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

🔎 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.

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;
    }
})

I think I get your point.

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

This issue has been marked as "Duplicate" and has seen no recent activity. It has been automatically closed for house-keeping purposes.

Mateusz Burzyński (@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;
}
})

Mateusz Burzyński (@Andarist) Why won't that work though?

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

Especially as s has noinfer which means it should infer from the return type that was explicitly given.

Harpush

Harpush this one could work, see: #60964

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Function Generic is `unknown` if argument has `NoInfer<Generic>` but returns the Generic #60922

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Function Generic is unknown if argument has NoInfer<Generic> but returns the Generic #60922

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🔎 Search Terms

🕗 Version & Regression Information

⏯ Playground Link

💻 Code

🙁 Actual behavior

🙂 Expected behavior

Additional information about the issue

Activity

Andarist commented on Jan 6, 2025

JeanMeche commented on Jan 6, 2025

typescript-bot commented on Jan 10, 2025

Harpush commented on Jan 10, 2025

Andarist commented on Jan 13, 2025

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Function Generic is `unknown` if argument has `NoInfer<Generic>` but returns the Generic #60922