[Bug] Conditional types throw error since TS 4.9.x

[jxn-30]

Bug Report

🔎 Search Terms

conditional types, 2339

🕗 Version & Regression Information

• This changed between versions 4.8.4 and 4.93
• This bug also occurs on nightly version 5.0.0-dev.20221121

⏯ Playground Link

Playground link with relevant code and test cases

💻 Code

// Errors in TS 4.9.3 but not in 4.8.4
interface A {}
type B = A extends Record<'foo', string> ? A['foo'] : string; // Error: Property 'foo' does not exist on type 'A'.(2339)

// Succeeds in TS 4.9.3 and 4.8.4
interface C { foo: number; }
type D = C extends Record<'foo', string> ? C['foo'] : string; // Correct: string

// Succeeds in TS 4.9.3 and 4.8.4
interface E { foo: string; }
type F = E extends Record<'foo', string> ? E['foo'] : string; // Correct: string

// Succeeds in TS 4.9.3 and 4.8.4
interface G { foo: 'bar'; }
type H = G extends Record<'foo', string> ? G['foo'] : string; // Correct: 'bar'

🙁 Actual behavior

interface A {}
type B = A extends Record<'foo', string> ? A['foo'] : string;

errors with 2339: Property 'foo' does not exist on type 'A'.

This change causes @vuejs/pinia to fail as one can see in vuejs/pinia#1814

🙂 Expected behavior

interface A {}
type B = A extends Record<'foo', string> ? A['foo'] : string;

returns with B being an alias for type string.

[MartinJohns]

Workaround: A extends Record<'foo', infer X extends string> ? X : string

[ahejlsberg]

This is working as intended. In cases where the outcome of a conditional type is known to always be the false branch, we no longer pretend that the type in the true branch has satisfied the extends check. In the example that produces an error above, the outcome of the conditional type is always the false branch string, and in the true branch we no longer pretend it is possible to write A['foo'] because that property is known not to exist. I think this is preferable to allowing nonsense like this:

type T0 = Number extends String ? Number["length"] : String["length"];  // This was previously ok but now errors

I understand there's a possible usage pattern where the conditional type is a form of conditional compilation based on types declared elsewhere in the program, but in those cases I think it is reasonable to require an explicit intersection:

type B = A extends Record<'foo', string> ? (A & Record<'foo', string>)['foo'] : string;

[Andarist]

@ahejlsberg what about interface augmentation though? With it, the type checker can't actually prove that the outcome of this conditional type is always its false branch. The conditional type can live in the library's code and the augmentation might happen in the consuming application. It seems that this is exactly how this was used here.

In general, I agree with your reasoning but when providing the fix, I assumed that this was meant to be supported exactly because of the interface augmentation.

[ahejlsberg]

what about interface augmentation though?

Yeah, although unintended, I agree that's meaningful scenario. Let's call this a regression then and merge your PR.