Question

I’m going through Write Yourself a Scheme and am struck at the exercise 4 on this page.

How do I go about this? I’ve got this far, but have no idea whatsoever where the readHex is supposed to go, must I liftM it ? Do you case match the parser ?

parseNumber = liftM (Number . read) $ choice [many1 digit, char '#' >> oneOf "hd" >>= a]
    where a f = case f of
        'h' -> many1 digit

Also, I don’t suppose you can apply <|> on Parser LispVal functions, right?

Answer 1

I’ve got this far, but have no idea whatsoever where the readHex is supposed to go, must I liftM it ?

Yes, since readHex is most likely a pure function and liftM lifts it into the monadic context of Parser.

Since I don’t quite understand what your local function a is good for, I’m leaving it for now and simply use the functions parseDecimal and parseHex. In that case, you could write parseNumber like so:

parseNumber :: Parser LispVal
parseNumber = liftM Number $ parseDecimal <|> parseHex
    where parseDecimal :: Parser Integer
          parseDecimal = liftM read $ many1 digit
          parseHex :: Parser Integer
          parseHex     = liftM readHex $ char '#' >> ... -- parse hex value

Of course you can omit the type signatures, I just added them for clarity.

Also, I don’t suppose you can apply <|> on Parser LispVal functions, right?

<|> works for every Parser a.

I recommend reading some material on parser combinators, i.e. the Parsec User Guide.