Typescript not recognizing if-statement-function-call in type-check [duplicate]

Question

I found this issue in typescript. I have following example:

type AirplaneDTO = {
    color?: string|null
    owner?: string|null
}

type Airplane = {
    color: string
    owner: string
}

function isValidAirplane(airplane: AirplaneDTO): boolean {
    return !!(airplane.color && airplane.owner);
}

function getValidAirplane(airplane: AirplaneDTO): Airplane|undefined {
    if (!isValidAirplane(airplane)) {
        return;
    }
    return {
        color: airplane.color,
        owner: airplane.owner,
    }

}

As you can see, I am converting data from AirplaneDTO to Airplane. I don't want to set Airplane object if some of the fields in AirplaneDTO are null or undefined.

As I am doing this check if all the fields inside are set, TS does not recognize the "isValidAirplane" function-call and gives me following Error: Type 'string | null | undefined' is not assignable to type 'string'. Type 'undefined' is not assignable to type 'string'.(2322)

You can try my code and Error here: https://www.typescriptlang.org/play?ssl=28&ssc=1&pln=1&pc=1#code/C4TwDgpgBAgglgJzAGwIYDsIBEAqB5KAXigG8AoKSqAYwHtlaEB+ALigGdgE50BzAH3QBXZMgpVaAd0zM2nbn0EixAXzJlQkWIhQZoxclRr1Gcrj17jKUmWYWW1ZAGZD01YHFroocdgDVUZDgAE3gkNEwAClQdCIg2MN1MXDwASjYAI1p6CAxSKygECGAhBG8AQnLo2L0AOjoGBCgAMmaoGPC6mwgEVIBuMkcXNw8vKF5igKDQmqiOpPjtTuT8dKWF-ldgiCceCGD8ozgnKEjy3ymQxLjq5YhU1MOjKiKSsoGjNSNX0u9DZ8oDVM7VmEHqJgQABoCkZugg2PM4rU4dDPupHOogA

Is this an known Typescript issue or did I do something wrong here?

If it's already a valid "Airplane", why not return it instead of undefined? — Pointy
– Pointy, Commented Jan 23, 2024 at 15:27
You're close, just make isValidAirplane a type predicate by changing the return type from boolean to airplane is Airplane. More here. — T.J. Crowder
– T.J. Crowder, Commented Jan 23, 2024 at 15:28
Just FWIW, you don't need to create a new object in getValidAirplane if you don't want to. You know the object that was passed in is a valid Airplane after the isValidAirplane call, so you could just return it. — T.J. Crowder
– T.J. Crowder, Commented Jan 23, 2024 at 15:30
@Pointy its not a valid Airplane if one of the fields inside are null or undefined. — Markus B
– Markus B, Commented Jan 23, 2024 at 15:44
@MarkusB - Here's your playground updated with the change I mentioned above. — T.J. Crowder
– T.J. Crowder, Commented Jan 23, 2024 at 15:50

mb21 · Accepted Answer · 2024-01-23 15:27:23Z

0

I would recommend to parse, not validate.

Thus instead of isValidAirplane you have the following function:

function parseAirplane(airplane: AirplaneDTO): Airplane | undefined {
    const { color, owner } = airplane
    if (color && owner) {
      return { color, owner }
    }
}

Instead of returning undefined, you can also return an Error type...

answered Jan 23, 2024 at 15:27

mb21

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3 Comments

Pointy Over a year ago

I don't agree; the word "parse" really means (or, should mean) to extract structure from plain text. Also, either "color" or "owner" could be empty in a valid airplane, for all we know.

mb21 Over a year ago

@Pointy did you read the linked article? (and according to OPs type definitions, a valid Airplane has no optional fields...

Markus B Over a year ago

This is very nice with just two parameters to check. but if there are much more fields and checks I want to make instead of if (color && owner) I want to extract it in an own function... And this is my problem, that TS doesn't recognize that.

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