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المصفوفات: القائمة المرتبة

An array is an ordered list of values. In JavaScript the elements can be of different types from each other (although in "good" code they tend to be homogeneous). It's written with square brackets:

const colori = ['rosso', 'verde', 'blu'];
const misti = [1, 'due', true, null];
const vuoto = [];

Reading and writing by index

Indices start at 0. The last valid index is array.length - 1.

const colori = ['rosso', 'verde', 'blu'];
colori[0]; // 'rosso'
colori[2]; // 'blu'
colori[99]; // undefined  ← no error, just undefined
colori.length; // 3

colori[1] = 'giallo'; // now colori is ['rosso', 'giallo', 'blu']
colori[colori.length] = 'x'; // manual push, not recommended

Adding and removing elements

The four classic methods (push / pop / unshift / shift) modify the array in place:

const a = [1, 2, 3];
a.push(4); // a = [1, 2, 3, 4],  returns the new length
a.pop(); // a = [1, 2, 3],     returns the removed element (4)
a.unshift(0); // a = [0, 1, 2, 3],  adds at the head
a.shift(); // a = [1, 2, 3],     removes from the head

Including, searching

['rosso', 'verde', 'blu'].includes('verde'); // true
['rosso', 'verde', 'blu'].indexOf('blu'); // 2
['rosso', 'verde', 'blu'].indexOf('giallo'); // -1

Try it

تمرين#js.m3.l1.e1

المحاولات: 0جارٍ التحميل…

Given the array `numbers = [10, 20, 30, 40]`, return as the last expression the last element using length.

جارٍ تحميل المحرر…

إظهار التلميح

The last valid index is length - 1.

الحل متاح بعد 3 من المحاولات

Review exercise

تمرين#js.m3.l1.e2