Trouble arises from the object data type not being properly acknowledged in TypeScript

In the code snippet provided, I am facing a challenge where I need to pass data to an if block with two different types. These types are handled separately in the if block. How can I make TypeScript understand that the selected object could be either of type CustomCar or CustomPlane? I want to avoid having separate actionGuard functions for each type and instead have one function that can handle both cases simultaneously. Currently, when I try to use the selected object in the if condition, TypeScript throws an error saying that selected could be either type, each with different keys and values, all without having to resort to casting.

import { CustomCar } from '../app/carSlice';
import { CustomPlane } from '../app/planeSlice';


export const actionGuard = (
  selected: CustomCar | CustomPlane 
  check: CustomCar[] | CustomPlane[] 
  term: string,
  dispatch: any,
  SetSnackBarMsg: any,
  action: string
) => {
  if (term === 'CustomCar' ) {
    if (array.includes((selected as CustomCar){
     // some logic 
    }
   
    if (term === 'CustomPlane') {
    if (array.includes((selected as CustomPlane ){
    // some logic 
    }
};

Answer №1

While it may seem necessary to ensure that the value passed into .includes matches the type of the array item, consider asserting the array itself as unknown[] instead. This way, you have the flexibility to pass any value into its .includes method.

if ((array as unknown[]).includes((selected){
  // execute some logic 
}

It's unfortunate that a type assertion (using as) is needed due to TypeScript's peculiar design decision with .includes. For more information on this issue, refer to this GitHub thread.

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