JavaScript Double Question Mark: Keep Valid Falsy Values with ??

The task

Apply default settings to three deliberately falsy values and two missing-value markers. Compare with || to expose why a valid zero can be replaced unintentionally.

Runnable Node 22 example

const assert = require('node:assert/strict');

const supplied = {retries: 0, enabled: false, caption: '', note: null};
assert.equal(supplied.retries ?? 3, 0);
assert.equal(supplied.enabled ?? true, false);
assert.equal(supplied.caption ?? 'untitled', '');
assert.equal(supplied.note ?? 'none', 'none');
assert.equal(supplied.missing ?? 'fallback', 'fallback');
assert.equal(supplied.retries || 3, 3);
console.log(`retries=${supplied.retries ?? 3}; enabled=${supplied.enabled ?? true}; note=${supplied.note ?? 'none'}`);

Expected output and explanation

The fixture treats zero retries as an explicit decision and false as an explicit disabled setting. Both survive the nullish fallback. Null and the missing property trigger the alternative because their resulting values are nullish.

The || comparison produces three retries because zero is falsy. That might be correct under a different rule, but it contradicts this setting contract. Choose the operator from what counts as missing rather than which notation looks shorter.

Pitfall to check

Parenthesize when mixing ?? with || or &&; the language does not accept every unparenthesized mixture. Also remember that a fallback supplies a value, not validation: an unexpected non-null value still passes through unchanged.