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Catch Off By One Errors When Using Indexing

Catch Off By One Errors When Using Indexing

Basics

Due to the way JavaScript indexes work firstFive has five elements but they are indexed from 0 to 4!

console.log(len); // 5
console.log(firstFive[0]); // 1
/**/
console.log(firstFive[4]); // 5
console.log(firstFive[5]); // undefined

That should give you enough to grasp the limits of firstFive. Direct your attention to the loop. What does it do? You could try debugging it to find out!

Debugging

You are given this code:

  for (let i = 1; i <= len; i++) {
    console.log(firstFive[i]);
  }

To debug this piece of code, use console.clear(). What would be the best place for it? The answer is right before the for statement!

  console.clear();
  for (let i = 1; i <= len; i++) {
    console.log(firstFive[i]);
  }

Console output:

  Console was cleared.
  2
  3
  4
  5
  undefined

Analysis

Examine the output. Under these conditions the loop first prints the element positioned at 1... which is 2! It also tries to print the element indexed at 5 which is undefined.

This can be considered the point of this challenge. Keep console.log() and console.clear() present. They will help you understand how your code works.

Solution

The most straightforward way to fix this is to alter the for() conditions. Make i start at 0. Also the loop should not be executed for i == 5. In other words, the relationship between i and len should be false when i == 5. That can be achieved by using i < len (Is 5 < len? false, and the loop won't be executed!).

  for (let i = 0; i < len; i++) {

Happy Coding! 💻

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