Wrong inference for defaulted generic in nested call

🔎 Search Terms

nested generic call inline inference contextual typing function calls default type parameter

🕗 Version & Regression Information

• I see the same issue from version 4.1.5 to nightly in TypeScript Playground

⏯ Playground Link

https://www.typescriptlang.org/play/?ts=6.0.0-dev.20260401#code/CYUwxgNghgTiAEAzArgOzAFwJYHtXwCsctUAeAFXhAA8MRVgBneRjGEgcwBp4BVK2vSYs2nHgDUBdBs1btUHeAF54AclUA+ABStYGAFzxyPIYd49GIAA4B+Q+ICUhgAYASAN7kAvh-E-3vF7OAFCgkLAIKOjYeAJW4BgU2gBuUBDIIIbkTvDu8Bg4AEIgWjTxmCDAWTnJxMDwXsHBZQlaRCRaqgASIBAQOKo8qgDqODAQwKoODgB0BcWdPX04o+OTDvAA9JvwAEZQwE1geKzwAO5YGAAWOMgYAMrWsFAFMMqExKiLvf2DaqsTKbBbbwUHwAB6Nma1HKGC0F2utweTxgLzGs3mJQAREt+gDgFiNiCODgcIdobC2p9vss-iMxoChmppnMiiVuj8cPB8VMtjsSWTgkA

💻 Code

declare function join<T extends string, U extends string, V extends string = ''>(start: T, end: U, sep?: V): `${T}${V}${U}`
declare function expect<T>(value: T): { toBe(expected: T): void }

expect(join('Hello', 'World')).toBe('HelloWorld') // bad

const withoutSeparator = join('Hello', 'World')
//    ^? const withoutSeparator: "HelloWorld"
expect(withoutSeparator).toBe("HelloWorld") // good

expect(join('Hello', 'World', ' ')).toBe('Hello World') // good

🙁 Actual behavior

TypeScript does not infer the return types correctly of a nested function call when an optional generic param is omitted, even if a default is provided. When the function call is not nested, it is inferred correctly. It is also inferred correctly when all params are provided.

🙂 Expected behavior

I would expect that both of these approaches would compile:

expect(join('Hello', 'World')).toBe('HelloWorld') // bad

const withoutSeparator = join('Hello', 'World')
expect(withoutSeparator).toBe("HelloWorld") // good

Additional information about the issue

There are very similar issues, such as #54184, #56714, and #62680 but they didn't quite match up to what I'm seeing here, particularly since the reproductions are quite a bit more complex.

Beginne damit, die verknüpfte TypeScript Playground-Reproduktion auszuführen, und vergleiche den verschachtelten Aufruf mit dem eigenständigen und dem vollständig spezifizierten Aufruf. Lies die verwandten Issues #54184, #56714 und #62680, um den bestehenden Inferenzkontext zu verstehen. Erledigt ist die Aufgabe, wenn beide Beispiele ohne angegebenes Trennzeichen mit dem erwarteten Literaltyp kompiliert werden, während der Fall mit angegebenem Trennzeichen weiterhin korrekt bleibt.