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Arrays: de geordende lijst

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

Oefening#js.m3.l1.e1

Pogingen: 0Laden…

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

Editor laden…

Toon hint

The last valid index is length - 1.

Oplossing beschikbaar na 3 pogingen

Review exercise

Oefening#js.m3.l1.e2