Mul rest
    "-" -> parseBinary Sub rest
    "/" -> parseBinary Div rest
    lit ->
      case readEither of
        Left err -> Left err
        Right lit' -> Right (Lit lit', rest)

First Post!

RebeccaSkinner

RebeccaSkinner

Author of Effective Haskell

Hi! It looks like this is a bug that has been fixed now. The correct example should be:

case readEither lit of
  ...

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