That's a useful operator! Unfortunately it does not play nice with $. Of
less importance: some syntactic constructs can not appear in the arguments
without parenthesis, let bindings for instance (although lambda abstraction
works parenthesis-free).
Also I'm not sure this can be used for defining trees or nested function
application since a nesting of the operator inevitably require parenthesis.
/J
On 26 May 2011 14:52, Daniel Fischer
On Thursday 26 May 2011 14:35:41, Neil Brown wrote:
foo is the function we want to apply, and eg shows how to apply it in do-notation with an argument on each line. I couldn't manage to remove the r$ at the beginning of each line, which rather ruins the whole scheme :-( On the plus side, there's no brackets, it's only two extra characters per line, and you can have whatever you like after the r$.
Wouldn't that be also achievable with
infixl 0 ?
(?) :: (a -> b) -> a -> b f ? x = f x
eg = foo ? 2 + 1 ? 'c' ? "hello" ++ "goodbye" ? 3.0
?
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