Type predicated should eliminate execution path

[ycmjason]

TypeScript Version: 3.5.1

Search Terms:

• type predicate
• path
• eliminate

Code

type Awesome = 'i' | 'am' | 'awesome'

const isActuallyAwesome = (x: Awesome): x is 'awesome' => x === 'awesome'

const isNotReallyAwesome = (x: Awesome): x is 'i' | 'am' => x !== 'awesome'


const happyFunction = (a: Awesome): number => {
    if (isActuallyAwesome(a)) {
        return 0
    }

    if (isNotReallyAwesome(a)) {
        // typescript not be able to infer `isNotReallyAwesome(a)` always evaluates to true at this point?
        return 1
    }

    throw 'i need to throw :('
}

Expected behavior:
isNotReallyAwesome has type signature: (x: Awesome) => x is 'i' | 'am'. Shouldn't TS be able to tell when x is passed in as 'i' | 'am' and therefore infer that it will always return true?

I hope this make sense.

Actual behavior:

Playground Link:

https://www.typescriptlang.org/play/index.html#code/C4TwDgpgBAgg7hAzgewLbQLxQOQEttQA+OAhqgcdiQiutgFD0DGyAdosFLojE8AK4kANkJDwkaTFAAUADwBcsGpICUi2V0SlldKBgB8UDRhPaJdRi3aduAOWTAAShGGjxtKXMXvV6zTnwiUnI9Qw0AQlMqHQgGSzYOKAALEjAwEAAxflY+XDY9GRJvGLUoVn5UACMIACdQqABveigWrgAzGW5eAVcxGOkSFRVG5taxmogBGtYoAAZRloBfRjHcDuk7B2den3QBoZGxsYB6Y6gWfiEAEyhQSEQmGtwwTmqoEkqhaGBkLlY22qaexOFwiPrmCD7d5COAkEBaCAAN2EgmASFuv2ANX40BInGASW4UDAyFwrGAAH4FkcJlMZgBGanLakEmrIOABMoQCA3H63JJsjnyaQMZlAA

[jack-williams]

TypeScript does infer that the conditional is always true.

const happyFunction = (a: Awesome): number => {
    if (isActuallyAwesome(a)) {
        return 0
    }

    if (isNotReallyAwesome(a)) {
        return 1
    }

    const whenAmIReachable: never = a
    return a;
}

But if you're looking for actual code elimination in the compiled output - that's never going to happen as per:

This could be implemented without emitting different JS based on the types of the expressions

[ycmjason]

i didn't mean code elimination in the compiled code.

I mean that typescript should be able to infer that whatever after the second conditional is dead code.

I should not need to write throw 'i need to throw :('

[ycmjason]

Let me give another example:

const coolFunction = (): number => {
    if (true) {
         return 1
    } 

     // typescript won't complain about the return type should be `number | undefined`
}

so shouldn't the my example have the same behaviour?

[jack-williams]

This issue explains why the literal true is treated differently: #29323.

[jack-williams]

Also covered here: #17358.

[ycmjason]

thanks @jack-williams !