TypeScript interface with XOR, {bar:string} xor {can:number} [duplicate]

Question

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Closed 3 months ago.

Does Typescript support mutually exclusive types? (4 answers)

How do I say that I want an interface to be one or the other, but not both or neither?

interface IFoo {
    bar: string /*^XOR^*/ can: number;
}

Answer 1

As proposed in this issue, you can use conditional types (introduced in Typescript 2.8) to write a XOR type:

type Without<T, U> = { [P in Exclude<keyof T, keyof U>]?: never };
type XOR<T, U> = (T | U) extends object ? (Without<T, U> & U) | (Without<U, T> & T) : T | U;

And you can use it like so:

type IFoo = XOR<{bar: string;}, {can: number}>;
let test: IFoo;
test = { bar: "test" } // OK
test = { can: 1 } // OK
test = { bar: "test",  can: 1 } // Error
test = {} // Error

Answer 2

You can use union types along with the never type to achieve this:

type IFoo = {
  bar: string; can?: never
} | {
    bar?: never; can: number
  };


let val0: IFoo = { bar: "hello" } // OK only bar
let val1: IFoo = { can: 22 } // OK only can
let val2: IFoo = { bar: "hello",  can: 22 } // Error foo and can
let val3: IFoo = {  } // Error neither foo or can

Answer 3

You can get "one but not the other" with union and optional void type:

type IFoo = {bar: string; can?: void} | {bar?:void; can: number};

However, you have to use --strictNullChecks to prevent having neither.

Answer 4

try this:

type Foo = {
    bar?: void;
    foo: string;
}

type Bar = {
    foo?: void;
    bar: number;
}

type FooBar = Foo | Bar;

// Error: Type 'string' is not assignable to type 'void'
let foobar: FooBar = {
    foo: "1",
    bar: 1
}

// no errors
let foo = {
    foo: "1"
}