Can somebody explain to me this JS code?

Question

1.00/5 (2 votes)

See more:

var a1 = [[0,1],[0,1],[1,1]],
    a2 = [[0,0,0,0,0,0],[0,0,0,0,0,0],[0,0,0,0,0,0]],
   res = a2.map((s,i) => s.map((n,j) => a1[i][j] !== void 0 ? a1[i][j] : n));

The code is supposed to overlay a1 onto a2, but how exactly does it do this?
What are the variables s, i, j, and n?
Code taken from answer to this question: How to impose a 2D array on top of another in JavaScript - Quora[^]

What I have tried:

I have tried putting the code in to jsfiddle where it finds some errors but it seems to work.

Posted 25-Feb-17 3:09am

Member 13023154

Updated 25-Feb-17 4:34am

Add a Solution

2 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

OriginalGriff · Answer 1 · 2017-02-25T03:59:00

Solution 1

See here: JavaScript's Lambda and Arrow Functions | Vinta Software[^]

Posted 25-Feb-17 3:59am

OriginalGriff

Comments

Maciej Los 26-Feb-17 8:32am

5ed!

Peter Leow · Answer 2 · 2017-02-25T04:34:00

Solution 2

To add to solution 1, this

res = a2.map((s,i) => s.map((n,j) => a1[i][j] !== void 0 ? a1[i][j] : n));

is the same as this:

res=a2.map(function(s,i){

	return s.map(function(n,j){

		if (a1[i][j] !== void 0){
			return a1[i][j];
		} else {
			return n;
		}
	
    })

})

in "normal" JavaScript. Learn the working of JavaScript Array map() Method[^]

Posted 25-Feb-17 4:34am

Peter Leow

Comments

Maciej Los 26-Feb-17 8:32am

5ed!

Peter Leow 26-Feb-17 21:26pm

Thank you, Maciej.