let arr = [1, 2, 3];

arr.slice().reverse().forEach(x => console.log(x))

will print:

3
2
1

arr will still be [1, 2, 3], the .slice() creates a shallow copy.

Simple, loop the other way. Instead of i++ use i--:

for(i = 0; i < array.length; i++){
    // do something with array[i]
}

// you go backwards:
for(i = array.length - 1; i >= 0; i--){
    // do something with array[i]
}

Because your forwards-condition has to receive the length property of your array each time, whilst the other condition only has to check for "greater then zero", a very fast task.

When your array length doesn't change during the loop, and you really look at ns-perfomance, you can use

for (var i=0, l=arr.length; i<l; i++)

BTW: Instead of for (var i = arr.length; i > 0; --i) you might use for (var i = arr.length; i-- > 0; ) which really runs through your array from n-1 to 0, not from n to 1.

Something like this would work:

var fruits = ["apple","banana","cherry"];
var direction = 1;  // or -1

var i = direction > 0 ? 0 : fruits.length - 1,
    stop = direction > 0 ? fruits.length : -1;
for (; i != stop; i += direction)
    console.log(i, fruits[i]);

var fruits = ["apple", "banana", "cherry"];

function iterate(direction) {
  var results = $("#results").empty();
  var i = direction > 0 ? 0 : fruits.length - 1,
    stop = direction > 0 ? fruits.length : -1;
  for (; i != stop; i += direction)
    $("<span>").text(i + ": " + fruits[i] + "\n").appendTo(results);
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.0.0/jquery.min.js"></script>
<button onclick="iterate(1)">Forward</button>
<button onclick="iterate(-1)">Backward</button>
<pre id="results" />

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--) {

You don't need to slice the array (which uses additional memory, as it creates a new array) to do that.

What you are describing is a loop that starts at index = 96 until it reaches 0, decreasing index one by one.

So you just need to change let i = keyToValue2.length[96] - 1 to let i = 96.

Here's an example using an array with 32 values and logging them backwards, starting at index 16. Just used these values because StackOverflow snippets limit the number of log entries:

// This creates a new array with 32 numbers (0 to 31, both included):
const array = new Array(32).fill(null).map((_, i) => `Element at index ${ i }.`);

// We start iterating at index 16 and go backwards until 0 (both included):
for (let i = 16; i >= 0; --i) {
  console.log(array[i])
}

If you want to make sure the index 96 actually exists in your array, then use let i = Math.min(96, keyToValue2.length - 1:

// This creates a new array with 32 numbers (0 to 31, both included):
const array = new Array(32).fill(null).map((_, i) => `Element at index ${ i }.`);

// We start iterating at index 31 (as this array doesn't have 64 elements, it has only 32)
// and go backwards until 0 (both included):
for (let i = Math.min(64, array.length - 1); i >= 0; --i) {
  console.log(array[i])
}