JavaScript: Find the number in an array which is closest to an given number

Question

It's actually an exercise from an Assembly-course I'm doing:

Find closest number.

Add the following into the data section:
nums dd 23h,75h,111h,0abch,443h,1000h,5h,2213h,433a34h,0deadbeafh
This is an array of numbers. 
Write a program that receives a number x as input, and finds the dword
inside the array nums, which is the closest to x. (We define the distance
between two numbers to be the absolute value of the difference: |a-b|).
Example:
For the input of 100h, the result will be 111h, because 111h is closer to
100h than any other number in the nums array. (|100h - 111h| = 11h).

Full exercise-sheet on GitHub.

Because it's easier for me to develop the algorithm in more familiar JavaScript and because it's a nice exercise I have done it in JavaScript first. Now the only thing left to do is to find an Assembly-specific way.

Anyway: Here's the JavaScript-code I've written to solve the task.

let nums = [ 0x23, 0x75, 0x111,
 0xabc, 0x443, 0x1000,
 0x5, 0x2213, 0x433a34,
 0xdeadbeaf
 ];
let given = 0x100; // { 256 }10
// --- The Relevant Part --------------------------------------
function findIndexClosestNumber(arrayToSearch, compareNumber) {
 'use strict';
 let closest = compareNumber;
 let indexClosest = 0;
 arrayToSearch.forEach((currentNumber, index) => {
 let currentDistance =
 Math.abs(compareNumber - currentNumber);
 if (currentDistance < closest) {
 indexClosest = index;
 closest = currentDistance;
 }
 });
 return indexClosest;
}
// -------------------------------------------------------------
// ------ Testing & Demo ---------------------------------------
console.log(findIndexClosestNumber(nums, given));
nums.length = 0;
for (let i = 0; i <= 90; i += 10) {
 nums.push(i);
}
console.log(nums);
for (let i = 0; i < 10; i++) {
 var randomNumber = Math.floor((Math.random() * 100));
 console.log('%s gives index %s', randomNumber,
 findIndexClosestNumber(
 nums, randomNumber));
}

• Can I expect my implementation to solve the task in a correct way?

• Was it a good idea to initialize the closest-variable with the value of the compareNumber-parameter?

• Was it a good idea to check for currentDistance < closest? Or should I have used currentDistance <= closest?

Currently the first occurrence-index is returned in case the number is contained multiple times. Using <= it would return the last occurrence-index.

I'm not sure what's the better practice.

Any other comments concerning structure, naming, used methods, algorithm appreciated too.

And if someone knows a cool JavaScript specific way to solve the task: That would interest me too. ;)

Answer 1

Your main questions

• Can I expect my implementation to solve the task in a correct way?

You should know that the answer is "yes" before posting a question on Code Review.

• Was it a good idea to initialize the 'closest'-variable with the value of the 'compareNumber'-parameter?

No, it's a mistake. The function incorrectly returns 0 for input [10, 9], 3. When looking for the minimum distance, you must choose as the default the largest possible number. For example if you target ES6, then Number.MAX_SAFE_INTEGER.

• Was it a good idea to check for 'currentDistance < closest'? Or should I have used 'currentDistance <= closest'?

That depends entirely on the specification. If you're looking for the first closest index, then use <. If the last, then use >.

If the specification was not clear, then ask to clarify.

Naming

The names are overly verbose. When short names are sufficient, they are better than long names.

For example, in a function to find the closest index to a number, the array parameter is obviously for searching, so instead of arrayToSearch, it could be simply nums, and compareNumber could be target.

I would have written like this:

function indexOfClosest(nums, target) {
 let closest = Number.MAX_SAFE_INTEGER;
 let index = 0;
 nums.forEach((num, i) => {
 let dist = Math.abs(target - num);
 if (dist < closest) {
 index = i;
 closest = dist;
 }
 });
 return index;
}