What is the best way to create an assertion function for validating a discriminated union type in my code?

Question

What is the best way to create an assertion function for validating a discriminated union type in my code?

I have a union type with discriminated properties:

type Status =
    { tag: "Active", /* other props */ }
  | { tag: "Inactive", /* other props */ }

Currently, I need to execute certain code only when in a specific state:

// At some point
const status: Status = …;

// Later on
if (status.tag !== "Active") {
  throw new AssertionError(…);
}

// Now I can be confident that `status` is of type "Active"

Is it possible to refactor the condition into an assert function?

// Somewhere
const assertStatus = (status: Status, tag: Status["tag"]) => ???

// And later:
assertStatus(status, "Active");

// Here I am sure `status` is of type "Active"

typescript

Answer 1

Answer №1

To achieve this, one can utilize assertion functions.

type State =
    { tag: "A", /* other properties */ }
  | { tag: "B", /* other properties */ }

function assertState (condition: boolean): asserts condition { 
  if (!condition) {
    throw new Error("message")
  }
}

function test(){
  let state: State = {} as any

  assertState(state.tag === 'A')

  state.tag // now only has type `A`
}

Playground

Answer 2

To achieve this, one can utilize assertion functions.

type State =
    { tag: "A", /* other properties */ }
  | { tag: "B", /* other properties */ }

function assertState (condition: boolean): asserts condition { 
  if (!condition) {
    throw new Error("message")
  }
}

function test(){
  let state: State = {} as any

  assertState(state.tag === 'A')

  state.tag // now only has type `A`
}

Playground

What is the best way to create an assertion function for validating a discriminated union type in my code?

Answer №1

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