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Untagged variant puts one typed array in another typed array's case

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Issue 类型
缺陷
描述清晰度
基本清楚
活跃度
活跃
技术栈
javascript
领域
compilers

调研方向

Start with rule 5 in compiler/ml/UntaggedVariants.md and review the linked context in #8952 and #8950, which changes how the types are recognized. Reproduce the example with the two typed arrays and trace how their type names are used for run-time case checks. Done means the invalid overlap is rejected or the mismatched value becomes a type error.

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描述

An untagged variant with cases for two typed arrays accepts a value of one typed array in the case of the other, and the value then matches the other case.

@unboxed type ta = I8(Int8Array.t) | U8(Uint8Array.t)
let f = v => switch v { | I8(_) => "i8" | U8(_) => "u8" }
Console.log(f(I8(Uint8Array.fromArray([1]))))

Output: u8.

The typed-array types are abbreviations of one type, so I8(Uint8Array…) type-checks. The cases are classified by the type name as written, so the run-time check is instanceof Uint8Array.

Expected: either the cases are distinct types, so that I8(Uint8Array…) is a type error, or the variant is rejected because two cases share one type.

The rule is compiler/ml/UntaggedVariants.md rule 5, in #8952. #8950 changes how these types are recognized.

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