How do I make `foo[1] ==1; foo[1][2]==3; foo[1][2][3]==6` work in JavaScript?

Question

How do I define foo to make the following code work correctly as expected (in JavaScript)?

foo[1] + 1             // 2
foo[1][2] + 1          // 4
foo[10][20][30] + 1    // 61
foo[100][200][300] + 1 // 601

This is an interview question I once met.

That a proxy is allowed is Valuable Information. – Dave Newton Apr 28 '22 at 02:20 — Dave Newton, Apr 28 '22 at 02:20
@DaveNewton It's a puzzle. Anything is "allowed" :-) – Bergi Apr 28 '22 at 02:34 — Bergi, Apr 28 '22 at 02:34

score 5 · Accepted Answer · answered Apr 28 '22 at 02:16

This sounds like a new take on the old variadic chainable sum function puzzle, with proxy member access instead of function calls. The possible approaches are still the same though, and the …+1 gives away that they are looking for the valueOf solution. In the same vein, foo+1 would be 1 and +foo/Number(foo) would be 0.

const handler = {
    get(target, propName, receiver) {
        if (typeof propName == 'string' && /^\d+$/.test(propName))
            return sumProxy(target + parseInt(propName, 10));
        else
            return Reflect.get(target, propName, receiver);
    }
};
function sumProxy(value) {
    return new Proxy({
        valueOf() { return value; }
    }, handler);
}
const foo = sumProxy(0);

console.log(foo + 1);                // 1
console.log(foo[1] + 1);             // 2
console.log(foo[1][2] + 1);          // 4
console.log(foo[10][20][30] + 1);    // 61
console.log(foo[100][200][300] + 1); // 601

How do I make `foo[1] ==1; foo[1][2]==3; foo[1][2][3]==6` work in JavaScript?

1 Answers1