4 ms·
Inside of type descriptions, literal values are always treated as literals. The problem only appears when you declare a variable _without_ an explicit type like
by phiresky 7y ago
Inside of type descriptions, literal values are always treated as literals. The problem only appears when you declare a variable _without_ an explicit type like `const foo = {k: 'x'}`, which the compiler then infers to be of type `foo: {k: string}` and not `foo: {k: 'x'}`.
Both of these are valid inferences, and it's impossible to know which one the user wants in the general case, which is why you have to explicitly say
`const foo = {k: 'x' as const}`.
- Dylan16807 7y agoSo the example of `false | 'x' | 'y' | undefined` turning into string is invalid? I feel like I'm missing something.
- phiresky 7y agohere's a full example. If you do this: function test(p: {k: false | 'x' | 'y' | undefined}) { // do stuff } const p = {k: 'x'} // inferred type: {k: string} //... test(p); // type error : Type 'string' is not assignable to type 'false | "x" | "y" | undefined' The problem is that the type of p.k is inferred to `string` not `x`. The compiler can't know this is fine because you could call `foo(p)` in between the declaration and the `test` call, that mutates p.k. If you add `as const` or an explicit type to p it is fine. (I think this is what the GGP was referring to)
- Dylan16807 7y agoThanks!
- Rapzid 7y agoDidn't realize "as const" worked; will have to try that next time.