Typescript generic functions order

Question

I would like to create a function that takes some partial props to populate a type, and returns another function, which in turn accepts exactly the remainging props to create an object of the type.

declare function creator<T extends U>(defaults: U) =>
  (fields: Pick<T, Exclude<keyof U, keyof T>>) => T

so that I can do:

type A = { a: number, b: string };
const make = creator<A>({ a: 1 }); // ({ b: string }) => A

I don't know how to let creator only accept T at call-time, but declare it extending U and infer U.
I don't know why I cannot "construct" T from U and Pick<T, Exclude<keyof U, keyof T>> with generics. It is possible with types directly:

type T = { a: number; b: string };
type A = { a: number };
type T1 = A & Pick<T, Exclude<keyof A, keyof T>>;
const t1: T = { a: 1, b: "b" };
const t2: T1 = t1;

Titian Cernicova-Dragomir · Accepted Answer · 2019-06-03 08:37:08Z

1

You can't have partial argument inference in typescript. This means you can't specify T and infer U in the same call. You can do it using function currying.

type Omit<T, K extends PropertyKey> = Pick<T, Exclude<keyof T, K>> // not needed in 3.5
function creator<T>() {
    return function <K extends keyof T>(defaults: Pick<T, K>) {
        return function (values: Omit<T, K>) :T {
            return Object.assign(values, defaults) as unknown as T
        }       
    }
}

type A = { a: number, b: string };
const make = creator<A>()({ a: 1 }); // ({ b: string }) => A
make({ b: "" })

Also inside the function where T is still not known typescript will not be able to do a lot of reasoning about what Omit<T, K> is. This is why assigning the result of Object. assign fails inside the function, while in similar conditions outside a generic function (where all type are known) it would succeed. Using a type assertion is the only way to get things to work.

answered Jun 3, 2019 at 8:37

Titian Cernicova-Dragomir

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1 Comment

Gipsy King Over a year ago

Thank you, creator<A>()({ a: 1 }); is fine.

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