union of pattern template literals intersected with optional brand mutually assignable to one with incompatible brand

🔎 Search Terms

brand, flavor, pattern template literal, placeholder, union, intersection, optional, assignable

🕗 Version & Regression Information

• This changed between versions 4.3.0-dev.20210330 and 4.3.0-dev.20210331
• This changed in commit or PR #43440, I think

⏯ Playground Link

Playground link

💻 Code

type TmpLitUnion = `0${string}` | `1${string}`
type BrandX = TmpLitUnion & { a?: "x" };
type BrandY = TmpLitUnion & { a?: "y" };
declare const brandY: BrandY;
const brandX: BrandX = brandY; // <-- no error

🙁 Actual behavior

BrandX and BrandY are mutually assignable, even though they have an incompatible optional a property.

🙂 Expected behavior

There should be an error when trying to assign a BrandY to a BrandX or vice versa.

Additional information about the issue

This is from a Stack Overflow question. Looks like optional brands (this is called... "flavored" I guess?) don't distinguish between unions-of-pattern-template-literals. If you get rid of the union, you get the expected error:

type TmpLit = `0${string}`;
type BrandX = TmpLit & { a?: "x" };
type BrandY = TmpLit & { a?: "y" };
declare const brandY: BrandY;
const brandX: BrandX = brandY; // <-- error

Or if you add something else to the union, you get the expected error:

type TmpLitUnionOrSomethingElse = `0${string}` | `1${string}` | "somethingElse"
type BrandX = TmpLitUnionOrSomethingElse & { a?: "x" };
type BrandY = TmpLitUnionOrSomethingElse & { a?: "y" };
declare const brandY: BrandY;
const brandX: BrandX = brandY; // <-- error

Or if the brand is not optional, etc.

Not sure if this is a bug or a design limitation, or somehow intentional.

在链接的 TypeScript Playground 中复现该赋值，然后将其与 issue 中不含 union 和不含可选 brand 的示例进行比较。调查与可选属性相交的模式模板字面量 union 的可赋值性；完成标准是 BrandX 和 BrandY 不再可以相互赋值，同时报告的比较用例仍然正确。