What steps should I take to resolve the issue where ""'T' could be instantiated with a different subtype of constraint - ts(2345)""?

I've been researching various resources to find a solution for the error I'm encountering, but haven't found one that fits my specific case.

My goal is to populate dummy rows in my web application so that I can use them as placeholders for the loading state when using MUI Skeleton.

I am aiming to create a function that takes an object with certain properties of a specified type and returns an array with IDs included.

export const generateDummyRows = <T extends { id: string }>(
  count: number,
  record: Omit<T, "id">,
): T[] => Array.from({ length: count }, (_, i) => ({ ...record, id: `${i}` }));

The Array.from expression triggers this type error message:

Type '(Omit<T, "id"> & { id: string; })[]' is not assignable to type 'T[]'.
  Type 'Omit<T, "id"> & { id: string; }' is not a valid assignment for type 'T'. 
    While 'Omit<T, "id"> &a`mT;<string> }' can be assigned to the constraints of type 'T', there might exist another subtype of constraint '{ id: string; }' that could be selected instead. (2322)

Do you have any suggestions on how to rectify this issue?

Here's an example of how to use the function:

const dummyRows = generateDummyRows<Student>(10, {
  firstName: string,
  lastName: string,
  createdAt: Date.now(),
});

UPDATE regarding @jcalz's comment: Initially, I had marked record as optional because I have a record type where all properties are optional except for id. This setup would allow for a call like generateDummyRows<Foo>(3) which would result in

[{id: '0'}, {id: '1'}, {id: '2'}]
.

I am open to alternative suggestions, such as avoiding the use of Omit, but I prefer not to resort to type assertion (as) since it seems to override the issue, as mentioned in @Alexis' response below. Is this correct?

Check out the minimal reproducible example here

Answer №1

The warning received pertains to a significant issue rather than a general one, emphasizing that

Omit<T, "id"> & {id: string}
cannot always be equated to T, particularly when T is constrained by {id: string}. This discrepancy arises from the potential for the id property in T to be narrower than string, leading to possible violations of the call signature with generateDummyRows(). It's essential to consider string literal types and unions of such types:

interface Foo {
    id: "x" | "y"
    bar: string;
}

const dummyRows = generateDummyRows<Foo>(10, { bar: "abc" });
//    ^? const dummyRows: Foo[]

In this scenario, a valid Foo must possess an id value of either "x" or "y", excluding other possibilities like "0" or "1". Despite specifying that generateDummyRows() takes type {bar: string} and yields an array of Foo, there's a notable flaw related to the id property.

dummyRows.forEach(v => ({ x: 0, y: 1 })[v.id].toFixed())

Due to the incorrect interpretation of it as an array of Foo[], issues arise during runtime since ({ x: 0, y: 1 })[v.id] results in undefined.

Refer to Why can't I return a generic 'T' to satisfy a Partial<T>? for further insight into challenges concerning specifics with generic types.


A recommended approach involves avoiding claims about input being Omit<T, "id"> while outputting an array of T. Instead, define T to represent the input (initially categorized as Omit<T, id>), where the resulting array corresponds to T & {id: string}. This adjustment aligns more closely with reality, though exceptions exist if T already features an id property. TypeScript recognizes object literal spread as generating intersections:

const generateDummyRows = <T extends object>(
    count: number,
    record: T,
): (T & { id: string })[] =>
    Array.from({ length: count }, (_, i) => ({ ...record, id: \`\${i}\` }));

This revised formulation compiles error-free, elucidating the operations clearly. By providing explicit return type annotations as clarity aids, the risk of misconstrued outcomes diminishes significantly:

const oops: Foo[] = generateDummyRows(10, { bar: "abc" }); // error!

You can now comfortably implement your original code as follows:

const dummyRows: Student[] = generateDummyRows(10, {
    firstName: 'John',
    lastName: 'Doe',
    createdAt: Date.now(),
});

Note the shift in interpreting

T</code's meaning, eliminating the need for calling <code>generateDummyRows<Student>(⋯)
. If necessary, opting for
generateDummyRows<Omit<Student, "id">>(⋯)
remains viable. Ultimately, leaving out manual specification of the generic type argument proves simpler, focusing instead on annotating the dummyRows variable. Any discrepancies should surface at that point.

Visit playground link for code exploration

Answer №2

One issue with Typescript is that it cannot automatically determine if the spread operator on record combined with the id property will perfectly match type T. A workaround for this is to make record mandatory, include a type assertion (as T) in the new object, and define the return type of the function.

export const createDummyItems = <T extends { id: string }>(
  amount: number,
  record: Omit<T, "id">,
): T[] => 
  Array.from(
    { length: amount }, 
    (_, index): T => ({ ...record, id: `${index}` } as T)
  );

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