I'm struggling to find the definition of the & operator in TypeScript. I have recently come across the following code:
type IRecord<T> = T & TypedMap<T>;
What does that operator do, and how is it different from the union type |?
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2 Answers
This looks like it's from the Intersection Types portion of the Language Specification. Specifically, the & is an intersection type literal. As for what it does:
Intersection types represent values that simultaneously have multiple types. A value of an intersection type A & B is a value that is both of type A and type B. Intersection types are written using intersection type literals (section 3.8.7).
The spec goes on to offer a helpful snippet to better understand the behavior:
interface A { a: number }
interface B { b: number }
var ab: A & B = { a: 1, b: 1 };
var a: A = ab; // A & B assignable to A
var b: B = ab; // A & B assignable to B
Because ab is both of type A and of type B, we can assign it to a and/or b. If ab were only of type B, we could only assign it to b.
The code you shared may be from this comment on GitHub, which mentions Intersection Types.
1 Comment
nick722
Looks like the links to the spec are outdated. Can be found here typescriptlang.org/docs/handbook/…
Worth noting that if you'd prefer to use interfaces over types (although they're largely similar) that you can typically use interface extension instead of type intersection like this:
// base type
interface Shape {
color: string;
}
// extension
interface Square extends Shape {
sideLength: number;
}
// intersection
type Square = Shape & {
sideLength: number;
}
See Also: Difference between extending and intersecting interfaces in TypeScript?