Combination of two generic interfaces creates a union type

I have been diving into the world of typescript and I encountered a challenge with the syntax of union types, specifically when using a generic interface:

interface ArrayElementError {
  kind: 'failure'
  reason: string
}

interface ArrayElementSuccess<T> {
  kind: 'success'
  value: T
}

type ArrayElement = ArrayElementError | ArrayElementSuccess
// or interface ArrayElement<T extends ArrayEmementError | ArrayElementSuccess> {}

When I tried to compile the code above, I received an error message:

TSError: ⨯ Unable to compile TypeScript:

Could someone please guide me on the correct syntax for extending two generic interfaces (or creating a new type using the type keyword)?

Answer №1

The recommended approach would be:

type ArrayElementType<T> = ArrayElementError | ArrayElementSuccess<T>;

Here, the ArrayElementType is a generic type with type parameter T, which can be assigned as the type argument for ArrayElementSuccess.

Check out this Playground link to code

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