TypeScript const assertions: how to use Array.prototype.includes?

The standard library signature for Array<T>.includes(u) assumes that the value to be checked is of the same or narrower type than the array’s elements T. But in your case you are doing the opposite, checking against a value which is of a wider type. In fact, the only time you would say that Array<T>.includes<U>(x: U) is a mistake and must be prohibited is if there is no overlap between T and U (i.e., when T & U is never).

Now, if you’re not going to be doing this sort of “opposite” use of includes() very often, and you want zero runtime efects, you should just widen CAPITAL_LETTERS to ReadonlyArray<string> via type assertion:

(CAPITAL_LETTERS as ReadonlyArray<string>).includes(str); // okay

If, on the other hand, you feel seriously enough that this use of includes() should be accepted with no type assertions, and you want it to happen in all of your code, you could merge in a custom declaration:

// global augmentation needed if your code is in a module
// if your code is not in a module, get rid of "declare global":
declare global { 
  interface ReadonlyArray<T> {
    includes<U>(x: U & ((T & U) extends never ? never : unknown)): boolean;

That will make it so that an array (well, a readonly array, but that’s what you have in this example) will allow any parameter for .includes() as long as there is some overlap between the array element type and the parameter type. Since string & CapitalLetter is not never, it will allow the call. It will still forbid CAPITAL_LETTERS.includes(123), though.

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