[swift-evolution] [Proposal]: Free the '$' Symbol!

[Félix Cloutier]

I'm routinely proven wrong, but if $ was allowed to be either an operator or an identifier, it seems to me that `a <$> b` could produce two different and (potentially) valid ASTs depending only on whether <$> exists as an operator. Some people don't like operator overloading, imagine if you told them that they can't even be sure that what they're looking at is an operator at all.

Félix

> Le 3 janv. 2016 à 21:02:40, Jacob Bandes-Storch via swift-evolution <swift-evolution at swift.org> a écrit :
> 
> Is it considered infeasible for any characters to be allowed in both identifiers and operators?
> 
> On Sun, Jan 3, 2016 at 1:23 PM, Chris Lattner via swift-evolution <swift-evolution at swift.org <mailto:swift-evolution at swift.org>> wrote:
> 
> > On Jan 2, 2016, at 11:53 PM, Brent Royal-Gordon via swift-evolution <swift-evolution at swift.org <mailto:swift-evolution at swift.org>> wrote:
> >
> >> Swift currently does not allow operators to use $ - I assume because the grammar reserves it in one place: `implicit-parameter-name`.  I don't see why an entire class of identifiers has been eliminated, so I propose $ instead be reclassified as an `operator-character` so it can be used mixed in with other such characters, but prevents the introduction of `$Identifier`-style declarations that might conflict with implicit parameters.
> >
> > I believe the reason you don't see any other $ variables is that they're reserved for the debugger and REPL.
> >
> >       brent at Brents-MacBook-Pro ~/D/Code> swift
> >       Welcome to Apple Swift version 2.1.1 (swiftlang-700.1.101.15 clang-700.1.81). Type :help for assistance.
> >         1> "foo"
> >       $R0: String = "foo"
> >         2> print($R0)
> >       foo
> 
> Right.  That said, our current operator space (particularly the unicode segments covered) is not super well considered.  It would be great for someone to take a more systematic pass over them to rationalize things.
> 
> -Chris
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