Repository navigation
Generic type inference prefers conditional types #31766
Description
TypeScript Version: 3.4.5 and 3.5.1
Search Terms: Conditional types, Type inference
Code
interface I {a: string}
type Cond<T> = T extends any ? Partial<T>: T;
declare function f<T>(x1: T, x2: Cond<T>): T;
function g(p1: I, p2: Cond<I>): I {
return f(p1, p2); // error here
}Expected behavior:
No errors. A trivial type inference is T=I. This is also the behavior of TypeScript 3.3.
Actual behavior:
error TS2322: Type 'Partial<I>' is not assignable to type 'I'.
From that error message the actual inference is T=Partial<I> (or Cond<I>), which seems like a regression to me.
Playground Link:
playground link
Related Issues:
There are some issues around conditional types but nothing quite a duplication, I think.
#31601 requires function overloads
#24085 is a problem with enums
#30489 is more about type inference on the conditional type itself, not the generic type when the conditional type is used
#30341 is similar but it's marked as fixed
Activity
RyanCavanaugh commentedon Jun 13, 2019
What is the intent of writing
T extends any?bowenni commentedon Jun 13, 2019
To create a conditional type. I think the problem only happens with a conditional type.
If the example is oversimplified and doesn't make sense, below is another example, without the
T extends any.RyanCavanaugh commentedon Jun 25, 2019
ahejlsberg commentedon Jul 1, 2019
This is caused by #30010. Part of why we inferred to a union type was to get lower priority candidates for inferences to naked type variables, but that got lost with the fix in #30010. We just need to reinstate that part.
weswigham commentedon Jul 2, 2019
To be fair, at the time I'm pretty sure we had done away with the naked type variable priority; we just didn't think to add it to the conditional inference bit when we added it back. :P