What is the method to access an interface or type alias that has not been explicitly exported in TypeScript type definitions?

Question

What is the method to access an interface or type alias that has not been explicitly exported in TypeScript type definitions?

I am looking to create a new class that inherits from Vinyl. The constructor in the superclass takes a single parameter of type ConstructorOptions.

export default class MarkupVinylFile extends Vinyl {

  public constructor(options: ConstructorOptions) {
    super(options);
  }

  // ...

}

However, I am facing an issue with importing the ConstructorOptions for use as a type annotation. In vinyl/index.d.ts, it is defined as:

interface ConstructorOptions {
    cwd?: string | undefined;
    // ...
}

I am unable to import it using

import Vinyl, { ConstructorOptions } from "vinyl";

. Can anyone suggest a way to import the ConstructorOptions?

📄 Vinyl Type Definitions

typescript typescript-typings

Answer 1

Answer №1

As mentioned by @TobiasS. in the comments of the question, there is an indirect way to refer to this specific type.

In this case, the ConstructorParameters built-in utility type can be utilized:

This utility constructs a tuple or array type based on the types found within a constructor function type.

// To access the "class" from merging variable, interface, and namespace,
// TS targeting the variable using `typeof` becomes necessary
type ConstructorOptions = ConstructorParameters<typeof Vinyl>[0];
//   ^? type ConstructorOptions = ConstructorOptions | undefined

// Alternatively, when more than one argument is involved,
// a rest operator along with the complete tuple could have been used:
export default class MarkupVinylFile extends Vinyl {

  public constructor(...args: ConstructorParameters<typeof Vinyl>) {
    super(...args);
  }

  // ...

}

Playground Link

If your constructor primarily involves calling super(), it can be omitted.

Please note that an error might occur when attempting to extends Vinyl:

All base constructors must have the same return type.(2510)

This happens because the vinyl types define multiple constructors with different return types.

Answer 2

As mentioned by @TobiasS. in the comments of the question, there is an indirect way to refer to this specific type.

In this case, the ConstructorParameters built-in utility type can be utilized:

This utility constructs a tuple or array type based on the types found within a constructor function type.

// To access the "class" from merging variable, interface, and namespace,
// TS targeting the variable using `typeof` becomes necessary
type ConstructorOptions = ConstructorParameters<typeof Vinyl>[0];
//   ^? type ConstructorOptions = ConstructorOptions | undefined

// Alternatively, when more than one argument is involved,
// a rest operator along with the complete tuple could have been used:
export default class MarkupVinylFile extends Vinyl {

  public constructor(...args: ConstructorParameters<typeof Vinyl>) {
    super(...args);
  }

  // ...

}

Playground Link

If your constructor primarily involves calling super(), it can be omitted.

Please note that an error might occur when attempting to extends Vinyl:

All base constructors must have the same return type.(2510)

This happens because the vinyl types define multiple constructors with different return types.

What is the method to access an interface or type alias that has not been explicitly exported in TypeScript type definitions?

Answer №1

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