Expand type references recursively in cache key #18207
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This means that `A<B<T, C<U>>>` will include the keys for `B` and `C` now.
4 is the limit.
ahejlsberg
requested changes
Sep 5, 2017
src/compiler/checker.ts
Outdated
| @@ -9782,15 +9782,15 @@ namespace ts { | |||
| return type.flags & TypeFlags.TypeParameter && !getConstraintFromTypeParameter(<TypeParameter>type); | |||
| } | |||
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| function isTypeReferenceWithGenericArguments(type: Type) { | |||
| return getObjectFlags(type) & ObjectFlags.Reference && some((<TypeReference>type).typeArguments, isUnconstrainedTypeParameter); | |||
| function isTypeReferenceWithGenericArguments(type: Type): type is TypeReference { | |||
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Technically this is not correct. A type might be a type reference even if this function returns false.
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Oops, I forgot that 'type is TypeReference' implies that type is not TypeReference if the return value is false.
I forgot that 'f(x): x is T' implies that x is *not* T if f returns false.
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This means that
A<B<T, C<U>>>will include the keys forBandCnow.Fixes the secondary cause of #17716 — when comparing
Bluebird<string>toBluebird<string>, the compilation ends up using two versions of Bluebird. And it turns out Bluebird is a very large type, and nearly every method returns another instance of Bluebird. This causes the compiler to run out of memory and crash. With this fix, compilation finishes in 2.9 seconds compared to 0.9 seconds when package deduping work correctly, skipping the structuralBluebird <-> Bluebirdcomparison.The fix works because many of Bluebird's methods return
Bluebird<T[]>. So extendinggetTypeReferenceIdto understand arrays allows caching to perform better. The intuition is that if we don't learn anything by relatingA<T> ==> B<T>when already relatingA<U> ==> B<U>, then we also don't learn anything by relatingA<T[]> ==> B<T[]>when already relatingA<U[]> ==> B<U[]>.This technique, in fact, generalises to any depth of nested type references. that is a type argument of a type reference. So if we are already relating
Foo<Bar<T, Baz<V>>> ==> Qux<Quid<T, Quam<V>>>then we will skipFoo<Bar<U, Baz<W>>> ==> Qux<Quid<U, Quam<W>>>.This works by generating the same key in
getTypeReferenceId. I add one additional case that checks if a type argument is itself a type reference with generic arguments. If so, thengetTypeReferenceIdcalls itself recursively on that type argument.Note that we could limit the depth of the expansion when generating the key by adding a depth parameter to
getTypeReferenceId.