Typescript type inference not recognizing value must be defined

Question

This might not be possible in typescript but it seems like the TS compiler should be able to figure this out some way. Example here


function validateFoo(val: string | undefined) {
    return val !== undefined ? true : false
}

let myVal: string | undefined

if (validateFoo(myVal)) {
    console.log("output: ", myVal.toLowerCase()) // Error: Object is possibly 'undefined'.
}

it seems like inside the if statement that myVal must not be undefined. However typescript says it may be. Is there some way to annotate that function similar maybe to:

if (!myVal) {
  throw new Error("myVal undefined")
}

console.log(myVal.toLowerCase())

would mark myVal as defined after the throw.

Federkun · Accepted Answer · 2021-02-06 00:09:57Z

2

That's what type guards does:

function validateFoo(val: string | undefined): val is string {

https://www.typescriptlang.org/docs/handbook/advanced-types.html#user-defined-type-guards

answered Feb 6, 2021 at 0:09

Federkun

37.2k8 gold badges82 silver badges93 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Jared Smith Over a year ago

Strikes again

john_ryan Over a year ago

Thanks for this - i didn't really know how to phrase the question to search the docs

john_ryan Over a year ago

@JaredSmith or @Federkun to avoid another duplicate question - using guard makes the type never in the else is there a way to annotate that? typescriptlang.org/play?#code/…

Collectives™ on Stack Overflow

Typescript type inference not recognizing value must be defined

1 Answer 1

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related