3 replies

Creating a Safe Discriminated Union in TypeScript Without Failing on Unimplemented Types

Checking if I'm missing something. Wanting a discriminated union over

type

type

type

type, however I do not want to fail by encountering a

type

type

type

type I did not implement yet. Is this the way to express?

const E = S.union(
  S.struct({type: S.literal('foo'), data: S.string}),
  S.struct({type: S.literal('bar'), data: S.number}),
  S.struct({type: S.string})
)

const E = S.union(
  S.struct({type: S.literal('foo'), data: S.string}),
  S.struct({type: S.literal('bar'), data: S.number}),
  S.struct({type: S.string})
)

const E = S.union(
  S.struct({type: S.literal('foo'), data: S.string}),
  S.struct({type: S.literal('bar'), data: S.number}),
  S.struct({type: S.string})
)

const E = S.union(
  S.struct({type: S.literal('foo'), data: S.string}),
  S.struct({type: S.literal('bar'), data: S.number}),
  S.struct({type: S.string})
)

Creating a Safe Discriminated Union in TypeScript Without Failing on Unimplemented Types

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