I write in TypeScript and also use the type-fest
npm package.
In my code, for documentation purposes, I sometimes like to add assertions that two types are indeed equal or unequal. E.g. like the following:
const b: IsEqual<{a: 1}, {a: 1}> = true;
console.log(b);
If the types passed to IsEqual
weren't equal the first line in the above code would fail during type-checking by the TS compiler. The second line is just to get rid of the TS error variable b is defined but not used
.
So far so good.
Now I want to create a small utility that would allow me to assert the equality of two types even more succinctly, like so:
function type_fest_is_equal_assertion<T, S>(): void {
let R: IsEqual<T, S> = true; // *
console.log(R);
}
// with the idea that I can then simply write:
type_fest_is_equal_assertion<{a: number}, {a: number}>()
Unfortunately, the above code fails during compilation at the line marked with the asterisk (*
) with:
Type 'boolean' is not assignable to type 'IsEqual<T, S>'.
Playground is here.
Is there a way to accomplish what I am seeking?
Tangentially, I am also seeking clarification for the above error message in this related question.
The only way this could work is if your generic function constrains one of its type parameters so that IsEqual<T, S>
is known to be true
. This runs the risk of making your constraints illegally circular, so we have to be careful. One approach looks like
function typeFestIsEqualAssertion<
T extends IsEqual<T, S> extends true ? unknown : never,
S extends IsEqual<T, S> extends true ? unknown : never
>(): void {
console.log(true);
}
which luckily compiles without error.
Essentially, both T
and S
are constrained to IsEqual<T, S> extends true ? unknown : never
. When you call the function and specify T
and S
, if IsEqual<T, S>
is true
, then this constraint looks like T extends unknown, S extends unknown
, which will always work because unknown
is a universal supertype in TypeScript. On the other hand, if IsEqual<T, S>
is not true
, then this constraint looks like T extends never, S extends never
, which will almost always fail because never
is the universal subtype in TypeScript. The only type assignable to never
is never
itself. This means you'll get your expected behavior as long as at least one of S
and T
is not specified as never
. And if both are never
then presumably IsEqual<never, never>
would be true
and you'd be happy that it's accepted:
typeFestIsEqualAssertion<{ a: 1 }, { a: 1, b: 2 }>(); // error
typeFestIsEqualAssertion<{ a: 1 }, { a: 1 }>(); // okay
typeFestIsEqualAssertion<{ a: 1 }, { b: 2 }>(); // error
typeFestIsEqualAssertion<{ a: 1 }, never>(); // error
typeFestIsEqualAssertion<never, { a: 1 }>(); // error
typeFestIsEqualAssertion<never, never>(); // okay
Note that the error messages will all say that the specified type doesn't satisfy the constraint never
, which might or might not be acceptable for your use case. If you need something more descriptive you'll need to work around the lack of "invalid types" or "throw types" in TypeScript. Look at microsoft/TypeScript#23689 for a discussion of that.