How do I use an existing type as the argument types of a function?

When working with TypeScript, it’s common to define custom types to represent specific data structures or objects. These types can then be used throughout your codebase to ensure type safety and improve code readability. However, there may be scenarios where you want to reuse an existing type as the argument types of a function. In this blog post, we will explore different solutions to achieve this.

Solution 1: Using the existing type directly

The simplest way to use an existing type as the argument types of a function is to reference the type directly in the function signature. Let’s say we have a type called Person that represents a person’s information:

type Person = {
  name: string;
  age: number;
  email: string;
};

We can then define a function that takes two Person objects as arguments:

function comparePeople(person1: Person, person2: Person) {
  // Function logic here
}

This way, the comparePeople function expects two arguments of type Person and will enforce type checking accordingly.

Solution 2: Creating a type alias

If you find yourself reusing the same type as argument types in multiple functions, you can create a type alias to simplify the code. Using the Person type from the previous example, we can create a type alias called PersonComparison:

type PersonComparison = (person1: Person, person2: Person) => void;

Now, instead of specifying the argument types explicitly in each function, you can use the PersonComparison type alias:

const comparePeople: PersonComparison = (person1, person2) => {
  // Function logic here
};

This approach improves code readability and makes it easier to update the argument types if needed.

Solution 3: Using an interface

If you prefer using interfaces instead of types, you can achieve the same result by defining an interface with the desired argument types. Let’s create an interface called PersonComparison:

interface PersonComparison {
  (person1: Person, person2: Person): void;
}

Now, you can use the PersonComparison interface to define your function:

const comparePeople: PersonComparison = (person1, person2) => {
  // Function logic here
};

This approach provides a more object-oriented approach and allows you to extend the interface if needed.

These are three different solutions to use an existing type as the argument types of a function in TypeScript. Choose the one that best fits your coding style and project requirements. Happy coding!

How do I use an existing type as the argument types of a function?