Function Generic is unknown if argument has NoInfer<Generic> but returns the Generic

[JeanMeche]

🔎 Search Terms

• Generic
• NoInfer
• Unknown

🕗 Version & Regression Information

Still the case in the latest 5.7.2.

⏯ Playground Link

https://www.typescriptlang.org/play/?#code/KYDwDg9gTgLgBAE2AYwDYEMrDgMwK4B2yMAlhAXKiQQNbAIDKJA5geqgDwMA0cAIgD4AFBDClyAZwBccAN4AoOErgSIeKMmAyhASjgBeAXAYBuRcuQQAtmDwx04gttXrNMgHIQAkgRzAoHIJ6hvxmAL46Mnxm8pYEEvASLGyoBpTUdIzJ7EIKyipqGlpwugZGAMzc5kqWNnYOZE4lEsEVcPIR8vJAA

💻 Code

export declare function linkedSignal<S, D>(options: {
    source: () => S;
    computation: (source: NoInfer<D>) => D;
}): D;

const signal = linkedSignal({
    source: () => 3,
    computation: (s) => 3 
})

🙁 Actual behavior

signal is unknown

🙂 Expected behavior

signal is number

Additional information about the issue

I wasn't sure, but it could be a duplicate of #60544.

[Andarist]

This, indeed, looks like a simpler version of #60544

I don't quite see how this could be implemented beyond some very basic support for a small set of scenarios. The return type of a function very often depends (even if indirectly) on its parameters types. Like, how would you determine what D should be here?

const signal = linkedSignal({
    source: () => 3,
    computation: (s) => s
})

Would unknown be a satisfiable inference here? What about this one?

const signal = linkedSignal({
    source: () => 3,
    computation: (s) => {
      return Math.random() > 0.5 > 1 : s;
    }
})

[JeanMeche]

I think I get your point.

number | unknown narrows down to unknown, so unknown is the reseanable inference.