How to compare Enums in TypeScript

Question

In TypeScript, I want to compare two variables containing enum values. Here\'s my minimal code example:

enum E {
  A,
  B
}

let e1: E = E.A
let e2: E = E.B

if (         


        

Answer 1

In typescript an example enum:

enum Example {
   type1,
   type2
};

is transformed to javascript into this object:

Example {
    '0': 'type1', 'type1': 0,
    '1': 'type2', 'type2': 1
}

I had many problems with comparison enums in typescript. This simple script solves the problem:

enum Example {
    type1 = 'type1',
    type2 = 'type2'
};

then in javascript, the object is transformed into:

Example {
    'type1': 'type1',
    'type2': 'type2'
}

If you don't need to use enums - it's better not to use. Typescript has more advanced types, more here: https://www.typescriptlang.org/docs/handbook/advanced-types.html You can use instead:

type Example = 'type1' | 'type2';

Answer 2

I would define values for Enum like this and compare with ===

const enum AnimalInfo {
Tiger = "Tiger",
Lion = "Lion"
}

let tigerStr = "Tiger";

if (tigerStr === AnimalInfo.Tiger) {
  console.log('true');
} else {
  console.log('false');
}

Answer 3

There is another way: if you don't want generated javascript code to be affected in any way, you can use type cast:

let e1: E = E.A
let e2: E = E.B


if (e1 as E === e2 as E) {
  console.log("equal")
}

In general, this is caused by control-flow based type inference. With current typescript implementation, it's turned off whenever function call is involved, so you can also do this:

let id = a => a

let e1: E = id(E.A)
let e2: E = id(E.B)

if (e1 === e2) {
  console.log('equal');
}

The weird thing is, there is still no error if the id function is declared to return precisely the same type as its agument:

function id<T>(t: T): T { return t; }