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Spread Operator – How Spread Works in JavaScript

The spread operator (...) helps you expand iterables into individual elements.

The spread syntax serves within array literals, function calls, and initialized properties objects to spread the values of iterable objects into separate items. So effectively, it does the opposite thing from the rest operator.

note

A spread syntax is effective only when used within array literals, function calls, or initialized properties objects.

So, what exactly does this mean? Let's see with some examples.

Spread Example 1: How Spread Works in an Array Literal​

const myName = ["Sofela", "is", "my"];
const aboutMe = ["Oluwatobi", ...myName, "name."];

console.log(aboutMe);

// The invocation above will return:
// ["Oluwatobi", "Sofela", "is", "my", "name."]

Try Editing It

The snippet above used spread (...) to copy myName array into aboutMe.

note
  • Alterations to myName will not reflect in aboutMe because all the values inside myName are primitives. Therefore, the spread operator simply copied and pasted myName's content into aboutMe without creating any reference back to the original array.
  • As mentioned by @nombrekeff in the DEV comment, the spread operator only does shallow copy. So, keep in mind that supposing myName contained any non-primitive value, the computer would have created a reference between myName and aboutMe. See info 3 and info 4 for more on how the spread operator works with primitive and non-primitive values.
  • Suppose we did not use the spread syntax to duplicate myName's content. For instance, if we had written const aboutMe = ["Oluwatobi", myName, "name."]. In such a case, the computer would have assigned a reference back to myName. As such, any change made in the original array would reflect in the duplicated one.

Spread Example 2: How to Use Spread to Convert a String into Individual Array Items​

const myName = "Oluwatobi Sofela";

console.log([...myName]);

// The invocation above will return:
//["O", "l", "u", "w", "a", "t", "o", "b", "i", " ", "S", "o", "f", "e", "l", "a"]

Try Editing It

In the snippet above, we used the spread syntax (...) within an array literal object ([...]) to expand myName's string value into individual items.

As such, "Oluwatobi Sofela" got expanded into [ "O", "l", "u", "w", "a", "t", "o", "b", "i", " ", "S", "o", "f", "e", "l", "a" ].

Spread Example 3: How the Spread Operator Works in a Function Call​

const numbers = [1, 3, 5, 7];

function addNumbers(a, b, c, d) {
  return a + b + c + d;
}

console.log(addNumbers(...numbers));

// The invocation above will return: 16

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In the snippet above, we used the spread syntax to spread the numbers array's content across addNumbers()'s parameters.

Suppose the numbers array had more than four items. In such a case, the computer will only use the first four items as addNumbers() argument and ignore the rest.

Here's an example:

const numbers = [1, 3, 5, 7, 10, 200, 90, 59];

function addNumbers(a, b, c, d) {
  return a + b + c + d;
}

console.log(addNumbers(...numbers));

// The invocation above will return: 16

Try Editing It

Here's another example:

const myName = "Oluwatobi Sofela";

function spellName(a, b, c) {
  return a + b + c;
}

console.log(spellName(...myName)); // returns: "Olu"
console.log(spellName(...myName[3])); // returns: "wundefinedundefined"
console.log(spellName([...myName])); // returns: "O,l,u,w,a,t,o,b,i, ,S,o,f,e,l,aundefinedundefined"
console.log(spellName({ ...myName })); // returns: "[object Object]undefinedundefined"

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Spread Example 4: How Spread Works in an Object Literal​

const myNames = ["Oluwatobi", "Sofela"];
const bio = { ...myNames, runs: "codesweetly.com" };

console.log(bio);

// The invocation above will return:
{0: "Oluwatobi", 1: "Sofela", runs: "codesweetly.com"}

Try Editing It

In the snippet above, we used spread inside the bio object to expand myNames values into individual properties.

So now that we know what the spread operator is, we can talk about what makes it different from the rest operator.

Rest vs. Spread Operator: What's the Difference?​

JavaScript uses three dots (...) for both the rest and spread operators. But these two operators are not the same.

The main difference between rest and spread is that the rest operator puts the rest of some specific user-supplied values into a JavaScript array. But the spread syntax expands iterables into individual elements.

For instance, consider this code that uses rest to enclose some values into an array:

// Use rest to enclose the rest of specific user-supplied values into an array:
function myBio(firstName, lastName, ...otherInfo) {
  return otherInfo;
}

// Invoke myBio function while passing five arguments to its parameters:
myBio("Oluwatobi", "Sofela", "CodeSweetly", "Web Developer", "Male");

// The invocation above will return:
// ["CodeSweetly", "Web Developer", "Male"]

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In the snippet above, we used the ...otherInfo rest parameter to encase "CodeSweetly", "Web Developer", and "Male" into an array.

Now, consider this example of a spread operator:

// Define a function with three parameters:
function myBio(firstName, lastName, company) {
  return `${firstName} ${lastName} runs ${company}`;
}

// Use spread to expand an array's items into individual arguments:
myBio(...["Oluwatobi", "Sofela", "CodeSweetly"]);

// The invocation above will return: "Oluwatobi Sofela runs CodeSweetly"

Try Editing It

In the snippet above, we used the spread operator (...) to spread ["Oluwatobi", "Sofela", "CodeSweetly"]'s content across myBio()'s parameters.

Important Stuff to Know about the Spread Operator​

Keep these four essential pieces of info in mind whenever you choose to use the spread operator.

Info 1: Spread operators can't expand object literal's values​

Since a properties object is not an iterable object, you cannot use the spread operator to expand its values.

However, you can use the spread operator to clone properties from one object into another.

Here's an example:

const myName = { firstName: "Oluwatobi", lastName: "Sofela" };
const bio = { ...myName, website: "codesweetly.com" };

console.log(bio);

// The invocation above will return:
{firstName: "Oluwatobi", lastName: "Sofela", website: "codesweetly.com"}

Try Editing It

The snippet above used the spread operator to clone myName's content into the bio object.

note
  • The spread operator can expand iterable objects' values only.
  • An object is iterable only if it (or any object in its prototype chain) has a property with a @@iterator key.
  • Array, TypedArray, String, Map, and Set are all built-in iterable types because they have the @@iterator property by default.
  • A properties object is not an iterable data type because it does not have the @@iterator property by default.
  • You can make a properties object iterable by adding @@iterator onto it.

Info 2: The spread operator does not clone identical properties​

When using the spread operator, suppose you clone properties from object A into object B. And suppose object B contains properties identical to those in object A. In such a case, B's versions will override those inside A.

Here's an example:

const myName = { firstName: "Tobi", lastName: "Sofela" };
const bio = { ...myName, firstName: "Oluwatobi", website: "codesweetly.com" };

console.log(bio);

// The invocation above will return:
{firstName: "Oluwatobi", lastName: "Sofela", website: "codesweetly.com"}

Try Editing It

Observe that the spread operator did not copy myName's firstName property into the bio object because bio already contains a firstName property.

Info 3: How does spread work on objects containing only primitive values?​

Suppose you used the spread operator on an object (or array) containing only primitive values. The computer will not create any reference between the original object and the duplicated one.

For instance, consider this code below:

const myName = ["Sofela", "is", "my"];
const aboutMe = ["Oluwatobi", ...myName, "name."];

console.log(aboutMe);

// The invocation above will return:
// ["Oluwatobi", "Sofela", "is", "my", "name."]

Try Editing It

Observe that every item in myName is a primitive value. Therefore, when we used the spread operator to clone myName into aboutMe, the computer did not create any reference between the two arrays.

As such, any alteration you make to myName will not reflect in aboutMe, and vice versa.

As an example, let's add more content to myName:

myName.push("real");

Now, let's check the current state of myName and aboutMe:

console.log(myName); // ["Sofela", "is", "my", "real"]
console.log(aboutMe); // ["Oluwatobi", "Sofela", "is", "my", "name."]

Try Editing It

Notice that myName's updated content did not reflect in aboutMe—because spread created no reference between the original array and the duplicated one.

Info 4: How does spread work on objects containing one or more non-primitive values?​

Suppose you used the spread operator on an object (or array) containing one or more non-primitives. In that case, spread will create a reference between the original non-primitive and the cloned one.

Here is an example:

const myName = [["Sofela", "is", "my"]];
const aboutMe = ["Oluwatobi", ...myName, "name."];

console.log(aboutMe);

// The invocation above will return:
// ["Oluwatobi", ["Sofela", "is", "my"], "name."]

Try Editing It

Observe that myName contains a non-primitive value.

Therefore, using the spread operator to clone myName's content into aboutMe caused the computer to create a reference between the two arrays.

As such, any alteration you make to myName's copy will reflect in aboutMe's version, and vice versa.

As an example, let's add more content to myName:

myName[0].push("real");

Now, let's check the current state of myName and aboutMe:

console.log(myName); // [["Sofela", "is", "my", "real"]]
console.log(aboutMe); // ["Oluwatobi", ["Sofela", "is", "my", "real"], "name."]

Try Editing It

Notice that myName's updated content is reflected in aboutMe—because spread created a reference between the original array and the duplicated one.

Here's another example:

const myName = { firstName: "Oluwatobi", lastName: "Sofela" };
const bio = { ...myName };

myName.firstName = "Tobi";

console.log(myName); // { firstName: "Tobi", lastName: "Sofela" }
console.log(bio); // { firstName: "Oluwatobi", lastName: "Sofela" }

Try Editing It

In the snippet above, myName's update did not reflect in bio because we used the spread operator on an object that contains primitive values only.

note

A developer would call myName a shallow object because it contains only primitive items. In other words, the depth to access myName's values is shallow.

Here's one more example:

const myName = {
  fullName: { firstName: "Oluwatobi", lastName: "Sofela" },
};

const bio = { ...myName };

myName.fullName.firstName = "Tobi";

console.log(myName); // { fullName: { firstName: "Tobi", lastName: "Sofela" } }
console.log(bio); // { fullName: { firstName: "Tobi", lastName: "Sofela" } }

Try Editing It

In the snippet above, myName's update is reflected in bio because we used the spread operator on an object that contains a non-primitive value.

note

We call myName a deep object because it contains a non-primitive item. In other words, the depth to access one or more of myName's values is deep.

Shallow Copy vs. Deep Copy: What's the Difference?​

You do shallow copy when you create references while cloning one object into another. For instance, consider this snippet:

const myName = {
  fullName: { firstName: "Oluwatobi", lastName: "Sofela" },
};

const bio = { ...myName };

myName.fullName.firstName = "Tobi";

console.log(myName); // { fullName: { firstName: "Tobi", lastName: "Sofela" } }

console.log(bio); // { fullName: { firstName: "Tobi", lastName: "Sofela" } }

Try Editing It

...myName produces a shallow copy of the myName object because whatever alteration you make in one will reflect in the other. In other words, bio's detachment from myName is shallow since they still have some connections.

You do deep copy when you clone objects without creating references. The JSON.parse(JSON.stringify()) and structuredClone() methods are tools you can use to create a deep copy.

Here's an example:

const myName = {
  fullName: { firstName: "Oluwatobi", lastName: "Sofela" },
};

const bio = JSON.parse(JSON.stringify(myName));

myName.fullName.firstName = "Tobi";

console.log(myName); // { fullName: { firstName: "Tobi", lastName: "Sofela" } }

console.log(bio); // { fullName: { firstName: "Oluwatobi", lastName: "Sofela" } }

Try Editing It

The snippet above used the JSON.parse(JSON.stringify()) method to clone myName into bio without creating any reference (deep copy). In other words, bio's detachment from myName is deep since they no longer have any connections.

Here's another example:

const myName = {
  fullName: { firstName: "Oluwatobi", lastName: "Sofela" },
};

const bio = structuredClone(myName);

myName.fullName.firstName = "Tobi";

console.log(myName); // { fullName: { firstName: "Tobi", lastName: "Sofela" } }

console.log(bio); // { fullName: { firstName: "Oluwatobi", lastName: "Sofela" } }

Try Editing It

The snippet above used the structuredClone() method to clone myName into bio without creating any reference (deep copy). In other words, bio's detachment from myName is deep since they no longer have any connections.

You can break off the reference between two connected objects by replacing any of the two with a new object.

Here's an example:

// Create a deep object:
const myName = {
  fullName: { firstName: "Oluwatobi", lastName: "Sofela" },
};

// Create a shallow copy of myName:
const bio = { ...myName };

// Change myName's firstName:
myName.fullName.firstName = "Tobi";

// Check myName's content:
console.log(myName); // { fullName: { firstName: "Tobi", lastName: "Sofela" } }

// Check bio's content:
console.log(bio); // { fullName: { firstName: "Tobi", lastName: "Sofela" } }

// Notice that myName and bio are linked. Let's break the connection:
myName.fullName = { firstName: "Oluwa", lastName: "Sofela" };

// Check myName's content:
console.log(myName); // { fullName: { firstName: "Oluwa", lastName: "Sofela" } }

// Check bio's content:
console.log(bio); // { fullName: { firstName: "Tobi", lastName: "Sofela" } }

Try Editing It

The snippet above disconnected the pointer between myName and bio's fullName objects by reassigning myName.fullName with a new object.

Overview​

This article discussed what the spread operator is. We also looked at how spread works in array literals, function calls, and object literals.

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