配列: 順序付きリスト

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

運動#js.m3.l1.e1

試行回数: 0読み込み中…

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

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ヒントを表示

The last valid index is length - 1.

3 回の試行後に解決策が利用可能になります

Review exercise

運動#js.m3.l1.e2

試行回数: 0読み込み中…

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

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ヒントを表示

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

3 回の試行後に解決策が利用可能になります