on Mon May 02 2016, Tony Allevato <allevato-AT-google.com> wrote:
> On Mon, May 2, 2016 at 1:20 PM Dave Abrahams via swift-evolution
> <swift-evolution@swift.org> wrote:
>
> Tony, thanks for writing this up!
>
> on Mon May 02 2016, Tony Allevato <swift-evolution@swift.org> wrote:
>
> > Other kinds of operators (prefix, postfix, assignment)
> >
> > Static operator methods have the same signatures as their global
> counterparts.
> > So, for example, prefix and postfix operators as well as assignment
> operators
> > would be defined the way one would expect:
> >
> > protocol SomeProtocol {
> > static func +=(lhs: inout Self, rhs: Self)
> > static prefix func ~(value: Self) -> Self
> >
> > // This one is deprecated, of course, but used here just to serve as an
> > // example.
> > static postfix func ++(value: inout Self) -> Self
> > }
> >
> > // Trampolines
> > func += <T: SomeProtocol>(lhs: inout T, rhs T) {
> > T.+=(&lhs, rhs)
> > }
> > prefix func ~ <T: SomeProtocol>(value: T) -> T {
> > return T.~(value)
> > }
> > postfix func ++ <T: SomeProtocol>(value: inout T) -> T {
> > return T.++(&value)
> > }
>
> How does one distinguish between calls to a static prefix operator and a
> static postfix operator with the same name?
>
> Ah, that's a tricky one that I don't have an immediate answer to, so I'm
> definitely open to creative thoughts here.
One possibility: just use “qualified operator” notation.
lhs T.+= rhs
T.++x
x T.++
> The first stab I would take at is, what if we included the token "prefix" or
> "suffix" before the operator name in the expression, like this?
>
> return T.prefix ++(&value)
> return T.postfix ++(&value)
>
> But that could start to look like an invocation of "++" on a static property
> "T.prefix". I haven't dug into the parser to determine if that would even be
> feasible or not.
These are not unreasonable either:
return prefix T.++(&value)
return postfix T.++(&value)
--
Dave
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