once: a closures and higher-order functions problem in JavaScript

Implement once(fn). It returns a new function, a wrapper, that calls fn at most one time.

The first call runs fn with that call's arguments and returns its result.
Every later call returns the result of the first call without running fn again. Their arguments are ignored.
fn runs with the same this the new function was called with.
The first result is reused even when it's undefined or another falsy value. A falsy value acts like false in an if, for example 0, '' or false.
Each call to once makes a new, independent wrapper. Wrappers don't share state.

Input: const init = once(() => { console.log('setup'); return 42; }); init(); init(); Output: logs 'setup' once; both calls return 42

Input: const f = once((a, b) => a + b); f(2, 3); f(10, 20); Output: 5, then 5The second call's arguments are ignored.