What happens when you create a Union type alias between two connected interfaces?

Recently, I encountered a peculiar situation that left me puzzled as I scoured the internet and documentation for answers but found none.

Let's delve into the following scenario:

interface A {
    foo: string;
    bar: string;
}

interface B extends A {
    baz: string;
}

export type C = B | A;

In my environment, utilizing type C mentioned above seems to default to type A, even when it is explicitly specified elsewhere. This has led me to question whether this behavior stems from the TypeScript compiler or the IDE favoring the base interface (A in this instance).

Consider the following code snippet as an example:

const fooBar = (props: C) => {
    const {
        foo, // valid
        bar, // valid
        baz // IDE displays an error as 'baz' is not recognized in 'A'
    } = props // requires casting with "as B"

    return(
        {baz ? baz : ""} // display 'baz' if it exists
    );
}

Answer №1

In the scenario where X represents a specific concept and Y extends X, combining Y | X essentially results in X. In other words, the amalgamation of X and Y, constructing the minimal type that includes both elements, simplifies down to X due to the fact that every instance of Y is inherently an X.

When accepting a variable with the type Y | X, it essentially needs to fulfill the requirements of X alone. If there's a need to mandate that a type adheres to both Y and X collectively, then utilizing Y & X, known as the intersection type, would be appropriate. However, given the existing subtyping dynamics, Y & X effectively boils down to Y.

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