When TypeScript's Exclude<UnionOfTypes, Interface> is used, the resulting type is always "never"

Question

When TypeScript's Exclude<UnionOfTypes, Interface> is used, the resulting type is always "never"

What causes Exclude<A,B> to resolve to the never type in the code snippet below? Shouldn't the typescript compiler be able to infer (through static analysis) that A and B are extending Parent, leading to Exclude<Choices, Parent> resolving to type C?

interface Parent {}

interface A extends Parent {}
interface B extends Parent {}
interface C {}

type Choices = A | B | C

type Test = Exclude<Choices, Parent> // = type "never"???

const c: C = {}
const d: Test = c // Type 'C' is not assignable to type 'never'

One possible workaround is to manually set Parent = A | B, but the necessity of this step remains unclear.

typescript typing union-types

Answer 1

Answer №1

The reason behind this is due to TypeScript's duck typing. In this case, because both C and Parent are of the same interface, C can be assigned to Parent.

This example illustrates this concept:

const c: C = {};
const p: Parent = c;

Even though C doesn't explicitly declare that it extends Parent, TypeScript still recognizes C as a type of Parent.

If you want to ensure this behavior, simply include a property in Parent that is not present in C.

interface Parent { foo: string }

interface A extends Parent {}
interface B extends Parent {}
interface C {}

type Choices = A | B | C

type Test = Exclude<Choices, Parent> // = type C

const c: C = {}
const d: Test = c // works!

Answer 2

The reason behind this is due to TypeScript's duck typing. In this case, because both C and Parent are of the same interface, C can be assigned to Parent.

This example illustrates this concept:

const c: C = {};
const p: Parent = c;

Even though C doesn't explicitly declare that it extends Parent, TypeScript still recognizes C as a type of Parent.

If you want to ensure this behavior, simply include a property in Parent that is not present in C.

interface Parent { foo: string }

interface A extends Parent {}
interface B extends Parent {}
interface C {}

type Choices = A | B | C

type Test = Exclude<Choices, Parent> // = type C

const c: C = {}
const d: Test = c // works!

Answer 3

Answer №2

I have developed a unique type tool in the following manner

// This tool functions as the inverse of [exclude].
type Contains<T, U> = U extends T ? U : never;

Here is an example of how to use it

    type X = {
      x : number;
      xx:number;
      xxx:number;
    }

    type Y = {
      y : number;
    }

    type Z = {
      z : number;
    }

    type W = {
      w : number;
    }


    // In this scenario, we are checking for the type X
    type Element_ContainsX = Contains<{
      x : number;
    }, X|Y|Z|W>

Answer 4

I have developed a unique type tool in the following manner

// This tool functions as the inverse of [exclude].
type Contains<T, U> = U extends T ? U : never;

Here is an example of how to use it

    type X = {
      x : number;
      xx:number;
      xxx:number;
    }

    type Y = {
      y : number;
    }

    type Z = {
      z : number;
    }

    type W = {
      w : number;
    }


    // In this scenario, we are checking for the type X
    type Element_ContainsX = Contains<{
      x : number;
    }, X|Y|Z|W>

When TypeScript's Exclude<UnionOfTypes, Interface> is used, the resulting type is always "never"

Answer №1

Answer №2

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