RequirementMachine: Protocol reduction order refactoring #39864
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Previously we said that if P1 inherits from P2, then [P1:T] < [P2:T].
However, this didn't generalize to merged associated type symbols;
we always had [P1&P2:T] < [P3:T] even if P3 inherited from both P1
and P2.
This meant that the 'merge' operation did not preserve order, which
fired an assertion in the completion logic.
Fix this by generalizing the linear order to compare the 'support'
of protocols rather than the 'depth', where the 'support' is
defined as the size of the transitive closure of the set under
protocol inheritance.
Then, if you have something like
protocol P1 {}
protocol P2 {}
protocol P3 : P1, P2 {}
The support of 'P1 & P2' is 2, and the support of 'P3' is 3,
therefore [P3:T] < [P1&P2:T] in this example.
Fixes <rdar://problem/83768458>.
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@swift-ci Please test source compatibility |
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The previous reduction order did not satisfy the property that a merged associated type pair is smaller than or equal to both original associated types. This would trigger an assertion failure in completion.
That is, if P3 : P2 and P3 : P1, we want
Because the merge of
[P3:T]with[P1&P2:T]is in fact[P3:T].In the previous order, a merged symbol was always smaller than a single associated type symbol, so we had
[P1&P2:T] < [P3:T].Also, storing the reduction order in a per-rewrite system protocol graph didn’t really make sense. Make it global on RewriteContext and fold what remains of the ProtocolGraph into the RewriteSystemBuilder, where it is used to compute the transitive closure of the set of protocol dependencies.
Fixes rdar://problem/83768458.