[Haskell-cafe] [Haskell] Rank-N types with (.) composition

[Tyson Whitehead]

On 11/02/15 04:39 AM, Simon Peyton Jones wrote:
> ($) has its own *typing rule*; it does not have a special type.  It's very ad hoc, but ($) is used so much to decrease parens that (e1 $ e2) is almost special syntax!

I do like the idea of being able to end a function application with a 
lambda, let, if, case, or do without having to use $ though.  Would 
something like the following grammar rules be problematic for parsing?

infixexp→	lexp qop infixexp
	|	- infixexp
	|	[fexp] lexp	(*)
	|	aexp		(*)

lexp	→	\ apat1 … apatn -> exp
	|	let decls in exp
	|	if exp [;] then exp [;] else exp
	|	case exp of { alts }
	|	do { stmts }
				(*)

fexp	→	[fexp] aexp

aexp	→	qvar
	|	gcon
	|	literal
	|	( exp )
	|	( exp1 , … , expk )
	|	[ exp1 , … , expk ]
	|	[ exp1 [, exp2] .. [exp3] ]
	|	[ exp | qual1 , … , qualn ]
	|	( infixexp qop )
	|	( qop⟨-⟩ infixexp )
	|	qcon { fbind1 , … , fbindn }
	|	aexp⟨qcon⟩ { fbind1 , … , fbindn }

https://www.haskell.org/onlinereport/haskell2010/haskellch3.html

were I've added (*) to where I have added or removed lines (basically 
I've moved function application from lexp to infixexp in order to allow 
the last exp in a function application to be an lexp).

> At the moment the *only* robust way to pass a polymorphic function to a polymorphic function (here, you are passing Wrap to (.)) is to wrap it in a newtype, much as Wrap does.

Good to know.

Thanks!  -Tyson