How to create extended type in a TypeScript?

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I want to show two seemingly identical situations, but they behave differently

1)

type User = { name: string, age: number };
const user1 = { name: 'Harry', age: 21, skills: 'backend', isAdmin: true }
const user2: User = user1; // NO ERROR

In this situation, TypeScript allows you to add additional properties to the object

2)

type User = { name: string, age: number };
const user2: User = { name: 'Alex', age: 23, skills: 'frontend' } // ERROR!

And in this case we will see an error, although the situations are quite similar

With this in mind, I have 2 questions:

1) How to explain the difference in the behaviors above?

2) How do I declare a type that will cause the object to AT LEAST have all the necessary properties from the User type, and the rest of the properties can be completely different?

typescript
2 Answers
  1. How to explain the difference in the behaviors above?

In the first line you are creating an anonymous object. So no type checking is applied. In the second line you are specifically telling typescript that your object is of type User, so compile-time check was applied.

  1. How do I declare a type that will cause the object to AT LEAST have all the necessary properties from the User type, and the rest of the properties can be completely different?

You can take help of Intersection typescript types. something like this

type User = { name: string, age: number };
type NewUser = User & {[key:string]:any}
const user1: NewUser = { name: 'Harry', age: 21, skills: 'backend', isAdmin: true }
user_10326550

For the second question, you can use

type User = { name: string; age: number; [key: string]: any };
user_181232720
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