Typescript & operator

Typescript & operator

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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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edited Nov 23, 2015 at 19:19

Sampson

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asked Nov 23, 2015 at 16:15

ppoliani

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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.

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edited Dec 20, 2021 at 19:32

KyleMit♦

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answered Nov 23, 2015 at 16:21

Sampson

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•  
  Looks like the links to the spec are outdated. Can be found here typescriptlang.org/docs/handbook/… 

  – nick722

   Oct 19, 2020 at 5:56

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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?