Home HTML Data Types DOM JavaScript JS Debugging

College Board Big Idea 1

Identifying and Correcting Errors (Unit 1.4)

Become familiar with types of errors and strategies for fixing them

  • Review CollegeBoard videos and take notes on blog
  • Complete assigned MCQ questions if applicable

Code Segments

Practice fixing the following code segments!

Segment 1: Alphabet List

Intended behavior: create a list of characters from the string contained in the variable alphabet

Code:

%%js

var alphabet = "abcdefghijklmnopqrstuvwxyz";
var alphabetList = [];

for (var i = 0; i < 10; i++) {
	alphabetList.push(i);
}

console.log(alphabetList);
<IPython.core.display.Javascript object>

Altered Code:

%%js

var alphabet = "abcdefghijklmnopqrstuvwxyz";
var alphabetList = [];

for (var i = 0; i < 10; i++) {
	alphabetList.push(alphabet[i]); // push the alphabet at position i and not index i
}

console.log(alphabetList);
<IPython.core.display.Javascript object>

What I Changed

The code block inside the loop was pushing the index i (of the iteration) rather than the alphabet character at index i. Fixed by replacing i with alphabet[i].

Segment 2: Numbered Alphabet

Intended behavior: print the number of a given alphabet letter within the alphabet. For example:

"_" is letter number _ in the alphabet

Where the underscores (_) are replaced with the letter and the position of that letter within the alphabet (e.g. a=1, b=2, etc.)

Code:

%%js

var alphabet = "abcdefghijklmnopqrstuvwxyz";
var alphabetList = [];

for (var i = 0; i < 10; i++) {
	alphabetList.push(alphabet[i]); // push the alphabet at position i and not index i
}

let letterNumber = 5

for (var i = 0; i < alphabetList; i++) {
	if (i === letterNumber) {
		console.log(letterNumber + " is letter number 1 in the alphabet")
	}
}

// Should output:
// "e" is letter number 5 in the alphabet
<IPython.core.display.Javascript object>

Altered Code:

%%js

var alphabet = "abcdefghijklmnopqrstuvwxyz";
var alphabetList = [];

for (var i = 0; i < 10; i++) {
	alphabetList.push(alphabet[i]); // push the alphabet at position i and not index i
}

var letterNumber = 5

for (var i = 0; i < alphabetList.length; i++) {
	if (i === letterNumber-1) {
		console.log(alphabetList[i] + " is letter number 5 in the alphabet")
	}
}

// Should output:
// "e" is letter number 5 in the alphabet
<IPython.core.display.Javascript object>

What I Changed

  • Changed the let to var (let was kinda suspicious)
  • the i iterator should be an integer, so i < alphabetList.length instead of list itself
  • changed to i == letterNumber-1 because of zero-indexing
  • log the actual element of the list at index i instead of letterNumber. Using letterNumber defeats the whole purpose of the loop

Segment 3: Odd Numbers

Intended behavior: print a list of all the odd numbers below 10

Code:

%%js

let evens = [];
let i = 0;

while (i <= 10) {
  evens.push(i);
  i += 2;
}

console.log(evens);
<IPython.core.display.Javascript object>

Altered Code:

%%js

let evens = [];
let i = 1; // changed to 1

while (i <= 10) {
  evens.push(i);
  i += 2;
}

console.log(evens);
<IPython.core.display.Javascript object>

What I Changed

The initial i-value should be 1 since we’re adding 2 each time (which preserves parity). We must start at the first odd number, not even. Otherwise the code is correct.

BELOW NOT EDITED

The intended outcome is printing a number between 1 and 100 once, if it is a multiple of 2 or 5

  • What values are outputted incorrectly. Why?
  • Make changes to get the intended outcome.
%%js

var numbers = []
var newNumbers = []
var i = 0

while (i < 100) {
    numbers.push(i)
    i += 1
}
for (var i of numbers) {
    if (numbers[i] % 5 === 0)
        newNumbers.push(numbers[i])
    if (numbers[i] % 2 === 0)
        newNumbers.push(numbers[i])
}
console.log(newNumbers) 


<IPython.core.display.Javascript object>

Altered Code:

%%js

var numbers = []
var newNumbers = []
var i = 0

while (i < 100) {
    numbers.push(i)
    i += 1
}
for (var i of numbers) {
    if (numbers[i] % 5 === 0)
        newNumbers.push(numbers[i])
    else if (numbers[i] % 2 === 0)
        newNumbers.push(numbers[i])
}
console.log(newNumbers) 
<IPython.core.display.Javascript object>

What I Changed:

Just changed the second if to else-if because we don’t want the conditional (divisible by 2) to run/print again if numbers[i] is already divisible by 5.

Challenge

This code segment is at a very early stage of implementation.

  • What are some ways to (user) error proof this code?
  • The code should be able to calculate the cost of the meal of the user

Hint:

  • write a “single” test describing an expectation of the program of the program
  • test - input burger, expect output of burger price
  • run the test, which should fail because the program lacks that feature
  • write “just enough” code, the simplest possible, to make the test pass

Then repeat this process until you get program working like you want it to work.

%%js

var menu =  {"burger": 3.99,
         "fries": 1.99,
         "drink": 0.99}
var total = 0

//shows the user the menu and prompts them to select an item
console.log("Menu")
for (var item in menu) {
    console.log(item + "  $" + menu[item].toFixed(2)) //why is toFixed used?
}
//ideally the code should support mutliple items
var item = "burger"

//code should add the price of the menu items selected by the user 
console.log(total)

Hacks

  • Fix the errors in the first three segments in this notebook and say what you changed in the code cell under “What I Changed” (Challenge is optional)