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Array: daftar yang diurutkan

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:

JS
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.

JS
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:

JS
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

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

Try it

Latihan#js.m3.l1.e1
Upaya: 0Memuat…

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

Memuat editor…
Tunjukkan petunjuk

The last valid index is length - 1.

Solusi tersedia setelah 3 upaya

Review exercise

Latihan#js.m3.l1.e2
Upaya: 0Memuat…

Start from `fruits = ['mela', 'pera']`. Add 'banana' at the end with push, then return the full array as the last expression: ['mela','pera','banana'].

Memuat editor…
Tunjukkan petunjuk

push mutates the array; you don't need to reassign.

Solusi tersedia setelah 3 upaya