Am Mittwoch, 4. März 2009 13:25 schrieb Michael Easter:
Folks,
I have a simple type called Beverage and an example that allows me to construct Maybe Beverage types.
Here are some type signatures for example functions:
request :: String -> Maybe Beverage
addMalt :: Beverage -> Maybe Beverage
I have defined a chain function like so:
chain :: (Maybe a) -> (a -> Maybe b) -> (Maybe b) chain = (>>=)
I can do this:
(chain (request "beer") addMalt)
and
request "beer" `chain` addMalt
I think I understand why, as I use the back-ticks for infix.
However, I don't have to do that for the true bind function, (>>=)
request "beer" >>= addMalt
I would like to use chain in this way -- that is without back-ticks. I'm not sure how...
You can't. Haskell has (infix) operators, whose names are composed of symbols (>, <, |, :, +, ...) and (prefix) functions, whose names are composed of letters, underscores (_) and primes ('). If you want to use a function infix or an operator prefix, you must indicate that to the compiler, which is done by enclosing a function name in backticks or an operator symbol in parentheses. If you could use a function name infix or prefix without indicating which you want, what would id const even be? Should it be id `const` even or (id const) even?
Is there something I'm missing?
thanks Mike
ps. Thanks to everyone for the great discussion on IO (re: previous question)