The type inference in TypeScript sometimes struggles to accurately determine the type of an iterable

Struggling to get TypeScript to correctly infer the underlying iterable type for the function spread. The purpose of this function is to take an iterable of iterable types, infer the type of the underlying iterable, and return a new iterable with that inferred type.

interface Output<T, R> {
    (i: Iterable<T>): Iterable<R>;
}

function pipe<T, A>(i: Iterable<T>, p0: Output<T, A>): Iterable<A>;
function pipe(i: Iterable<any>, ...p: Output<any, any>[]): Iterable<any> {
    return []; // dummy iterable
}

// Attempting to implement the following function:
function spread<T extends Iterable<R>, R>(): Output<T, R> {
    return (iterable: Iterable<T>) => ({
        [Symbol.iterator](): Iterator<R> {
             return null as any; // dummy, for now
        }
    });
}

pipe(['text'], spread());
// expected types: spread => Output<string, string>, pipe => Iterable<string>
// actual types: spread => Iterable<string, unknown>, pipe => Iterable<unknown>

pipe([[1, 2], [3, 4]], spread());
// expected types: spread => Output<number[], number>, pipe => Iterable<number>
// actual types: spread => Output<number[], unknown>, pipe => Iterable<unknown>

While the template signature for the function spread seems correct in theory, I'm puzzled by why the iterable type resolves to unknown in the examples above.

As a result, spread infers as <T, unknown>, causing pipe to fail in type inference as well.


This issue arose while working on the iter-ops library. While I have successfully implemented various operators, I hit a roadblock with the spread operator's type inference.

UPDATE

With guidance from Titian Cernicova-Dragomir, I was able to successfully finalize the spread operator :)

Answer №1

My understanding of the code is that spread will convert an

Iterable<Iterable<T>>
into an Iterable<T>.

Based on this, the function returned by spread should have a signature of

(iterable: Iterable<Iterable<T>>) => Iterable<T>
. In terms of types, it would look like
Output<Iterable<T>, T>
. Therefore, there is no requirement for the R argument since the input and output are interconnected.

interface Output<T, R> {
    (i: Iterable<T>): Iterable<R>;
}

function pipe<T, A>(i: Iterable<T>, p0: Output<T, A>): Iterable<A>;
function pipe(i: Iterable<any>, ...p: Output<any, any>[]): Iterable<any> {
    return []; // dummy iterable
}

function spread<T>(): Output<Iterable<T>, T> {
    return (iterable: Iterable<Iterable<T>>) => ({
        [Symbol.iterator](): Iterator<T> {
            return null as any; // dummy, for now
        }
    });
}

let r = pipe(['text'], spread()); // Iterable<string>

let r2 = pipe([[1, 2], [3, 4]], spread()); // Iterable<number>

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