Interface type check with Typescript

Matheus Mello
Matheus Mello
September 2, 2023
Cover Image for Interface type check with Typescript

📝 Interface Type Check with TypeScript: Clarifying the Mystery!

Are you struggling to find out if a variable of type any implements an interface in TypeScript? 🤔 Don't worry, you're not alone! Many developers face this issue and find it perplexing. But fear not, in this blog post, we'll dive deep into this problem, provide easy solutions, and reveal a hidden TypeScript gem! So, let's get started! 💪

🔍 Understanding the Problem

Before we get into the solutions, let's analyze the issue at hand. The code snippet you provided seems fine on the surface, but TypeScript raises an error, claiming that "The name A does not exist in the current scope." 😱 However, the name A does exist in the current scope, so what's going on here? Let's dig deeper!

When you enter the code in the TypeScript playground, it transpiles to JavaScript, and things start to make sense. The generated JavaScript code doesn't retain any representation of A as an interface. Therefore, runtime type checks for interfaces are not possible. 😟

💡 Easy Solutions

  1. Solution 1: Use a Class

Since JavaScript has no concept of interfaces, one option is to use a class instead. However, this may cause the problem of not being able to create instances with object literals. But worry not, as we have other solutions to explore! 😉

  1. Solution 2: implements Keyword

Here comes the hidden TypeScript gem! 🙌 TypeScript actually offers an implements keyword. You mentioned the TypeScript playground's autocompletion revealed this method. Let's see how you can use it to overcome this issue.

interface A {
    member: string;
}

class AImplementation {
    member: string = "foobar";
}

var a: A = new AImplementation();

if (a instanceof A) {
    alert(a.member);
}

In this example, we create a new class, AImplementation, that implements the A interface. We then create an instance of AImplementation, assigning it to the variable a of type A. Finally, using the instanceof operator, we can perform the desired type check.

📢 Call-to-Action: Engage and Share!

Now that you have learned how to tackle the interface type check issue in TypeScript, it's time to put your knowledge into practice! Share this blog post with your developer friends and colleagues who might be struggling with the same problem. Let's spread the word and help fellow developers overcome this stumbling block! 💪🚀

Have you encountered any other TypeScript conundrums? Let us know in the comments section below. Let's foster a community of shared knowledge and provide solutions to all the TypeScript mysteries out there! 🔍✨

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