How to access the difference member for merged interface in typescript?

Consider we have two interface A and B, how can开发者_开发问答 I get the difference member for merged interface?

interface A {
    from: string,
    items: number[],
}

interface B {
    to: string,
    items: number[],
}

type C = A | B;

function foo(fromOrTo: C) {
    const result = fromOrTo.from || fromOrTo.to
    // Get error: "from" or "to" don't exist in C
}

I know another way to implement this:

interface C {
    from?: string,
    to?: string,
    items: number[],
}

function foo(fromOrTo: C) {
    const result = fromOrTo.from || fromOrTo.to;
    //  This is a one of solutions.
}

You can use a type guard to check for the existence of the property you are interested in, and maintain type-safety downstream

interface A {
    from: string,
    items: number[],
}

interface B {
    to: string,
    items: number[],
}

type C = A | B;

function isFrom(fromOrTo: C): fromOrTo is A {
    return (fromOrTo as A).from !== undefined
}
function isTo(fromOrTo: C): fromOrTo is B {
    return (fromOrTo as B).to !== undefined
}

function someFn() {
    if (isFrom(fromOrTo)) {
        fromOrTo // Type 'A'
    }
    else {
        fromOrTo // Type 'B'
    }
}

function foo(fromOrTo: C) {
    const result = isFrom(fromOrTo) ? fromOrTo.from : fromOrTo.to
}

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