[swift-evolution] Default Generic Arguments

Thu Jan 26 15:26:46 CST 2017

Very interesting point, Alexis. So can you reiterate again which of the
four options you outlined earlier support this use case? And if there are
multiple, which would be the most consistent with the rest of the language?

And Srdan, could you incorporate that information into your discussion?
On Thu, Jan 26, 2017 at 12:59 Srđan Rašić <srdan.rasic at gmail.com> wrote:

> That's a very good point Alexis and makes sense to me. I'll updated the
> proposal with that in mind and revise my examples.
>
> On Thu, Jan 26, 2017 at 7:06 PM, Alexis <abeingessner at apple.com> wrote:
>
>
> On Jan 25, 2017, at 8:15 PM, Xiaodi Wu <xiaodi.wu at gmail.com> wrote:
>
> Srdan, I'm afraid I don't understand your discussion. Can you simplify it
> for me by explaining your proposed solution in terms of Alexis's examples
> below?
>
> ```
> // Example 1: user supplied default is IntegerLiteralConvertible
>
> func foo<T=Int64>(t: T) { ... }
>
> foo(22)
> //  ^
> //  |
> //  What type gets inferred here?
> ```
>
> I believe that it is essential that the answer here be `Int` and not
> `Int64`.
>
> My reasoning is: a user's code *must not* change because a library *adds*
> a default in a newer version. (As mentioned in several design docs, most
> recently the new ABI manifesto, defaults in Swift are safe to add without
> breaking source compatibility.)
>
>
> I don’t agree: adding a default to an *existing* type parameter should be
> a strict source-breaking change (unless the chosen type can avoid all other
> defaulting rules, see the end of this email).
>
> Type Parameter Defaults, as I know them, are a tool for avoiding breakage
> when a *new* type parameter is introduced. That is, they allow you to
> perform the following transformation safe in the knowledge that it won’t
> break clients:
>
> func foo(input: X)
> func foo<T=X>(input: T)
>
> For this to work, you need to make the <T=X> default have dominance over
> the other default rules.
>
> Specifically you want this code to keep working identically:
>
>
> // before
> func foo(input: Int64)
> foo(0)  // Int64
>
> // after
> func foo<T=Int64>(input: T)
> foo(0) // Int64
>
>
>
> This is in direct conflict with making the following keep working
> identically:
>
>
> // before
> func foo<T>(input: T)
> foo(0) // Int
>
> // after
> func foo<T=Int64>(input: T)
> foo(0) // Int
>
>
>
> You have to choose which of these API evolution patterns is most
> important, because you can’t make both work. To me, the first one is
> obviously the most important, because that’s the whole point of the
> feature. The reason to do the second one is to try to make a common/correct
> case more ergonomic and/or the default. But unlike function argument
> defaults, type parameters can already have inferred values.
>
>
>
> Note that source breaking with adding defaults can be avoided as long as
> long as the chosen default isn’t:
>
> * XLiteralConvertible (pseudo-exception: if the default is also the
> XLiteralType it’s fine, but that type is user configurable)
> * A supertype of another type (T?, T!, SuperClass, Protocol, (…,
> someLabel: T, ...), [SuperType], [SuperType1:SuperType2], (SuperType) ->
> SubType, and probably more in the future)
>
> Concretely this means it’s fine to retroactively make an existing generic
> parameter default to MyFinalClass, MyStruct, MyEnum, and
> collections/functions/unlabeled-tuples thereof. Arguably,
> Int/String/Bool/Array/etc are fine, but there’s a niche situation where
> using them can cause user breakage due to changing XLiteralType.
>
> In practice I expect this will be robust enough to avoid breakage — I
> expect most defaults will be MyStruct/MyEnum, or an XLiteralType. Even if
> it’s not, you need to end up in a situation where inference can actually
> kick in and find an ambiguity *and* where the difference matters. (e.g.
> SubClass vs SuperClass isn’t a big deal in most cases)
>
>
>
> Here, if version 1 of a library has `func foo<T>(t: T) { ... }`, then
> `foo(22)` must infer `T` to be `Int`. That's just the rule in Swift, and it
> would be severely source-breaking to change that. Therefore, if version 2
> of that library has `func foo<T=Int64>(t: T) { ... }`, then `foo(22)` must
> still infer `T` to be `Int`.
>
> Does your proposed solution have the same effect?
>
> ```
> // Example 2: user supplied default isn't IntegerLiteralConvertible
>
> func bar<T=Character>(t: T) { ... }
>
> bar(22)
> //  ^
> //  |
> //  What type gets inferred here?
> ```
>
> By the same reasoning as above, this ought to be `Int`. What would the
> answer be in your proposed solution?
>
>
> On Wed, Jan 25, 2017 at 2:07 PM, Srđan Rašić <srdan.rasic at gmail.com>
> wrote:
>
> That's a good example Alexis. I do agree that generic arguments are
> inferred in a lot of cases, my point was that they should not be inferred
> in "type declarations". Not sure what's the right terminology here, but I
> mean following places:
>
> (I) Variable/Constant declaration
>
>   ```
>   let x: X
>   ```
>
> (II) Property declaration
>
>   ```
>   struct T {
>     let x: X
>   }
>   ```
>
> (III) Function declaration
>
>   ```
>   func a(x: X) -> X
>   ```
>
> (IV) Enumeration case declaration
>
>   ```
>   enum E {
>     case x(X)
>   }
>   ```
>
> (V) Where clauses
>
>   ```
>   extensions E where A == X {}
>   ```
>
> In those cases `X` should always mean `X<Int>` if it was defined as
> `struct X<T = Int>`. That's all my rule says. Sorry for not being clear in
> the last email :)
>
> As for the other cases, mostly those where an instance is created,
> inference should be applied.
>
> Let's go through your examples. Given
>
> struct BigInt: Integer {
>   var storage: Array<Int> = []
> }
>
> func process<T: BinaryInteger>(_ input: BigInt<T>) -> BigInt<T> { ... }
>
> what happens with `let val1 = process(BigInt())`? I think this is
> actually the same problem as what happens in case of `let x = BigInt()`.
>
> In such case my rule does not apply as we don't have full type
> declaration. In `let x = BigInt()` type is not defined at all, while in `func
> process<T: BinaryInteger>(_ input: BigInt<T>) -> BigInt<T> { ... }` type
> is explicitly weakened or "undefaulted" if you will.
>
> We should introduce new rule for such cases and allowing `Storage=Int`
> default to participate in such expressions would make sense. As you said,
> it also solves second example: let val2 = process(0).
>
> I guess this would be the problem we thought we were solving initially and
> in that case I think the solution should be what Doug suggested: if you
> can’t infer a particular type, fill in a default.
>
> Of course, if the default conflicts with the generic constraint, it would
> not be filled in and it would throw an error.
>
> For the sake of completeness,
>
> func fastProcess(_ input: BigInt<Int64>) -> BigInt<Int64> { ... }
> let val3 = fastProcess(BigInt())
>
> would certainly infer the type from context as my rule does not apply to
> initializers. It would infer BigInt<Int64>.
>
> As for your last example, I guess we can't do anything about that and
> that's ok.
>
>
> On Wed, Jan 25, 2017 at 7:50 PM, Alexis <abeingessner at apple.com> wrote:
>
> Yes, I agree with Xiaodi here. I don’t think this particular example is
> particularly compelling. Especially because it’s not following the full
> evolution of the APIs and usage, which is critical for understanding how
> defaults should work.
>
>
> Let's look at the evolution of an API and its consumers with the example
> of a BigInt:
>
>
> struct BigInt: Integer {
>   var storage: Array<Int> = []
> }
>
>
> which a consumer is using like:
>
>
> func process(_ input: BigInt) -> BigInt { ... }
> let val1 = process(BigInt())
> let val2 = process(0)
>
>
> Ok that's all fairly straightforward. Now we decide that BigInt should
> expose its storage type for power-users:
>
>
> struct BigInt<Storage: BinaryInteger = Int>: Integer {
>   var storage: Array<Storage> = []
> }
>
>
> Let's make sure our consumer still works:
>
>
> func process(_ input: BigInt) -> BigInt { ... }
> let val1 = process(BigInt())
> let val2 = process(0)
>
>
> Ok BigInt in process’s definition now means BigInt<Int>, so this still all
> works fine. Perfect!
>
>
> But then the developer of the process function catches wind of this new
> power user feature, and wants to support it.
> So they too become generic:
>
>
> func process<T: BinaryInteger>(_ input: BigInt<T>) -> BigInt<T> { ... }
>
>
> The usage sites are now more complicated, and whether they should compile
> is unclear:
>
>
> let val1 = process(BigInt())
> let val2 = process(0)
>
>
> For val1 you can take a hard stance with your rule: BigInt() means
> BigInt<Int>(), and that will work. But for val2 this rule doesn't work,
> because no one has written BigInt unqualified. However if you say that the
> `Storage=Int` default is allowed to participate in this expression, then we
> can still find the old behaviour by defaulting to it when we discover
> Storage is ambiguous.
>
> We can also consider another power-user function:
>
>
> func fastProcess(_ input: BigInt<Int64>) -> BigInt<Int64> { ... }
> let val3 = fastProcess(BigInt())
>
>
> Again, we must decide the interpretation of this. If we take the
> interpretation that BigInt() has an inferred type, then the type checker
> should discover that BigInt<Int64> is the correct result. If however we
> take stance that BigInt() means BigInt<Int>(), then we'll get a type
> checking error which our users will consider ridiculous: *of course* they
> wanted a BigInt<Int64> here!
>
> We do however have the problem that this won’t work:
>
>
> let temp = BigInt()
> fastProcess(temp) // ERROR — expected BigInt<Int64>, found BigInt<Int>
>
>
> But that’s just as true for normal ints:
>
>
> let temp = 0
> takesAnInt64(temp) // ERROR — expected Int64, found Int
>
>
> Such is the limit of Swift’s inference scheme.
>
>
>
>
>
>
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