The... Spread Operator
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The second thing weird thing we need to talk about is much more important. It's
called the spread operator. Open up play.js
and clear everything out. Create a
new function called printThreeThings(thing1, thing2, thing3)
. Inside,
console.log()
our three things:
let printThreeThings = function(thing1, thing2, thing3) { | |
console.log(thing1, thing2, thing3); | |
}; | |
// ... lines 4 - 8 |
Easy enough!
Below that, create a new array called yummyThings
, which of course naturally will
include pizza, gelato, and sushi:
let printThreeThings = function(thing1, thing2, thing3) { | |
console.log(thing1, thing2, thing3); | |
}; | |
let yummyThings = ['pizza', 'gelato', 'sushi']; | |
// ... lines 6 - 8 |
All delicious, but maybe not if you eat them all at the same time.
Here's the question:
How can I pass the three yummy things as the first, second, and third argument to
printThreeThings()
?
Well, we could say printThreeThings()
with yummyThings[0]
, yummyThings[1]
and yummyThings[2]
. But! Not cool enough! In ES2015, you can use the spread operator:
printThreeThings(...yummyThings)
:
let printThreeThings = function(thing1, thing2, thing3) { | |
console.log(thing1, thing2, thing3); | |
}; | |
let yummyThings = ['pizza', 'gelato', 'sushi']; | |
printThreeThings(...yummyThings); |
Woh.
If we try that, it prints our three things! That's crazy! The ...
is called the
"spread operator": it's always ...
and then your array. When you do this, it's
almost as if somebody went through and literally just wrote out all of the items
in your array manually, instead of us doing it by hand.
But of course, there are a lot of yummy things in the world, and since I'm from the
US, my yummyThings
should probably have a cheeseburger:
let printThreeThings = function(thing1, thing2, thing3) { | |
console.log(thing1, thing2, thing3); | |
}; | |
let yummyThings = ['pizza', 'gelato', 'sushi', 'cheeseburger']; | |
printThreeThings(...yummyThings); |
What will happen now? The array has 4 things, but we only have 3 arguments? Let's find out! Run the script!
It's the same result! Thanks to the spread operator, pizza
is passed as the
first argument, then gelato
, sushi
and finally, cheeseburger is passed as the 4th
argument. Since the function doesn't have a 4th argument, it's just ignored!
Spread Operator as an Array Merge
Ok great! But you're probably asking yourself: how could this ever be useful in any real world scenario?
Dang, good question. Actually, there are two really great use-cases. One is in ReactJS. I won't talk about it now... just trust me! The second deals with merging arrays together.
Suppose we need to create another array called greatThings
. And we decide that
swimming
is super great, and sunsets
are the best. They are the best. We also
decide that anything "yummy" must also be great, so I want to add all four of these
yummyThings
into the greatThings
array. In PHP, we might use array_merge()
.
In JavaScript, one option is the spread operator. Add a comma - as if we were going
to add another entry - and then say ...yummyThings
. We could even keep going and add
something else great, like New Orleans
. Because New Orleans
is a really great place:
let yummyThings = ['pizza', 'gelato', 'sushi', 'cheeseburger']; | |
let greatThings = ['swimming', 'sunsets', ...yummyThings, 'New Orleans']; | |
// ... lines 4 - 6 |
Tip
There are often many ways to do the same thing in JavaScript, especially with
arrays and objects. In this case, greatThings.concat(yummyThings)
is also an
option.
Ok, console.log(greatThings)
to see if it works!
let yummyThings = ['pizza', 'gelato', 'sushi', 'cheeseburger']; | |
let greatThings = ['swimming', 'sunsets', ...yummyThings, 'New Orleans']; | |
console.log(greatThings); |
It does: swimming
, sunset
, 4 yummy things, and New Orleans
at the bottom.
Spread Operator for Creating a new Array
But the spread operator has one more cool trick. At the bottom, add another variable:
let copyOfGreatThings = greatThings
:
let yummyThings = ['pizza', 'gelato', 'sushi', 'cheeseburger']; | |
let greatThings = ['swimming', 'sunsets', ...yummyThings, 'New Orleans']; | |
let copyOfGreatThings = greatThings; | |
// ... lines 6 - 10 |
Now, add something else to this new variable: use copyOfGreatThings.push()
to add something
that we all know is great: summer
:
let yummyThings = ['pizza', 'gelato', 'sushi', 'cheeseburger']; | |
let greatThings = ['swimming', 'sunsets', ...yummyThings, 'New Orleans']; | |
let copyOfGreatThings = greatThings; | |
copyOfGreatThings.push('summer'); | |
// ... lines 7 - 10 |
At the bottom, console.log(copyOfGreatThings)
:
let yummyThings = ['pizza', 'gelato', 'sushi', 'cheeseburger']; | |
let greatThings = ['swimming', 'sunsets', ...yummyThings, 'New Orleans']; | |
let copyOfGreatThings = greatThings; | |
copyOfGreatThings.push('summer'); | |
console.log(greatThings); | |
console.log(copyOfGreatThings); |
Here's the question: we know summer
now lives in copyOfGreatThings()
. But does
it also now live inside of greatThings
? Try it! It does! Summer lives in both
arrays! And this makes sense: arrays are objects in JavaScript, and just like
in PHP, objects are passed by reference. In reality, greatThings
and copyOfGreatThings
are identical: they both point to the same array in memory.
As we get more advanced, especially in React.js, the idea of not mutating objects will become increasingly important. What I mean is, there will be times when it wil be really important to copy an object, instead of modifying the original.
So how could we make copyOfGreatThings
a true copy? The spread operator has
an answer: re-assign copyOfGreatThings
to [...greatThings]
:
let yummyThings = ['pizza', 'gelato', 'sushi', 'cheeseburger']; | |
let greatThings = ['swimming', 'sunsets', ...yummyThings, 'New Orleans']; | |
let copyOfGreatThings = [...greatThings]; | |
// ... lines 6 - 10 |
And that is it! This will create a new array, and then put each item from greatThings
into it, one-by-one.
Try it! Yes! We can see summer in the copy, but we did not modify the original array.
Ok, let's move on something that's instantly useful: template strings!