TS1338: 'infer' declarations are only permitted in the 'extends' clause of a conditional type

TypeScript is a powerful programming language that builds on JavaScript, adding static types and other features that help developers create robust applications. Types in TypeScript provide a way to define the shape of data, enabling better tooling, error checking, and documentation. By using TypeScript, you can catch errors before they become actual bugs in your code, making your development process much smoother.

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What is a Superset Language?

A superset language is a language that includes all the features of another language, along with additional features. JavaScript is the foundation that TypeScript extends upon. This means any valid JavaScript code is also valid TypeScript code. TypeScript adds optional static typing, interfaces, enums, generics, and more, which enhance the developer experience without losing the power of JavaScript.

Understanding TS1338: 'infer' declarations are only permitted in the 'extends' clause of a conditional type

The TypeScript error TS1338 arises when an 'infer' declaration is used improperly outside the context of an extends clause within a conditional type. The 'infer' keyword allows you to create a type that extracts information from another type when used correctly.

Example of TS1338 Error

Consider the following code snippet:

type ExtractType<T> = T extends infer U ? U : never;

const value: ExtractType<string> = "Hello"; // Error: TS1338

This code will result in the TS1338 error. TypeScript is informing us that we cannot use the 'infer' declaration outside of the extends clause.

How to Fix the Error

To correctly use 'infer', you need to define it within the extends clause of a conditional type. Here's how you can modify the above example:

type ExtractType<T> = T extends infer U ? U : never;

type Result = ExtractType<string>; // Now Result is of type string

In the corrected version, we properly define U within the context of the extends clause, which resolves the TS1338 issue.

Important to Know!

1. The infer Keyword: 'Infer' is used to capture a type within a conditional type. You can’t just use it anywhere; it must be part of a conditional type declaration.

2. Conditional Types: They are defined with the syntax A extends B ? C : D, allowing different types to be selected based on conditions.

More Code Examples

Here are a couple of more examples that illustrate correct usage of infer:

Valid Usage of infer

type ReturnType<T> = T extends (...args: any[]) => infer R ? R : never;

type ExampleFunction = () => number;
type ResultType = ReturnType<ExampleFunction>; // ResultType is number

In this code, ReturnType uses infer correctly within the extends clause, allowing it to extract the return type of a function.

FAQ Section

Q: What is the purpose of TypeScript?

A: TypeScript adds static types to JavaScript, which helps in catching errors during development rather than at runtime.

Q: Can I use infer outside of the extends clause?

A: No, you cannot. As per the TS1338 error message, 'infer' declarations are only permitted in the 'extends' clause of a conditional type.

Important Things to Know

• Always define infer within a conditional type's extends clause to avoid TS1338 errors.
• Familiarize yourself with conditional types to leverage TypeScript's capabilities effectively.
• Use code snippets and examples to practice the valid uses of infer and other TypeScript features.

Conclusion

Understanding and resolving the TS1338 error is crucial for effective TypeScript development. Remember, TS1338: 'infer' declarations are only permitted in the 'extends' clause of a conditional type. By ensuring that you properly utilize the infer keyword, you can make the most of TypeScript's type system and improve your code quality. Happy coding!