I'm currently diving into Type Relations in TypeScript. Can someone help explain what happens when we intersect the types represented by these two expressions:
{a:number}[] & {b:string}[]
Does this result in
{a:number, b:string}[]
? Any clarification is appreciated.
Indeed, the documentation does not provide a clear explanation of operator precedence in the type definition. However, one can easily test it:
type Z = { a: number }[] & { b: string }[];
let a: Z = [];
a = [{ a: 1, b: "q" }]; // This assignment is invalid because `b` does not exist in `{ a: number }`.
a = [{ a: 1 }]; // This assignment is also invalid because `a` does not exist in `{ b: string }`.
a = [{ b: "foo" }]; // Another invalid assignment due to `b` not existing in `{ a: number }`.
a = []; // This assignment is valid.
let x: { a: number, b: string }[] = a; // This assignment is also invalid.
The results show that the assumption it resolves to { a:number,b:string } is incorrect. Initially, I assumed it would resolve to
({ a: number } | { b: string })[]The intersection denotes "an array of elements where each element satisfies the type { a: number } and simultaneously satisfies the type { b: string }. Since a is absent in the second type and b is missing in the first, no element can fulfill the intersection. Therefore, only empty arrays can be assigned to variables typed with the intersection.
To achieve the expected outcome, a different intersection is required:
type Z2 = { a: number, b?: string }[] & { a?: number, b: string }[];
let a2: Z2 = [];
a2 = [{ a: 1, b: "q" }];
This will work as anticipated because the first type in the intersection allows elements that may optionally have b, while the second type allows elements that may optionally have a.
declare type T1={a:number}[];
declare type T2={b:string}[];
declare type common =T1&T2;
let x:common;
let y=[{a:1,b:'2'}];
x=y;
therefore, your answer is accurate!
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