Type guards can't deal with unknown type as default

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

Search Terms:

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

Riproduci il comportamento di narrowing nel TypeScript Playground collegato usando le dichiarazioni di f, g e readonly number[] a, quindi esamina la logica di narrowing dei type guard. Il lavoro è completato quando g(a) restringe a a unknown[] come previsto e il comportamento è coperto da un test di regressione; la issue correlata #17002 può fornire contesto.