Conditionally make a type not undefined | TypeScript Community | Page 1

jovial lake Jun 7, 2024, 7:43 AM

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How can I make sure my value is not undefined after checking the error, since they can't both be undefined.

function run<V>(fn: () => V): [V | undefined, Error | undefined] {
    let value
    let error
    try {
      value = fn()
    } catch (e) {
      error = e as Error
    }
    return [value, error]
  }
  
  const [value, error] = run(() => 1)
  if (error) return

  value: number | undefined // expected "number"

cyan hollow Jun 7, 2024, 7:51 AM

#

You want a DU.

#

!hb disc

tawdry dragonBOT Jun 7, 2024, 7:51 AM

#

Get Started / TypeScript for Functional Programmers

Discriminated Unions

cyan hollow Jun 7, 2024, 7:51 AM

#

So [V, undefined] | [undefined, Error] would work.

jovial lake Jun 7, 2024, 7:52 AM

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That's what I tried, turns out i just had to cast the result to it as well thanks

cyan hollow Jun 7, 2024, 7:53 AM

#

You don't have to.

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You just need to change your code into:

function run<V>(fn: () => V): [V, undefined] | [undefined, Error] {
    try {
        return [fn(), undefined]
    } catch (e) {
        return [undefined, e as Error]
    }
}

#

(As a side note e as Error is unsafe, since JS allows you to throw anything and not limited to only Errors. But if you think that's a safe assumption to make then sure)

#Conditionally make a type not undefined