The behavior of the separator operator in TypeScript is puzzling

Highlighted below is the code snippet:

declare function test1<T extends unknown[]>(value: T): T
const res1 = test1([1, 2, 3]) // type is number[]


declare function test2<T extends unknown[]>(value: [...T]): T
const res2 = test2([1, 2, 3]) // type is [number,number,number]

Playground

Upon examination, it's noticeable that by utilizing the ... separator operator, a completely different result type is obtained. This raises questions about whether [...T] holds the same properties as T, given that T inherits an array structure.

Need to consider whether this distinction is a noteworthy trait to contemplate moving forward.

typescript

Answer №1

While not widely known, it's actually a notable feature. For more details, you can refer to this PR discussing Variadic tuple types. One interesting section to look at is Spreads in array literals.

The type [...T], with T being an array-like type parameter, serves as a handy way to hint towards inference of tuple types:

declare function ft1<T extends unknown[]>(t: T): T;
declare function ft2<T extends unknown[]>(t: T): readonly [...T];
declare function ft3<T extends unknown[]>(t: [...T]): T;
declare function ft4<T extends unknown[]>(t: [...T]): readonly [...T];

ft1(['hello', 42]);  // (string | number)[]
ft2(['hello', 42]);  // readonly (string | number)[]
ft3(['hello', 42]);  // [string, number]
ft4(['hello', 42]);  // readonly [string, number]

This feature offers the ability to treat arrays as tuples without needing to specify as const (which enforces read-only status on both the array and its elements).

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