How to throw a type error during Typescript type checking

Question

I have 2 types that should never intersect. Is there a way to make the typechecker flag when they do? Ideally, I'd like to do this purely in the types world without declaring any redundant variables.

Example:

type A = 1 | 2 // Must be different from B
type B_OK = 3
type B_FAIL = 2 | 3

// What I want (pseudo Typescript)
type AssertDifferent<X,Y> = Extract<X,Y> extends never ? any : fail // Fail if the types intersect

// Expected result (pseudo Typescript)
AssertDifferent<A,B_OK> // TS is happy
AssertDifferent<A,B_FAIL> // Fails type check

Answer 1

Your best bet is to make your conditional type return true or false, and then try to assign true to the result. Like so:

type A = 1 | 2 // Must be different from B
type B_OK = 3
type B_FAIL = 2 | 3

type AssertTrue<T extends true> = T;

type IsDifferent<X,Y> = Extract<X,Y> extends never ? true : false

type result1 = AssertTrue<IsDifferent<A, B_OK>>; // OK
type result2 = AssertTrue<IsDifferent<A, B_FAIL>>; // Error

You can use the new @ts-expect-error comments feature in version 3.9 on the second line to enforce that the error always throws.