[simulate-me]

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]

That'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]

For 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]

But 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]

Yes, that's correct. In the GP example, c was const, so the type can be determined at compile time.

[simulate-me]

I 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]

Ah, I'm sorry. You're right of course!