Typeguard should use conditional types

[mkulke]

TypeScript Version: 2.9.0-dev.20180428

Search Terms:

conditional types, type guards

Code

type Pred<T, U extends T> = (v: T) => v is U;
function someFn<T, U extends T>(pred: Pred<T, U>, val: T): void {
 if (pred(val)) {
 const u: U = val;
 console.log(u);
 return;
 }
 const notU: Exclude<T, U> = val; // Type 'T' is not assignable to type 'Exclude<T, U>'.
 console.log(notU);
}

Expected behavior:

The code should compile, the predicate type-guard sets val to U in the "true" branch, and to Exclude<T, U> in the "false" branch.

Actual behavior:

In the "false" branch val is inferred to be T:

Type 'T' is not assignable to type 'Exclude<T, U>'.

Playground Link: https://bit.ly/2HAtMpF

Related Issues:

[mkulke]

this works:

function doSth<T, U extends T>(pred: Pred<T, U>, val: T): void {
 const antiPred = (val: T): val is Exclude<T, U> => !pred(val);
 if (pred(val)) {
 const u: U = val;
 console.log(u);
 return;
 }
 if (antiPred(val)) {
 const notU: Exclude<T, U> = val;
 console.log(notU);
 }
}

[mkulke]

I made it work by altering val: T to val: U | Exclude<T, U>, which does not feel too wrong. but my intuition says that by stating <T, U extends T>, T would be equivalent to U | Exclude<T, U>. 🤔

type Pred<T, U extends T> = (v: T) => v is U;
function someFn<T, U extends T>(pred: Pred<T, U>, val: U | Exclude<T, U>): void {
 if (pred(val)) {
 const u: U = val;
 console.log(u);
 return;
 }
 const notU: Exclude<T, U> = val;
 console.log(notU);
}

if my intuition is off here, i'll gladly close the issue.

[mhegazy]

Related to #22348

[RyanCavanaugh]

my intuition says that by stating <T, U extends T>, T would be equivalent to U | Exclude<T, U>

This isn't the case.

[mkulke]

Ryan Cavanaugh (@RyanCavanaugh): alright i close the issue then. if you find the time, you can maybe elaborate. My naive reasoning would be: Given there is "Animal" and "Dog", stating: "Animal" would be be the same as "Dog and every Animal but Dog".