Generic Conditional Type resolved to `any` when passed an error `any` type about typescript HOT 8 CLOSED

Both are just the same for the compiler. It still feels to me that you could move this discussion to the other issue and try to advocate there that your use case is different than the one presented there (IMHO it's still the same but that's beyond this argument :P).

As Ryan said there, once you have an error "all bets are off" 😉

A conditional type by definition should return from either the true or false branch.

This isn't technically true. Sometimes it returns both :D

type Test = any extends string ? 1 : 2
//   ^? type Test = 1 | 2

EDIT:// I now realized the other issue is closed, so I'll give you that. I think it doesn't make sense to comment much on closed issues so perhaps a new one is appropriate here. I anticipate it will be closed with the same conclusion though.

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@lucaimbalzano The playground I linked to is a simplified version of a downstream bug (see nuxt/nuxt#29021). The intention was to show that the implicit any created from an "No export error" changes the behaviour of the conditional type.

If "some-module" exports a default value with any (but the type isn’t explicitly declared), TypeScript might not treat it exactly the same as any.

In this scenario, "some-module" does not have default exports. Also "some-module" would be any number of third-party modules, explicitly defining any is not an option

@Andarist

This isn't technically true. Sometimes it returns both :D

Thank you, I stand corrected. However, the bug still persists for a conditional like type IsTwo<T> = T extends 2 ? true : false and IsTwo<ImplicitAny> is supposed to be a boolean but becomes any (see this playgroud)

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Related to #58960, possible cause outlined here (#58960 (comment))

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The problem stems from how TypeScript infers types from modules. If "some-module" exports a default value with any (but the type isn’t explicitly declared), TypeScript might not treat it exactly the same as any.

To resolve this, you should explicitly declare the default export as any in your module.
Here's how:

1. Explicitly Declare the Module's Default Export as any:

declare module "some-module" {
  const _default: any;
  export default _default;
}

2. Type Checking with IsAny: Now, when you check IsAny, it should correctly detect the any type:

type IsAny<T> = 0 extends 1 & T ? true : false;

type ImplicitAny = typeof import("some-module").default;

type Result = IsAny<ImplicitAny>; // true
type ExpectedResult = IsAny<any>; // true

⏯ Here's you can check by ts live-code :

// Declare the module "some-module" with an explicit 'any' type for the default export
declare module "some-module" {
  const _default: any;
  export default _default;
}

// A utility type to check if...

Playground Link

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@Eazash you have already found the other issue about it, this is no different from it. I think it would be best to close this issue to keep the discussion focused in the original thread about this change.

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@Andarist In my opinion, there is a difference. The other issue is referring to the "Unresolved any" when a type doesn't exist, and how that exhibits different behavior. This issue is describing how an error any behaves differently from a regular any. Unlike unresolved there is no label/comment on hover, yet it breaks the contract of a conditional type. A conditional type by definition should return from either the true or false branch. By returning any it is quietly introducing a third branch.

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The propagating behavior of the error any is intentional, so that when you have one type error, it doesn't result in (potentially) tens of thousands of downstream errors.

You should write this instead if you'd prefer the other behavior

type PossibleDefault<T> = T extends { default: infer U } ? U : any;
type ImplicitAny = PossibleDefault<typeof import("some-module")>;

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This issue has been marked as "Not a Defect" and has seen no recent activity. It has been automatically closed for house-keeping purposes.

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