Determining the accurate object from a nested type in Typescript is essential

I'm currently facing a challenge with Typescript in inferring the correct object within a switch case. I find it puzzling why my scenario works as expected with a union type but not with an object.

Here is the code from the playground link:

interface Base {
  id: number;
  name:  "a" | "b" | "c" | "d"; // This is type-guarding correctly
  type: {
    id: number;
    name: "a" | "b" | "c" | "d"; // This is NOT type-guarding correctly
  }
}

// More code snippets omitted for brevity

function test(t: C) {
  // If you switch the switch case to 't.name'
  switch (t.type.name) {
    case "a":
      console.log(t.a)
      break;
    // More cases omitted for brevity
  }
}

The scenario involves one object that can have different properties based on the value of type.name. My aim is to use a switch case to determine the type.name and allow Typescript to infer the accessible properties accordingly.

The issue arises when pinpointing the type.name within the switch case, unlike when using name which functions correctly.

Playground Link

If you exclude using as or unraveling the object, are there any alternative options available?

Answer №1

One limitation of TypeScript is the lack of support for "nested" discriminated unions, where a subproperty is used to discriminate among members. For more information on this issue, refer to microsoft/TypeScript#18758. While type C is considered a discriminated union with name as the discriminant, attempting to use type.name as the discriminant will not work as expected. This eliminates the ability to utilize features designed to handle discriminated unions, such as using a switch statement based on the discriminant property.

If you need the compiler to perform type guarding on subproperties without altering your object structure or resorting to type assertions, one possible solution is to implement a user-defined type guard function. An example of how this can be achieved:

function hasNestedTypeName<
  T extends { type: { name: string } },
  K extends string
>(t: T, k: K): t is
  T extends { type: { name: infer N } }
  ? K extends N ? T : never
  : never {
  return t.type.name === k
}

By calling hasNestedTypeName(t, k) and receiving a true result, the compiler will narrow down t to only those union members whose type.name property matches the provided value k. It's necessary to refactor the test function to use if/else conditions instead of a switch due to the boolean return type of type guard functions. Here is an updated version:

function test(t: C) {
  if (hasNestedTypeName(t, "a")) {
    // function hasNestedTypeName<C, "a">(t: C, k: "a"): t is A
    t // A
    console.log(t.a)
  } else if (hasNestedTypeName(t, "c")) {
    // function hasNestedTypeName<B | Generic, "c">(t: B | Generic, k: "c"): t is Generic
    t // Generic
    console.log(t.name)
  } else if (hasNestedTypeName(t, "b")) {
    t // B 
    console.log(t.b)
  }

This approach offers similar functionality to your previous code, although additional care must be taken when using

hasNestedTypeName(t, "c")
to ensure it behaves correctly. It's advisable to avoid nested discriminated unions with union-based discriminants, as it may lead to unpredictable behavior. Ultimately, whether implementing such changes is worthwhile depends on the specific use case. Refactoring the object structure to have the discriminant at the top level might be a more effective strategy, leveraging TypeScript's strengths rather than trying to work around its limitations.

For a live demonstration of the code in action, check out this Playground link.

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