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Rane Wallin
Rane Wallin

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This simple math hack lets you create an image carousel without any if statements

If you are a web developer or a web developer student, you have probably made at least one image carousel in your career. In fact, you've probably made a few. While there are plenty of image carousel libraries out there, sometimes you want (or need) to make it from scratch.

Most image carousels are made from arrays of image data. When some event triggers a change (a timeout, button click, etc) the current image data is replaced with the next element in the array. The tricky part for many comes when they reach the end of the array. Now what? If you've been writing complex if statements trying to check for this condition, I'm here to tell you there's a better way.

Observe the code below:

const imageData = [ 'image1.png', 'img2.png', 'img3.png' ];
let currentImage = 0;

const handleImageChange = () => {
  currentImage = (currentImage + 1) % imageData.length;
}

That's it. That's the whole thing. (Explanation below amazed Keanu.)

Keanu Reeves looking dumbfounded, or maybe just dumb

How it works

Let's assume we have an array of 10 elements. Modulo division (what happens when you use the % operator) returns the remainder of the division. If currentImage is 0, then (0 + 1) % 10 is the remainder of 1/10, which is 1. This is because we cannot actually divide 1 by 10, so the whole dang 1 is left over. The same is true of 2 - 9. None of these numbers can be divided by 10, so the number itself is the remainder. The magic happens when we get to 10.

Since our array is zero-index, there is no tenth element. This works in our favor! When you divide a number by itself, the remainder is 0, which means our currentImage will be set to 0. This means that as soon as we get past the end of our array, its going to go back to the beginning. Nifty, yeah?

In computer science, this is known as a circular array. The array itself is just a plain ole array, but we use this math trick to allow it to loop indefinitely.

But, wait! What if we want to go the other way? Don't worry, I got you!

We can do the same thing in reverse. The formula for this is (currentValue - 1 + totalElements) % totalElements. If we add this to the above example, it could look something like this.

const imageData = [ 'image1.png', 'img2.png', 'img3.png' ];
const currentImage = 0;

const handleImageChange = (direction) => {
  if (direction == 'forward')
    currentImage = (currentImage + 1) % imageData.length;
  else
    currentImage = (currentImage - 1 + imageData.length) % imageData.length;
}

I know, I know, I said there wouldn't be any if statements, and there aren't, at least not for actually moving forward and back through the elements. We just need to know which direction to go.

This isn't just great for image carousels. Any time you need to increment through an array one element at a time, this will eliminate any condition checking to see if you are at the end.

Cover image by Michael and Sherry Martin (flickr)

Discussion (19)

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andrearaujo profile image
André Araújo (Andrev) • Edited

Amazing!

You could remove the "if" splitting the handler into 2 functions or using a hashmap;

const imageData = [ 'image1.png', 'img2.png', 'img3.png' ];
let currentImage = 0;

const nextImage = () => currentImage = (currentImage + 1) % imageData.length;

const prevImage = () => currentImage = (currentImage - 1 + imageData.length) % imageData.length;
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ranewallin profile image
Rane Wallin Author

Yes, though if the we are still keeping the code DRY we would just be offsetting the if somewhere else. We wouldn't want two separate buttons that only different by which handler they call, so the button would need some logic in it to decide which handler is the right handler. That's the cool thing about programming, though. So many different ways to solve every problem.

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andrearaujo profile image
André Araújo (Andrev) • Edited

Yeah, I wasn't saying it's wrong, and I know that the point of this post is to show the math, I just showed one of the possible solutions, but you confused me a little bit now...

What would be the difference in the pseudo-code below and why the second one wouldn't follow the DRY principle?

// some logics in the button...
carousel.handleImageChange('forward');
...
// some logics in the button...
carousel.handleImageChange('other direction');
...
// some logics in the button...
carousel.nextImage()
...
// some logics in the button...
carousel.prevImage()
...
Thread Thread
ranewallin profile image
Rane Wallin Author

Nothing wrong with it at all. My point was that you still have an if statement, it's just in a different place. The only way to eliminate it would be if there were two completely different button components, one for forward and one for back, that called different code, which would then be less dry. I wasn't saying there was anything wrong with what you posted :).

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jeancarl profile image
JeanCarl

Once you figure out the modulus operator, it's a pretty cool operator! A ternary operator can get rid of that last if:

const imageData = [ 'image1.png', 'img2.png', 'img3.png' ];
let currentImage = 0;

const handleImageChange = (direction) => {
    currentImage = (direction == 'forward') ? 
      (currentImage + 1) % imageData.length : 
      (currentImage - 1 + imageData.length) % imageData.length;
}

(also changed currentImage to let (otherwise "Assignment to constant variable." error)

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skhmt profile image
Mike S

I wouldn't say a ternary is better than an if, especially for multi-line things like this.

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ranewallin profile image
Rane Wallin Author

Thanks. I fixed it. I do that all the time.

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jonrandy profile image
Jon Randy • Edited
const imageData = [ 'image1.png', 'img2.png', 'img3.png' ];
let currentImage = 0;

const handleImageChange = (direction) => {
    currentImage = (currentImage + imageData.length + ((direction == 'forward') ? 1 : -1)) % imageData.length;
}
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mrwensveen profile image
Matthijs Wensveen

If you create a direction enum like so:

const direction = { FORWARD: 1, BACKWARD: -1 },

const handleImageChange = offset => {
  currentImage = (currentImage + imageData.length + offset) % imageData.length;
}

// call with enum value:
handleImageChange(direction.BACKWARD);

And you're golden. Well, except when you call handleImageChange(-7)

Solution: currentImage = (currentImage + imageData.length + (offset % imageData.length)) % imageData.length
😊

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ranewallin profile image
Rane Wallin Author

I don't think this will work as written. You only add the number of elements to the first part if you are going backwards. This looks like it's adding it either way. If I am at the last element (2) in this example then this would return in the next index being 3 instead of 0.

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jonrandy profile image
Jon Randy • Edited

Wrong. Adding either way works. 3 modulo 3 is zero, as is 6 modulo 3, 9 modulo 3 etc.

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rkichenama profile image
Richard Kichenama

What if instead of passing a string to the function, you gave it an integer number of images to move?

const imageData = [ 'image1.png', 'img2.png', 'img3.png', ... ];
let currentImage = 0;
____
const handleImageChange = (imageShift) => {
  currentImage = Math.max(
    0,
    Math.min(
      imageData.length - 1,
      (currentImage + imageShift) % imageData.length
    )
  );
}
____
const firstImage = () => handleImageChange(-imageData.length);
const prevImage = () => handleImageChange(-1);
const nextImage = () => handleImageChange(1);
const lastImage = () => handleImageChange(imageData.length);
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cecilelebleu profile image
Cécile Lebleu

Great tip!
I think you mistyped css in the tags and added cs instead

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ranewallin profile image
Rane Wallin Author

Thanks. I meant cs, like computer science. :)

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cecilelebleu profile image
Cécile Lebleu

Oh! I thought of C sharp, but that makes more sense 😅

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felipperegazio profile image
Felippe Regazio

Pretty cool :)

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daveskull81 profile image
dAVE Inden

This is really cool. It is always fun to see the power of simple math operations applied to the behavior of things like this and what it can drive code to do. Great post!

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drmandible profile image
DrMandible

Pretty nifty. Previewing the next element after the modulo would be a big challenge to try without ifs.

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ranewallin profile image
Rane Wallin Author

You could just do the same thing. I.e. nextImg = (currentImage + 1) % imageData.length;