find, some, every

Three compact methods to search in an array without writing the loop by hand.

`find`: the first one that matches

const utenti = [
  { nome: 'Anna', eta: 30 },
  { nome: 'Luca', eta: 12 },
  { nome: 'Sara', eta: 18 },
];

utenti.find((u) => u.eta >= 18);
// { nome: 'Anna', eta: 30 }

utenti.find((u) => u.nome === 'Marco');
// undefined  ← if none matches

findIndex variant: returns the index (or -1).

`some`: at least one

const nums = [1, 2, 3, 4, 5];

nums.some((n) => n > 4); // true
nums.some((n) => n > 100); // false

[].some(() => true); // false  ← on an empty array it is always false

`every`: all of them

const nums = [1, 2, 3, 4, 5];

nums.every((n) => n > 0); // true
nums.every((n) => n > 2); // false

[].every(() => false); // true  ← on an empty array it is always true!

The asymmetry of some and every on empty arrays is intentional and logically consistent with the "exists" (∃) and "for all" (∀) logical operators.

Try it

Exercise#js.m6.l3.e1

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Define `firstAdult(users)`: given an array of { name, age }, return the name of the first one with age >= 18, or null if none. Use find.

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users.find(...); if found return u.name, otherwise null.

Solution available after 3 attempts

Review exercise

Exercise#js.m6.l3.e2