Spread syntax (...)

Spread syntax can be used when all elements from an object or array need to be included in a new array or object, or should be applied one-by-one in a function call's arguments list. There are three distinct places that accept the spread syntax:

• Function arguments list (myFunction(a, ...iterableObj, b))
• Array literals ([1, ...iterableObj, '4', 'five', 6])
• Object literals ({ ...obj, key: 'value' })

Although the syntax looks the same, they come with slightly different semantics.

Only iterable objects, like Array, can be spread in array and function parameters. Many objects are not iterable, including all plain objects that lack a Symbol.iterator method:

const obj = { key1: 'value1' };
const array = [...obj]; // TypeError: obj is not iterable

On the other hand, spreading in object literals enumerates the own properties of the object. For typical arrays, all indices are enumerable own properties, so arrays can be spread into objects.

const array = [1, 2, 3];
const obj = { ...array }; // { 0: 1, 1: 2, 2: 3 }

When using spread syntax for function calls, be aware of the possibility of exceeding the JavaScript engine's argument length limit. See Function.prototype.apply() for more details.

Replace apply()

It is common to use Function.prototype.apply() in cases where you want to use the elements of an array as arguments to a function.

function myFunction(x, y, z) {}
const args = [0, 1, 2];
myFunction.apply(null, args);

With spread syntax the above can be written as:

function myFunction(x, y, z) {}
const args = [0, 1, 2];
myFunction(...args);

Any argument in the argument list can use spread syntax, and the spread syntax can be used multiple times.

function myFunction(v, w, x, y, z) {}
const args = [0, 1];
myFunction(-1, ...args, 2, ...[3]);

Apply for new operator

When calling a constructor with new, it's not possible to directly use an array and apply(), because apply() calls the target function instead of constructing it, which means, among other things, that new.target will be undefined. However, an array can be easily used with new thanks to spread syntax:

const dateFields = [1970, 0, 1];  // 1 Jan 1970
const d = new Date(...dateFields);

A more powerful array literal

Without spread syntax, to create a new array using an existing array as one part of it, the array literal syntax is no longer sufficient and imperative code must be used instead using a combination of push(), splice(), concat(), etc. With spread syntax this becomes much more succinct:

const parts = ['shoulders', 'knees'];
const lyrics = ['head', ...parts, 'and', 'toes'];
//  ["head", "shoulders", "knees", "and", "toes"]

Just like spread for argument lists, ... can be used anywhere in the array literal, and may be used more than once.

Copy an array

const arr = [1, 2, 3];
const arr2 = [...arr]; // like arr.slice()

arr2.push(4);
//  arr2 becomes [1, 2, 3, 4]
//  arr remains unaffected

const a = [[1], [2], [3]];
const b = [...a];

b.shift().shift();
//  1

//  Oh no!  Now array 'a' is affected as well:
a
//  [[], [2], [3]]

A better way to concatenate arrays

Array.prototype.concat() is often used to concatenate an array to the end of an existing array. Without spread syntax, this is done as:

let arr1 = [0, 1, 2];
const arr2 = [3, 4, 5];

// Append all items from arr2 onto arr1
arr1 = arr1.concat(arr2);

With spread syntax this becomes:

let arr1 = [0, 1, 2];
const arr2 = [3, 4, 5];

arr1 = [...arr1, ...arr2];
// arr1 is now [0, 1, 2, 3, 4, 5]

Array.prototype.unshift() is often used to insert an array of values at the start of an existing array. Without spread syntax, this is done as:

const arr1 = [0, 1, 2];
const arr2 = [3, 4, 5];

//  Prepend all items from arr2 onto arr1
Array.prototype.unshift.apply(arr1, arr2);

//  arr1 is now [3, 4, 5, 0, 1, 2]

With spread syntax, this becomes:

let arr1 = [0, 1, 2];
const arr2 = [3, 4, 5];

arr1 = [...arr2, ...arr1];
//  arr1 is now [3, 4, 5, 0, 1, 2]

Shallow-cloning (excluding prototype) or merging of objects is possible using a shorter syntax than Object.assign().

const obj1 = { foo: 'bar', x: 42 };
const obj2 = { foo: 'baz', y: 13 };

const clonedObj = { ...obj1 };
// Object { foo: "bar", x: 42 }

const mergedObj = { ...obj1, ...obj2 };
// Object { foo: "baz", x: 42, y: 13 }

Note that Object.assign() can be used to mutate an object, whereas spread syntax can't.

const obj1 = { foo: 'bar', x: 42 };
Object.assign(obj1, { x: 1337 });
console.log(obj1); // Object { foo: "bar", x: 1337 }

In addition, Object.assign() triggers setters on the target object, whereas spread syntax does not.

const objectAssign = Object.assign({ set foo(val) { console.log(val); } }, { foo: 1 });
// Logs "1"; objectAssign.foo is still the original setter

const spread = { set foo(val) { console.log(val); }, ...{ foo: 1 } };
// Nothing is logged; spread.foo is 1

You cannot naively re-implement the Object.assign() function through a single spread operator:

const obj1 = { foo: 'bar', x: 42 };
const obj2 = { foo: 'baz', y: 13 };
const merge = (...objects) => ({ ...objects });

const mergedObj1 = merge(obj1, obj2);
// Object { 0: { foo: 'bar', x: 42 }, 1: { foo: 'baz', y: 13 } }

const mergedObj2 = merge({}, obj1, obj2);
// Object { 0: {}, 1: { foo: 'bar', x: 42 }, 2: { foo: 'baz', y: 13 } }

In the above example, the spread syntax does not work as one might expect: it spreads an array of arguments into the object literal, due to the rest parameter. Here is an implementation of merge using the spread operator, whose behavior is similar to Object.assign(), except that it doesn't trigger setters, nor mutates any object:

const obj1 = { foo: 'bar', x: 42 };
const obj2 = { foo: 'baz', y: 13 };
const merge = (...objects) => objects.reduce((acc, cur) => ({ ...acc, ...cur }));

const mergedObj1 = merge(obj1, obj2);
// Object { foo: 'baz', x: 42, y: 13 }

• Rest parameters
• Rest property
• Function.prototype.apply()