[swift-evolution] Default Generic Arguments

Alexis abeingessner at apple.com
Wed Jan 25 12:50:28 CST 2017

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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