optional parameter for generics is being ignored

[plusgut]

Bug Report

🔎 Search Terms

generics, optional parameter, default generic type, generic constraint

🕗 Version & Regression Information

• This changed between versions 4.1.5 and 4.2.3

⏯ Playground Link

Playground link with relevant code

💻 Code

interface SomeInterface<T> {
    value: T
}

declare function factory<T extends string = string>(optionalParameter?: T): SomeInterface<T>;
declare function echo<K extends {[key: string]: SomeInterface<any>}>(opt: K ): K;

const fallbackToConstraintInsteadOfDefault = echo({ foo: factory() } as const);

🙁 Actual behavior

the return type of echo is not fully of type K, but falls back to the constraint of K.

🙂 Expected behavior

The return type is from the default type of T.

Notes

This behaviour does not occur in 4.1.5
This behaviour does not occur when no constraint for K is given.
This behaviour does not occur when factory gets called with a parameter.
This behaviour does not occur when the return value of factory is first stored in a variable.

[ahejlsberg]

Appears to be caused by #39384 and #41475. With those PRs we now produce contextual types for expressions in as const assertions and for that reason we make an inference from SomeInterface<any> (the element type in the constraint of K) to SomeInterface<T> (the return type of the factory() invocation). We do the same thing in both 4.1 and 4.2 if you remove as const. Things work as expected if you change the SomeInterface<any> constraint to SomeInterface<unknown> (since unknown isn't assignable to string we then pick string).

[plusgut]

A constraint is altering the type, and weakining it through const? That doesn't seem to be working as intended to me.

Especially since that doesn't always happen but only when the optionalParameter is not given, and therefore the default of T is assigned. In all the other cases it does not happen.

Your idea with unknown is interesting and does indeed work in this very simple scenario. But if I have a slightly more complex interface it will break. Since unknown is not assignable to string

interface SomeInterface<T> {
    value: (opt: T) => T
}

declare function factory<T extends string = string>(optionalParameter?: T): SomeInterface<T>;

declare function echo<K extends {[key: string]: SomeInterface<unknown>}>(opt: K ): K;


const fallbackToConstraintInsteadOfDefault = echo({ foo: factory() });

https://www.typescriptlang.org/play?jts=4.2.3&ssl=1&ssc=1&pln=12&pc=1#code/JYOwLgpgTgZghgYwgAgMoHsC2ECS5rxIA8AKgHzIDeAUMncgG5wA2ArhAFzIAU6ADmC4kAlMgC8FEtQC+1agBMICZnCgoYrEAjDB0IZITDooAT1LIIAD0gh5AZ2R2wUUAHNxj527K8BukCwACqpw2JBQAPxCwlwY2HjhhBCkZADccorKquqa2v4WCAAW6EQA0hbWELYOlADaANYQJlxOLiCuALqxWLj4sIjJmvUg6ADuIGTSPvyCyOUxc+lyCHpOBizMAEaI9SToAMKrznCgYHhOEHDyAPIwACIQ8KzMYB5KxdyUBujoXIbGJm4omkwnSKxAaxGIGufj0LBwdgAYqAUGICh8vjAfn9EEZTNwAETbABeBOBoOWR2QajArCgIAAaix2B5-viKeC1ooni9zpArrdDhDjqdERttgh6m8iuhPt9ftSILT6Uy2CgQekgA

[typescript-bot]

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