Looping backwards through javascript array with array.length using a for-loop

Almost. You have to:

• Decrease i instead of increasing it, and
• Start from array.length-1, because array indexes start from 0, not 1.

So use instead:

for (let i = array.length-1; i >=0 ; i--) {

It occurs cause arrays in Javascirpt are not limited to the size that they were declared. So, it implies that if you try to access an empty position you will receive undefined. For example, in this case, if you do array[5] you will receive undefined and not a ArrayIndexOutOfBoundsException as it would happen in Java.

Also, you should take a look in your code. Since you're iterating the array backwards, you sohuld decrement i with i-- instead of incrementing it.

Because its like this:

const array = [1,2,3,4];
for(let i=10;i>=0;i--){
  console.log(array[i]);
}

You can for cycle what ever you want. In javascript everything is an object, including array. Accessing something that does not exist in object with [] returns undefined.

for(let i=array.length-1;i>=0;i--){
  //do something 
}

you need to set i to the length of the array -1 (because array index starts at 0). Then each time you go through the loop you need to subtract one from i (the i--)

you code works but will run through the loop a huge amount of times until the number overflows into a negative. Try adding a console.log(i) to your loop to see what i mean. (it will likely freeze chrome if you do it in the inspector)

The length of the Array is equal to the number of entries assuming they were all full. Sparse arrays do not have a length of the number of elements in them, but based on the maximum index in them

For example:

let a = [];
a[10] = 1;
console.log(a.length);
console.log(a);

The length is 11 (0 to 10) and you can see that there are values of undefined for index 0 to 9.

Even setting the last value to undefined does not affect the length since there is still a value of undefined in that position.

let a = [];
a[9] = 1;
a[10] = 1;
console.log(a.length);

a[10] = undefined
console.log(a.length);

Using delete still does not affect the length.

let a = [];
a[9] = 1;
a[10] = 1;
console.log(a.length);

delete a[10];
console.log(a.length);

The only way to change the length is to create a new array as a subset of the original:

let a = [];
a[9] = 1;
a[10] = 1;
console.log(a.length);
console.log(a);

a = a.slice(0,9);
console.log(a.length);
console.log(a);