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memoize: cache results

Write a function memoize(fn) that returns a cached version of fn: a repeated call with the same arguments returns the cached result without calling fn.

Arguments are primitives, and there can be several. m(1) and m("1") are different calls. An undefined result must be cached too.

function memoize(fn) {
  // ...
}

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Tests (5)

const f = fn((n) => n * 2);
const m = memoize(f);
expect(m(2)).toBe(4);
expect(m(2)).toBe(4);
expect(f).toHaveBeenCalledTimes(1);

const f = fn((a, b) => a + b);
const m = memoize(f);
m(1, 2); m(1, 2); m(2, 1);
expect(f).toHaveBeenCalledTimes(2);

const f = fn((x) => typeof x);
const m = memoize(f);
expect(m(1)).toBe('number');
expect(m('1')).toBe('string');
expect(f).toHaveBeenCalledTimes(2);

const f = fn(() => undefined);
const m = memoize(f);
m(5); m(5);
expect(f).toHaveBeenCalledTimes(1);

const a = memoize((x) => x + 1);
const b = memoize((x) => x + 2);
expect(a(1)).toBe(2);
expect(b(1)).toBe(3);