ts-pattern

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Describe the bug In the code below, all occurrences of v1 and v2 are inferred as string | undefined but in fact they should be string because the undefined cases are already handled.

const getOptionalString: () => string | undefined = () => Math.random() > 5 ? undefined : 'foo';

const value = match([getOptionalString(), getOptionalString()])
  .with([undefined, undefined], () => 0)
  .with([P.select(), undefined], (v1) => 1) // BAD: v1 is inferred as string|undefined but should be just string
  .with([undefined, P.select()], (v2) => 2)
  .otherwise(([v1, v2]) => 3);

In fact, it works if the pattern is a single value instead of a tuple:

const value = match(getOptionalString())
  .with(undefined, () => 0)
  .otherwise((v) => 1); // GOOD: v is inferred as string without undefined

Code Sandbox with a minimal reproduction case https://codesandbox.io/p/sandbox/crazy-galileo-3zsfmh

Versions

• TypeScript version: 5.1.6
• ts-pattern version: 5.0.6
• environment: N/A (it happens at dev time)

This behavior is expected. In order to support this we would need to distribute unions in the input type even without using .exhaustive() and this is pretty expensive in terms of type checking time.

See the discussion in issue https://github.com/gvergnaud/ts-pattern/issues/145 for some additional context on how type narrowing and exhaustive work in ts-pattern