Conditional type inference does not work with a function's generic argument

[samijaber]

TypeScript Version: 3.0.0-dev.201xxxxx

Search Terms: conditional type inference generic

Code

type Get<T> = T extends {} ? true : false;

const fn = <T extends {}>(t: T) => {
    // Expected `true`
    // Actual `T extends {} ? true : false`
    type Result = Get<T>;
}

Expected behavior: correct type inference: type Result = true

Actual behavior: type inference not performed at all

Playground Link: link

[jack-williams]

This conditional type is distributive because there is a naked type parameter in the check clause. A distributive conditional type will evaluate to never when the check type is instantiated to never, so it is not possible to immediately resolve this to true.

To get the desired behaviour make the conditional type non-distributive like so:

type Get<T> = [T] extends [{}] ? true : false;

const fn = <T extends {}>(t: T) => {
    const x: Get<T> = true; // ok, Get<T> = true
}

[samijaber]

Ah, I see. Thanks @jack-williams! Will close this then.