5 ms·
Another very common case in TypeScript (and other languages) is union types. Other languages have constructs for checking the underlying type of a union type di
by simulate-me 5y ago
Another very common case in TypeScript (and other languages) is union types. Other languages have constructs for checking the underlying type of a union type directly, but typescript is smart enough to figure it out based off of the properties of the constituent types and what you've checked so far in your code.
e.g.
interface A { type: 'a' }
interface B { type: 'b' }
type C = A | B;
const c: C = ...;
if (c.type === 'a') { /* c is of type A here */ }
- jakelazaroff 5y agoThat's basically case 1, right? The compiler knows which type is being used at each call site, so it can generate a separate function for each type in the union and eliminate the type check/dead branches. I guess the exception would be if you have a non-homogenous array that you try to map over. In that case there's probably no way around boxing the values.
- xg15 5y agoFor consts and function arguments yes, I think. But you could have some mutable variable or field whose value depends on runtime state. In that case, you could have an actual polymorphic type.
- epolanski 5y agoBut the previous example isn't that case. if (c.type === 'a') { /* c is of type A here */ } This is dynamic typing, this code is checked at runtime, and it's leveraged statically.
- xg15 5y agoYes, that's correct. In the GP example, c was const, so the type can be determined at compile time.
- simulate-me 5y agoI was writing TypeScript, where const just means the variable is not reassigned. For instance this is valid: const c: C = Math.random() < 0.5 ? { type: 'a' } : { type: 'b' }.
- xg15 5y agoAh, I'm sorry. You're right of course!
- deleted 5y ago[deleted]