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Type guards can't deal with unknown type as default #37905

@falsandtru

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@falsandtru

We can't know the failures of type inference in type guards because of this problem. Maybe a is any[] should also return unknown[].

TypeScript Version: 3.7.x-dev.20200410

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Code

declare function f(a: any): a is any[];
declare function g(a: any): a is unknown[];
declare const a: readonly number[];

f(a) && a; // any[]
g(a) && a; // not unknown[]

Expected behavior:
a's type narrowed by g is unknown[].
Actual behavior:
a's type narrowed by g is readonly number[] & unknown[].

Playground Link: https://www.typescriptlang.org/play/?ts=3.9.0-dev.20200410&ssl=1&ssc=1&pln=6&pc=11#code/CYUwxgNghgTiAEAzArgOzAFwJYHtVIAooAueKVATwEpSp4sBnMygbQF0BuAKFElgRTpseeAHMitSjTL0maANaocAd1TtuvaHHhg8DDGVJwowPBArxUyALYAjEDHVcuiIlXgAyD2W7io7rx8gA

Related Issues: #17002

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