Arrow style user defined type guard?

Question

The typescript handbook on user defined type guards defines an example type guard as

function isFish(pet: Fish | Bird): pet is Fish {
    return (<Fish>pet).swim !== undefined;
}

Is there a corresponding syntax for arrow functions?

Answer 1

TypeScript supports arrow-function type guards (which may have not been the case in 2017). The following now works as expected. I've used this style frequently in production code:

const isFish = (pet: Fish | Bird): pet is Fish => 
  (pet as Fish).swim !== undefined;

Answer 2

Yes, just as you would have returned boolean, you simply return pet is Fish:

const isFish: (pet: Fish | Bird) => pet is Fish = pet => (pet as Fish).swim !== undefined

Arrow functions are declared as a single type ((pet: Fish | Bird) => pet is Fish) rather than the parameters and the return type separately.

Answer 3

Use type assertion instead of type declaration:

const isFish = (pet => !!pet.swim) as (pet) => pet is Fish 

However, given the fact that it is more verbose, I would prefer to write type guards as normal functions, unless you really need the this binding, but that is probably a code smell.

Answer 4

as an alternative, this also works

const isFish = (pet: any): pet is Fish => !!pet.swim;

or if you prefer to be more explicit

const isFish = (pet: any): pet is Fish => pet.swim === 'function';

while the test can be made against properties as well

const isFish = (pet: any): pet is Fish => pet.hasOwnProperty('genus');