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
Home | HTML | Data Types | DOM | JavaScript | JS Debugging |
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 < alphabet.length; i++) {
alphabetList.push(alphabet[i]);
}
console.log(alphabetList);
<IPython.core.display.Javascript object>
What I Changed
I changed the loop condition from (var i = 0; i < 10; i++) to (var i = 0; i < alphabet.length; i++) to create a list for every character in the alphabet instead of just 10. Then I modified the code to push alphabet[i] instead of just i so it would add the characters into the list array. This pushes the character that corresponds with the i index in the string.
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 < alphabet.length; i++) {
alphabetList.push(alphabet[i]);
}
let letterNumber = 5;
for (var i = 0; i < alphabetList.length; i++) {
if (i === letterNumber - 1) {
console.log(alphabetList[i] + " is letter number " + letterNumber + " in the alphabet");
}
}
// Should output:
// "e" is letter number 5 in the alphabet
<IPython.core.display.Javascript object>
What I Changed
I changed the loop condition from i < alphabetList to i < alphabetList.length so that the loop will iterate through the whole length of the list. I also changed if (i === letterNumber) to if (i === letterNumber - 1) to account for the index starting at 0
Segment 3: Odd Numbers
Intended behavior: print a list of all the odd numbers below 10
Code:
%%js
let odds = [];
let i = 1;
while (i <= 10) {
odds.push(i);
i += 2;
}
console.log(odds);
<IPython.core.display.Javascript object>
What I Changed
I changed i to equal the first odd number and for each iteration, I then added the current value of i to the odds array and then incremented i by 2 to get the next odd number.
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)
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)