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.

まず、リンク先の TypeScript Playground の再現コードを実行し、ネストされた呼び出しを単独の呼び出しおよび完全に指定された呼び出しと比較してください。既存の推論コンテキストについては、関連する issue #54184、#56714、#62680 を読んでください。完了条件は、区切り文字を省略した両方の例が期待されるリテラル型でコンパイルされ、区切り文字を指定したケースが引き続き正しいことです。