[SR-12692] Conditional conformance of generics and specialized algorithms is broken. #55136
Comments
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Here's an even simpler example, that shows how it trips up expert generic programmers: extension Sequence {
var array: Array<Element> {
print("preallocating", self.underestimatedCount)
return Array(self)
}
}
_ = Array(0..<1000).reversed().array // preallocating 1000
_ = repeatElement(1..<10, count: 200).joined().array // preallocating 0Because the result of `joined()` is a type that only conditionally conforms to `Collection`, `Array.init` will not find the `underestimatedCount` for collections, and instead of preallocating storage for all elements as it is supposed to, it will make log(N) allocations growing the array's buffer exponentially. Oh, wow, it's worse than a trip-up: there's no way to fix this! Even if `FlattenSequence` implemented `underestimatedCount` in its conditional conformance to `Collection`, it wouldn't get used. /cc kylemacomber (JIRA User) |
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Comment by Daryle Walker (JIRA) From the first comment, what if you overrode |
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@swift-ci create |
Additional Detail from JIRA
md5: 822582f2db61889e5af168490d06a19e
Issue Description:
If I am the author of a generic library—e.g. the Swift Standard Library—and I have an algorithm for one of my protocols (e.g.
Collection) that I know can have a lower complexity bound for a refined protocol (e.g.RandomAccessCollection), the prescription is to make the algorithm a requirement of the base protocolCollectionand implement it in extensions of bothCollectionandRandomAccessCollection. I can then document the complexity of the algorithm as dependent on the conformance of the model toRandomAccessCollection, right?Wrong!! I have to say the complexity depends on the conformance to
RandomAccessCollectionunless the model is a generic type and its most general form conforms toCollection, in which case the complexity depends on whether the conformance toRandomAccessCollectionis statically known at the point where the algorithm is invoked. In other words, there's nothing I can say about it that's both intelligible and accurate.Furthermore, in this case, as far as I can tell, there's no way at all to access the specialized version of the algorithm from generic code or for the author of the conditionally-conforming type to be explicit about getting the specialized algorithm implementation.
This gist demonstrates.
IMO this model is untenable, and is clearly incompatible with Swift's goals for supporting generic programming.
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