JavaScript shift: Remove the First Item from a Small Queue

The task

Consume two print-job labels in arrival order, then inspect an empty queue. This is a small behavior demonstration rather than a performance claim for a large queue.

Runnable Node 22 example

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

const pending = ['poster', 'leaflet'];
const first = pending.shift();
assert.equal(first, 'poster');
assert.deepEqual(pending, ['leaflet']);
const second = pending.shift();
assert.equal(second, 'leaflet');
assert.equal(pending.length, 0);
assert.equal(pending.shift(), undefined);
console.log(`first=${first}; second=${second}; remaining=${pending.length}`);

Expected output and explanation

The first call returns poster while pending becomes a one-item array. The second returns leaflet and leaves no elements. The third does not return another label; its result is undefined. Keeping the return and remaining container separate makes the lifecycle clear.

FIFO means this example consumes the arrival order rather than the newest item. Use pop for a last-in-first-out stack. Reading pending[0] alone would inspect the next item without consuming it.

Pitfall to check

An array can also contain an actual undefined element, so a returned undefined is not a universally sufficient “queue was empty” signal. Check the queue contract or length before consuming. Repeated front removal also should not be assumed constant-time for large workloads.