Typescript: How to typecheck against a complex type?

Question

I have some complex types:

type odds: 1 | 3 | 5 | 7 | 9;
type evens: 2 | 4 | 6 | 8 | 0

...and some function which takes those complex types:

function(digit: odds | evens) { ... }

I would like to check which type I'm getting, but none of the following work:

if (digit isntanceof odds) // error: odds refers to a type but is being used as a value
if (typeof digit === ???) // issue: no single value for typeof

.

How can I go about checking if digit is odd using types?

Answer 1

Types don't exist at compiler time so typeof will not work, you need some other type of check to test the type at runtime. Typescript has support for this using a custom type-guards

type odds = 1 | 3 | 5 | 7 | 9;
type evens = 2 | 4 | 6 | 8 | 0;
function isOdd(v: odds | evens) : v is odds {
    return v % 2 != 0
}

declare let n: odds | evens;
withEven(n) // invalid here no check has been performed
withOdd(n) // invalid here no check has been performed
if (isOdd(n)) {
    n // is typed as odd
    withEven(n) // invalid here
    withOdd(n) // valid here we checked it is odd
} else {
    n // is typed as even
    withEven(n) // valid here we checked it is not odd
    withOdd(n) // invalid here
}
function withEven(n: evens) { }
function withOdd(n: odds) { }