5 Uses for the Spread Operator

The spread operator is a favorite of JavaScript developers. It's a powerful piece of syntax that has numerous applications.

So many, in fact, that it's often hard to keep track of them all. In this post we're going to review 5 of the most common uses for the spread operator.

Copying an array

This is one of the most common uses of the spread operator. Taking the contents of an array and "spreading it out" to fill another array.

let arr = [1,2,3,4]
let copy = [...arr]
// copy is [ 1, 2, 3, 4 ]

Looked at in a different way, the spread operator is selecting each individual element inside the arr array and placing each of those elements in a new array structure.

Note that this is different than putting an array inside another array.

let arr = [1,2,3,4]
let copy = [arr]
// copy is [ [1, 2, 3, 4] ]

That option gives you a multidimensional array.

Concatenate arrays

Building on the previous example, it turns out that you're able to take multiple arrays and spread them out in a new array. One after another.

let arr1 = [1,2,3,4]
let arr2 = [5,6,7,8]
let concat = [...arr1, ...arr2]
// concat is [ 1, 2, 3, 4, 5, 6, 7, 8 ]

If we break it down as we did in the previous example, the spread operator is extracting each element in the initial arrays and placing it in the new array.

Pass arguments as arrays

This is where the spread operator starts showing off its versatility. In this example, we're passing three arguments into a function. The spread operator is used in front of an array with three elements inside of it.

function dev(x, y, z) { }

var args = [0, 1, 2]

dev(...args) // call function

A good way to make sense of this is to look at our previous examples. What would happen if we used the spread operator on an array and never placed it inside a new array?

Each element in the array would stand on its own. This is that intermediate transformation. Each element stands on its own and hasn't been placed in a new data structure. Therefore, all three elements can be passed in as arguments to the function individually.

Copy an object

Not only can the spread operator be used for arrays, it can be used for objects as well. Just like copying the array before, we can copy an object.

let obj = {a: 1, b: 2, c: 3}
let copy = {...obj}
// copy is {a: 1, b: 2, c: 3}

In this example, the spread operator extracts each key-value pair from obj and places them in a new object, copy.

And just like the array example, it's worth noting that this is different than putting an object inside another object.

let obj = {a: 1, b: 2, c: 3}
let copy = {obj}
// copy is { {a: 1, b: 2, c: 3} }

Merge object

We can also merge two objects together using the spread operator.

let obj1 = {a: 1, b: 2, c: 3}
let obj2 = {d: 4, e: 5, f: 6}

let merge = {...obj1, ...obj2}
// merge is {a: 1, b: 2, c: 3, d: 4, e: 5, f: 6}

Again, we're extracting all the key-value pairs from the initial objects and placing them in a new object data structure.

Bonus - An Error!

Despite the fact that the spread operator works on both arrays and objects, you can't mix and match these data types.

let obj = {a:1, b:2, c:3}
let copy = [...obj] // this won't work!

This makes sense if we think about it because when working on an array the spread operator is handling elements, and for an object, it's key-value pairs.

And there you have it!

That's a whole lot of uses for the spread operator, and those aren't even all of them. If you're looking for a full set, take a look at the mozilla docs.

And if you're interested in other JavaScript syntax that can help you write clean code, check out these articles!

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Comments

[Tom Butterwith]

Great examples!
Note that for merging objects, the keys in the later objects take precedence. e.g.

let obj1 = { a:1, b:2, c:3 };
let obj2 = { a:2 };

let obj3 = {...obj1, ...obj2};
// obj3 is { a:2, b:2, c:3 }

[Drew Town]

This is usually how I do optional options in an object.

If I have a function that accepts an object of options I'll spread them into the default options. It works great because all of the latter keys take precedence like you stated.

[Jordan Kicklighter]

This totally works, but have you tried using default values in the function signature? I believe that was added in ES6, but I can't remember for sure.

function (name: 'John', id: 5) {
  ...
}

[Drew Town]

I have and it works well for simple function signatures. But, when you reach a point where you have 5, 6, 7 optional parameters and you need to change the last one this pattern becomes a real pain. The problem is there is no way to invoke named function parameters, you have to do it by order.

[Jordan Kicklighter]

If you add curlies, you can still specify the arguments in the signature but treat it as an object. Best of both worlds.

[Felippe Regazio]

i do the same ;P

[Laurie]

Absolutely important to point out! Thanks for mentioning it.

[Douglas Barbosa de Lima]

I like to use the Set Object with Spread Operator to display non-repeat data into Array:

const arr = [1, 2, 3, 1, 2, 3];
const newArr = [...new Set(arr)]; // [1, 2, 3]

Spread is a awesome feature.
Congratulations for the post!

[Milos Protic]

Note that this only works if you have an array of primitive types

[Will Vincent]

Was going to leave the same comment :)

[Laurie]

Nice one!

[Fernando Doglio]

Nice quick round up! Thanks! A quick word of warning, when copying an array, it does a shallow copy, so doing:

let myarr = [[1],[2]]
let copyArr = [..myarr]
copyArr[0][0] = 12

This will change myarr as well.

[Laurie]

Definitely. My understanding is that this only applies because the example above is multidimensional. So the first level is copied, but the deeper levels are referenced. If it's one-dimensional it is a deep copy.

[Jonathan Silvestri]

Hey Laurie!

Good post, but just letting you know that your error example does work and will not throw an error.

  let arr = [1,2,3,4]
  let copy = {...arr}

Arrays are objects, where the index is the key. What you end up with in this example is:

  console.log(copy) // { 0: 1, 1: 2, 2: 3, 3: 4 }

[Laurie]

Oops! I think I meant to flip that around. Will fix, thanks for catching it.

[Jonathan Silvestri]

No problem!

[Lukáš Doležal]

Great stuff. Note that spread operator works also in deconstruction too, not only construction:

Take first (or n first) fields of an array:

let [first, ...rest] = [1, 2, 3, 4, 5];
// first == 1
// rest == [2, 3, 4, 5]

Or as function definition:

function giveMeAllTheParams(...params) {
  console.log(`You gave me ${params.length} params`);
}

giveMeAllTheParams(1, 2, 3, 4, 5);
// You gave me 5 params

[Atul Rawat]

Good one😍

[Greg Bacchus]

Likewise for objects, can be used to create a new one that omits certain properties.

const {a, ...rest} = {a: 1, b: 2, c: 3};

[Laurie]

All great examples!

[Jordan Kicklighter]

I always forget this since there seem to be more places to use the construction functionality. Thanks for the reminder!

[Miguel Rodriguez]

Hi Lauri, thank you for this post, I'm just learning JS and this really helped me to understand the spread operator.

As I'm a beginner in JS I have a question: in the "Pass arguments as arrays" part, I understand that x=0, y=1, and z=2. What would happen if my function is expecting 3 parameters but for some reason, the array has 2 or 4 elements? Or we would simply try to make sure that the size of the array and the parameters expected by the function are equal?

[Laurie]

The name of a function and the number of parameters it takes (the placeholders for arguments) represent something called a function signature. This is unique within the current scope of the program. If you were to pass too many arguments it would throw an error because it would not find a corresponding function signature, i.e. a function with that name taking 2 (or 4) elements.

[Aaron Klein]

That may be the case in other languages, but not in JS. In JS, if you pass too many arguments, the extras will be ignored (but accessible through the variable named arguments). If you pass too few arguments, the missing arguments will be given the value undefined (which may result in an error, but not definitely).

[Laurie]

You're correct, I put my Java hat back on and shouldn't have!
Looks like this has some good explanations. stackoverflow.com/questions/126940...

[Miguel Rodriguez]

Got it, thank you!

[J3m5]

You can also use it to conditionally add properties to objects.

const obj = {
  a:1,
  ...true && {b:2},
  ...false && {c:3}
}

// {a:1,b:2}

[Robin Kretzschmar]

I am using it in the React environment quite often and I'm curious if you have a clue about how the performance is compared to the equal methods like copying an array with spreading vs copying it via slice()?

[Laurie]

It depends on the JS engine running, i.e. chrome vs safari. My understanding is that slice used to be more performant but that gap has been cut significantly and on many browsers, the spread operator is the same, or possibly better performance wise.

And since slice() only works for arrays, the spread operator is a more powerful piece of syntax.

[Robin Kretzschmar]

Thanks for the answer, wasn't aware of it but a quick research confirmed it :)

[Jordan]

Also, here's a nice quick way to go from Map to array of entries.

let map = new Map([['a', 1], ['b', 2], [{foo: 'bar'}, 3]]);
let entries = [...map]; // [['a', 1], ['b', 2], [{foo: 'bar'}, 3]]

// maybe useful to quickly stringify a map?
JSON.stringify([...map])

[Ken Bellows]

Great summary! Two extra thoughts:

• Another handy use is to convert iterable objects to Arrays when you need the Array methods:

  /* Filtering a Set */
  const fruit = new Set(['apple','banana','avocado','watermelon'])
  // No no no!
  fruit.filter(f => f.startsWith('a')) // TypeError: fruit.filter is not a function
  // Instead...
  [...fruit].filter(f => f.startsWith('a')) // ["apple", "avocado"]

  /* Mapping over a generator's output */
  // counts from low to high-1
  function* range(low, high) {
    for (let i=low; i<high; i++) {
      yield i
    }
  }
  // No no no!
  range(1,10).map(n => n**2) // TypeError: range(...).map is not a function
  // Instead...
  [...range(1,10)].map(n => n**2) // [1, 4, 9, 16, 25, 36, 49, 64, 81]

• Just a thought I had: I sort of wish that your error example worked like Object.entries, and in fact that Objects were iterable as entries the way that Maps are. I wish I could do stuff like this (but to be clear, this does NOT currently work):

  const o = {'a': 10, 'b': 20}
  [...o] // [["a", 10], ["b", 20]]
  for (const entry of o) console.log(entry)
  // ["a", 10]
  // ["b", 20]

Sigh... but I dream...