Typescript: create a custom type using a variable's value

Issue

I'm new to TS and I would like to create a custom type

I want to have a custom type (Numbers_2) that accepts some numbers plus the value stored in some variable (n1) and have it work the same way Numbers_1 would

let n1 = 40;
type Numbers_1 = 10 | 20 | 30 | 40;
type Numbers_2 = 10 | 20 | 30 | n1; // 'n1' refers to a value, but is being used as a type here. Did you mean 'typeof n1'?
let n2: Numbers_2 = 1230; // This line should signal a problem

Help much appreciated!

Solution

You can, by using TypeScript's typeof type-operator (which is not the same thing as JavaScript's typeof operator), but you then also need to change let n1 to const n1 because when n1 is mutable the typeof n1 is number instead of the literal value.

Like so:

const n1 = 40;
type Numbers_1 = 10 | 20 | 30 | 40;
type Numbers_2 = 10 | 20 | 30 | typeof n1;
let n2: Numbers_2 = 1230;

...which results in an error on line 4, which is what your post expects:

Type '1230' is not assignable to type 'Numbers_2'.

---

If you want or need n1 to be mutable, then what you're asking is impossible, by way of demonstration:

let n1 = parseInt( prompt( "What is the answer to the Ultimate Question of Life, The Universe, and Everything?" ), 10 );

type Numbers_1 = 10 | 20 | 30 | 40;
type Numbers_2 = 10 | 20 | 30 | typeof n1; // `n1: number`

let n2: Numbers_2 = 1230; // OK

typeof n1 has to be number because parseInt returns number - which in-turn means that type Numbers_2 is 10 | 20 | 30 | number which means that let n2: Numbers_2 = 1230; is valid and okay.

If you need a number-typed "input" variable but still want Numbers_2 to be constrained to a small set of known, possible values then you'll need to explicitly narrow it, like so:

type TheAnswer = 42;
type Numbers_2 = 10 | 20 | 30 | TheAnswer;

function isNumbers2( x: number ): x is Numbers_2 {
    switch( number ) {
    case 10:
    case 20:
    case 30:
    case 42:
        return true;
    default:
        return false;
    }
}

//

// `n1` has type `number`
let n1 = parseInt( prompt( "What is the answer to the Ultimate Question of Life, The Universe, and Everything?" ), 10 );

if( isNumbers2( n1 ) ) {
    // Only within the scope of this `if` statement, `n1` has type `Numbers_2`
}

Answered By - Dai