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Broken narrowing of types intersecting with unique symbols #24846
Description
TypeScript Version: 3.0.0-dev.20180609 & 2.9.1
Search Terms:
Narrowing unique symbol union
Code
declare const phantom: unique symbol;
type Phantom = typeof phantom;
type PhantomString = string & Phantom;
const fn = (arg?: PhantomString) => arg;
const fn2 = (arg: Array<PhantomString | null>) => arg;Expected behavior:
fn and fn2 should be reported as:
const fn: (arg?: PhantomString) => PhantomString | undefined;
const fn2: (arg: Array<PhantomString | null>) => Array<PhantomString | null>;Actual behavior:
fn and fn2 are reported as:
const fn: (arg?: undefined) => undefined;
const fn2: (arg: null[]) => null[];Playground Link:
Appears to be partially related to strictNullChecks
Activity
jack-williams commentedon Jun 11, 2018
Related #23672 ?
mhegazy commentedon Jun 11, 2018
The compiler reduces empty intersection types to
neverwhen used in a union. an empty intersection type is one whose constituents are all unit types. so1 & 2"a" & falseor"s" & unique symbol.so
PhantomString | nullis justnull. see #18438 for more details.leemhenson commentedon Jun 12, 2018
Ah ok, I see. Using
type PhantomString = string & { "phantom-string": any };seems to work. I can see there are a few issues suggesting true nominal typing support so I'll add my 👍 to those.Thanks.