# [swift-evolution] [Pitch] [Phase 2] New `permuting` keyword for protocols

Mon Dec 26 10:17:16 CST 2016

Xiaodi's point is really important—being able to express the notions
simultaneously that "T has method a()" and "T does not have method a()"
would break the type system.

Instead of focusing on the proposed syntax, let's consider the problem
you're trying to solve. It sounds like what you're asking for could be
expressed more cleanly with a richer protocol algebra that supported
subtraction. It wouldn't be quite as automatic as what you propose, but it
would feel like a more natural extension of the type system if you could do
something like below, and would avoid combinatorial explosion of protocol
types (you go from O(n!) to O(n) things you actually have to define
concretely):

```
protocol WidthConstrainable {
func width(_ v: CGFloat) -> Self – WidthConstrainable
}
protocol HeightConstrainable {
func height(_ v: CGFloat) -> Self – HeightConstrainable
}
protocol XConstrainable {
func x(_ v: CGFloat) -> Self – XConstrainable
}
protocol YConstrainable {
func y(_ v: CGFloat) -> Self – YConstrainable
}
struct Constraint: WidthConstrainable, HeightConstrainable, XConstrainable,
YConstrainable {
...
}
```

If a type X is just a union or protocols (for example, X:
WidthConstrainable & HeightConstrainable & XConstrainable &
YConstrainable), the subtraction (X – something) is easy to define. It's
either valid if the subtrahend is present in the set, or it's invalid (and
detectable at compile time) if it's not.

But there are still some rough edges: what does it mean when a concrete
type is involved? Let's say you have T: P1 & P2 & P3, and you write (T –
P1). That could give you a type that contains all the members of T except
those in P1, which would be the members in P2, P3, and any that are defined
directly on T that do not come from protocol conformances.

But what is the relationship between types T and (T – P1)? (T – P1) being a
supertype of T seems fairly straightforward—any instance of T can be
expressed as (T – P1). But if I have an instance of type (T – P1), should I
be able to cast that back to T? On the one hand, why not? I can obviously
only get (T – P1) by starting with T at some point, so any instance of (T –
P1) must *also* be an instance of T. So that implies that T is a supertype
of (T – P1). In other words, they're supertypes of each other, without
being the same type? That would be a first in Swift's type system, I
believe. And if we allow the cast previously mentioned, that effectively
circumvents the goal you're trying to achieve. (We could argue that you'd
have to use a force-cast (as!) in this case.)

This could be worked around by forbidding subtraction from concrete types
and reducing T to the union of its protocols before performing the
subtraction. In that case, (T – P1) would equal P2 & P3. But that
relationship is still a little wonky: in that case, (T – P1) would also not
contain any members that are only defined on T, even though the expression
(T – P1) implies that they should. You would have to make that reduction
explicit somehow in order for that to not surprise users (require writing
something like `#protocols(of: T) – P1`?), and it leaves a certain subset
of possible type expressions (anything that wants members defined on T
without members of a protocol) unexpressible.

I actually glossed over this earlier by writing "..." in the struct body.
If I defined `width(_:)` there, what would my return type be? We currently
forbid `Self` in that context. Would `Self – WidthConstrainable` be
allowed? Would I have to use the new protocol-reduction operator above?
More details that would have to be worked out.

Protocol inheritance would pose similar questions. If you have this:

```
protocol P1 {}
protocol P2: P1 {}
```

What is the subtype/supertype relationship between P2 and (P2 – P1)? It's
the same situation we had with a concrete type. Maybe you just can't
subtract a super-protocol without also subtracting its lowest sub-protocol
from the type?

My PL type theory knowledge isn't the deepest by any means, but if
something like this was workable, I think it would be more feasible and
more expressive than the member permutation approach. And that being said,
this is probably a fairly narrow use case that wouldn't warrant the
complexity it would bring to the type system to make it work.

On Mon, Dec 26, 2016 at 7:03 AM Xiaodi Wu via swift-evolution <
swift-evolution at swift.org> wrote:

> Should the following compile?
>
> let bar = foo.a()
> func f(_ g: T) {
> _ = g.a()
> }
> f(bar)
>
> If so, your proposal cannot guarantee each method is called only once. If
> not, how can bar be of type T?
>
> On Mon, Dec 26, 2016 at 06:30 Adrian Zubarev <
>
> I think I revise what I said about value semantics in my last post.
>
> let chain: T = foo.a()
>
> let new = chain
> new. // should not see `a` here
>
> It’s more something like a local scoped chain. I’m not sure how to call it
> correctly here. I’m not a native English speaker. =)
>
>
>
> --
> Sent with Airmail
>
> Am 26. Dezember 2016 um 12:11:23, Adrian Zubarev (
>
> By ‘calling once’ I meant, calling once at a single permutation chain. If
> the chain is escaped or followed by a non-permuting member that returns the
> same protocol, you’d have the ability to use all members at the starting
> point of the new chain.
>
> permuting protocol T {
>     func a()
>     func b()
>     func c()
>     func d()
> }
>
> var foo: T = …
>
> func boo(_ val: T) -> U {
>     // Here val escapes the chain and creates a new one
>     // That means that you can create a local permutation chain here again
>
>     val.a() // we can use `a` here
>     return …
> }
>
> boo(foo.a()) // a is immediately invoked here
>
> I imagine this keyword to follow value semantics, so that any possible
> mutation is handled locally with a nice extra ability of permutation member
> chaining.
>
> Did I understood your point correctly here?
> ------------------------------
>
> Sure the idea needs to be more fleshed out, but I’m curious if that’s
> something that we might see in Swift one day. :)
>
>
> --
> Sent with Airmail
>
> Am 26. Dezember 2016 um 11:50:50, Xiaodi Wu (xiaodi.wu at gmail.com) schrieb:
>
> Given `foo: T` and methods a(), b(), c(), d(), each of which can only be
> called once, how can the return value of these methods be represented in
> the type system?
>
> That is, if `foo.a()` can be passed as an argument to an arbitrary
> function of type `(T) -> U`, either the function cannot immediately invoke
> a(), in which case foo is not of type T, or it can immediately invoke a(),
> in which case your keyword does not work.
>
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>
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