What is the process for combining an object and a primitive type to create a union type?

Question

What is the process for combining an object and a primitive type to create a union type?

Having a tricky time with Typescript and finding the correct typing for my variable. What seems to be the issue?

The variable selected in my code can either be of type DistanceSplit or number. I have an array that looks like this:

[-100, DistanceSplit, DistanceSplit, DistanceSplit, -100]

While everything works fine when selected is of type any, I am now aiming to remove all instances of any from my code for better clarity. However, doing so leads to some compilation errors.

Specifically, errors like

Type 'number' is not assignable to type 'DistanceSplit'

or

selected.name is not assigned to type number

start popping up.

I understand the reason behind these errors, but my question is: how do I go about resolving them in Typescript?

export const TRANSPONDER_Distance = -100;

export interface DistanceSplit {
    ID: number;
    name:string;
    content:string
}

// working
public selected: any;

// not working
public selected: DistanceSplit | number;


function() {
    this.selected.name = "bla";  // problem
}

Your assistance is much appreciated!

typescript typescript-typings

Answer 1

Answer №1

Ensure that your selected variable is properly typed in your code. It is important to validate whether the selection is a number or a DistanceSplit within your function.

To inform TypeScript that selected should be treated as type DistanceSplit before assigning the name property, utilize a Type Guard.

For your specific scenario, you can achieve this as follows:

function() {
  if (typeof this.selected !== 'number') {
    this.selected.name = 'foo';
  }
}

This approach instructs TypeScript to interpret selected as a DistanceSplit when it is not a number.

If more precise validation is required, consider implementing a type guard function like so:

function isDistanceSplit(selected: number | DistanceSplit): selected is DistanceSplit {
  // Implement necessary logic to verify that selected is indeed of type DistanceSplit.
  return (typeof selected !== 'number');
}

You can now incorporate this function into your code like this:

function() {
  if (isDistanceSplit(this.selected)) {
    this.selected.name = 'foo';
  }
}

Answer 2

Ensure that your selected variable is properly typed in your code. It is important to validate whether the selection is a number or a DistanceSplit within your function.

To inform TypeScript that selected should be treated as type DistanceSplit before assigning the name property, utilize a Type Guard.

For your specific scenario, you can achieve this as follows:

function() {
  if (typeof this.selected !== 'number') {
    this.selected.name = 'foo';
  }
}

This approach instructs TypeScript to interpret selected as a DistanceSplit when it is not a number.

If more precise validation is required, consider implementing a type guard function like so:

function isDistanceSplit(selected: number | DistanceSplit): selected is DistanceSplit {
  // Implement necessary logic to verify that selected is indeed of type DistanceSplit.
  return (typeof selected !== 'number');
}

You can now incorporate this function into your code like this:

function() {
  if (isDistanceSplit(this.selected)) {
    this.selected.name = 'foo';
  }
}

Answer 3

Answer №2

To effectively manage the typings of a union type, it is essential to differentiate between when it represents a number and when it represents a DistanceSplit type.

One way to accomplish this is by explicitly defining the typings as shown below:

function() {
    (this.selected as DistanceSplit).name = "bla";
}

For further insights, you can refer to https://www.typescriptlang.org/docs/handbook/advanced-types.html#type-guards-and-differentiating-types

Answer 4

To effectively manage the typings of a union type, it is essential to differentiate between when it represents a number and when it represents a DistanceSplit type.

One way to accomplish this is by explicitly defining the typings as shown below:

function() {
    (this.selected as DistanceSplit).name = "bla";
}

For further insights, you can refer to https://www.typescriptlang.org/docs/handbook/advanced-types.html#type-guards-and-differentiating-types

What is the process for combining an object and a primitive type to create a union type?

Answer №1

Answer №2

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