Typescript: Accessing property in union of object types fails for properties not defined on all union members

The doc says :

To get the same code working, we’ll need to use a type assertion:

let pet = getSmallPet();

if ((<Fish>pet).swim) {
    (<Fish>pet).swim();
}
else {
    (<Bird>pet).fly();
}

http://www.typescriptlang.org/docs/handbook/advanced-types.html

Then in your sample:

if ((<A>AorB).a) {
    // use AorB.a
}

The issue here is that because B doesn't declare an a property, it might at run-time have an a property of any possible type (because you can assign an object with any set of properties to a B as long as it has a b property of type string). You can make it work explicitly declaring an a: undefined property in B (thus ensuring that B won't have some random a property):

type A = { a: string; } 
type B = { b: string; a: undefined }
type AorB = A | B;

declare const AorB: AorB;

if (AorB.a) {
    // Ok
}

Makes perfect sense. Brain fart.

If you define a: undefined on type B, you'll have to set it to undefined when you create/pass a variable.
To get around that, you can declare a as a?: undefined, and typescript will be happy if you omit it.

If I understand the comments correctly, this should work (but throws; tested on playground): Is this a bug or not? 🤔

type LinkProps = {
    to: string;
    onClick?: undefined;
    // Link specific props:
    target: string;
}

type ButtonProps = {
    to?: undefined;
    onClick: Function;
    // Button specific props:
    disabled: boolean;
}

type ActionProps = LinkProps | ButtonProps;

const Action = (props: ActionProps) =>
    props.to ?
        'Link with target: ' + props.target // Error not on ButtonProps
    :
        'Button with disabled: ' + props.disabled; // Error: not on LinkProps

Action({
  to: 'dssd',
  target: '_blank'
});

I don't understand this at all. Isn't if ((<A>AorB).a) the same as using if (A.a) because you're forcing it back to type A?

If you define a: undefined on type B, you'll have to set it to undefined when you create/pass a variable.
To get around that, you can declare a as a?: undefined, and typescript will be happy if you omit it.

might be better a?: never, now you can not assign undefined by mistake

What if you don't give a name to each of the members of the union? (can't make a type assertion as above without a name for the asserted type, can I?)

type u = "str" | {prop:"val"};
function f(arg:u){return arg.prop} // TypeScript: Property 'prop' does not exist on type 'u'

Any solution?

i find a solution on book :

interface A { x:  number;} 
interface B { y:  string;}
function doStuff ( q: A | B ) {
  if ( 'x'  in q) {
    //  if type A...
  }
  else {
    // if type B...
  }
}

Excerpt From: Basarat Ali Syed. “TypeScript Deep Dive.” Apple Books.

a property on an object and can be used as a type guard, and the TypeScript can know which type you used.

type ColorItemType = {
    colorId: number,
    colorName: string,
}

type NumItemType = {
    numId: number,
    numName: string
}

type ResType = {
    itemId: number,
    // 0 color 1 num
    type: number,
    itemInfo: {
        colorList: Array<ColorItemType>
        numList: Array<NumItemType>
    }
}

const request = () => {
    return [{
        itemId: 1,
        type: 0,
        itemInfo: {
            colorList: [{
                colorId: 1,
                colorName: 'blue'
            }],
            numList: []
        }
    }];
};

const dataSource: Array<ResType> = request();

const formatData = dataSource.map(dataItem => {
    const list: Array<ColorItemType | NumItemType> = dataItem.type === 1 ? dataItem.itemInfo.numList : dataItem.itemInfo.colorList;
    return list.map(listItem => {
        return {
            // An error will be reported here
            value: listItem.numId || listItem.colorId,
            label: listItem.numName || listItem.colorName
        };
    });
});

The issue here is that because B doesn't declare an a property, it might at run-time have an a property of any possible type (because you can assign an object with any set of properties to a B as long as it has a b property of type string). You can make it work explicitly declaring an a: undefined property in B (thus ensuring that B won't have some random a property):

type A = { a: string; } 
type B = { b: string; a: undefined }
type AorB = A | B;

declare const AorB: AorB;

if (AorB.a) {
    // Ok
}

Does not work for me 07.10.2020

And I think that this behavior of TS is not good because it does not help a developer much...