TypeScript-All-In-One(A.R)

Literal Type Guard

You can use === / == / !== / != to distinguish between literal values

type TriState = 'yes' | 'no' | 'unknown';

function logOutState(state:TriState) {
  if (state == 'yes') {
    console.log('User selected yes');
  } else if (state == 'no') {
    console.log('User selected no');
  } else {
    console.log('User has not made a selection yet');
  }
}

This even works when you have literal types in a union. You can check the value of a shared property name to discriminate the union e.g.

type Foo = {
  kind: 'foo', // Literal type 
  foo: number
}
type Bar = {
  kind: 'bar', // Literal type 
  bar: number
}

function doStuff(arg: Foo | Bar) {
    if (arg.kind === 'foo') {
        console.log(arg.foo); // OK
        console.log(arg.bar); // Error!
    }
    else {  // MUST BE Bar!
        console.log(arg.foo); // Error!
        console.log(arg.bar); // OK
    }
}