Fix generic mapped type relationships#19564
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@mhegazy This fix makes |
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| else if (isGenericMappedType(target) && !isGenericMappedType(source) && getConstraintTypeFromMappedType(<MappedType>target) === getIndexType(source)) { |
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There's a branch lower down that checks whether target is a generic mapped type. Can this be performed there?
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No because that particular section of code is only entered if source is an object or intersection type.
mhegazy
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Oct 30, 2017
sandersn
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Recursive mapped types usually lead to the error "excess stack depth comparing types" because of a new type relation rule added in #19564 which says that "A source type T is related to a target type { [P in keyof T]: X } if T[P] is related to X". Unfortunately, with self-recursive mapped types like ```ts D<T> = { [P in keyof T]: D<T[P]> } ``` we get infinite recursion when trying to assign a type parameter T to D<T>, as T[P] is compared to D<T[P]>, T[P][P] is compared to D<T[P][P]>, and so on. We can avoid many of these infinite recursions by replacing occurrences of D in the template type with its type argument. This works because mapped types will completely cover the tree, so checking assignability of the top level implies that checking of lower level would succeed, even if there are infinitely many levels. For example: ```ts D<T> = { [P in keyof T]: D<T[P]> | undefined } <T>(t: T, dt: D<T>) => { dt = t } ``` would previously check that `T[P]` is assignable to `D<T[P]> | undefined`. Now, after replacement, it checks that `T[P]` is assignable to `T[P] | undefined`. This implementation suffers from 3 limitations: 1. I use aliasSymbol to detect whether a type reference is a self-recursive one. This only works when the mapped type is at the top level of a type alias. 2. Not all instances of D<T> are replaced with T, just those in intersections and unions. I think this covers almost all uses. 3. This doesn't fix #21048, which tries to assign an "off-by-one" partial-deep type to itself. Mostly fixes #21592. One repro there has a type alias to a union, and a mapped type is a member of the union. But this can be split into two aliases: ```ts type SafeAnyMap<T> = { [K in keyof T]?: SafeAny<T[K] }; type SafeAny<T> = SafeAnyMap<T> | boolean | string | symbol | number | null | undefined; ```
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Fixes #18201.