Finding myself at a standstill with the first step of the Codecademy FizzBuzz app

Question

Finding myself at a standstill with the first step of the Codecademy FizzBuzz app

Let me share my code from the FizzBuzz lesson on Codecademy:

var i;
for ( i = 1; i > 20; i++ ) {
  "hello"
  if ( i % 3 === 0 ) {
    if ( i % 5 === 0 ) {
      "FizzBuzz";
    }
    else {
      "Fizz";
    }
  }
  else if ( i % 5 === 0 ) {
    "Buzz";
  }
 else {
    i;
  }
}

In my code, I aim to check if a number (i) is divisible by 3 before determining if it's also divisible by 5. If both conditions hold true, it should output "FizzBuzz". For cases where only the first condition applies, it's supposed to print "Fizz". If neither are divisible, but i is divisible by 5, it will display "Buzz". And as a last resort, it simply shows the number itself.

However, upon running the code, the results were not what I expected at all! Can you spot any glaring mistakes that I might have overlooked?

javascript fizzbuzz

Answer 1

Answer №1

To begin, your loop is not starting as intended:

for ( i = 1; i > 20; i++ )

because the initial condition is already false. It seems like you meant to do this instead:

for ( i = 1; i <= 20; i++ )

"FizzBuzz";

is merely a string that JavaScript is disregarding. You must find a way to display this string:

console.log("FizzBuzz");

Furthermore, this portion

else {
    i;
  }

does not have any purpose. Were you trying to show numbers that are not divisible by 3 or 5?

else {
    console.log(i);
  }

Also, why is there a "hello" at the start of the loop?

On a brighter note, I appreciate that you are using strict equality:

if ( i % 5 === 0 )

This practice is commendable since the non-strict equality operator == can lead to unexpected implicit conversions. Always opt for strict equality unless you intentionally want those conversions to occur.

Answer 2

To begin, your loop is not starting as intended:

for ( i = 1; i > 20; i++ )

because the initial condition is already false. It seems like you meant to do this instead:

for ( i = 1; i <= 20; i++ )

"FizzBuzz";

is merely a string that JavaScript is disregarding. You must find a way to display this string:

console.log("FizzBuzz");

Furthermore, this portion

else {
    i;
  }

does not have any purpose. Were you trying to show numbers that are not divisible by 3 or 5?

else {
    console.log(i);
  }

Also, why is there a "hello" at the start of the loop?

On a brighter note, I appreciate that you are using strict equality:

if ( i % 5 === 0 )

This practice is commendable since the non-strict equality operator == can lead to unexpected implicit conversions. Always opt for strict equality unless you intentionally want those conversions to occur.

Answer 3

Answer №2

The main issues you are facing include misunderstanding of how the for loop works and the misconception that statements like "somestring" or i actually execute any actions. The correct approach is to output the information to the console (or another output stream) depending on the runtime environment of your Javascript.

It's worth noting that numbers divisible by both three and five are multiples of 15, which can simplify the code implementation.

To streamline your code, consider using a structure like this:

for each number in a specified range:
    if the number is divisible by 15:
        print "FizzBuzz"
        continue looping

    if the number is divisible by 3:
        print "Fizz"
        continue looping

    if the number is divisible by 5:
        print "Buzz"
        continue looping

    print the number itself

Some may argue against having multiple exit points in a loop, but as long as you maintain clear control flow within a concise structure, it eliminates the risk of creating spaghetti code.

Here's the corresponding Javascript code embedded in a webpage for testing purposes:

<html><head></head><body><script type="text/javascript">
    var i;
    for (i = 1; i <= 20; i++) {
        if (i % 15 === 0) {
            document.write ("FizzBuzz<br>");
            continue;
        };  

        if (i % 3 === 0) {
            document.write ("Fizz<br>");
            continue;
        };  

        if (i % 5 === 0) {
            document.write ("Buzz<br>");
            continue;
        };  

        document.write (i + "<br>");
    }   
</script></body></html>

When run, the script outputs the expected sequence:

1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz
16
17
Fizz
19
Buzz

Answer 4

The main issues you are facing include misunderstanding of how the for loop works and the misconception that statements like "somestring" or i actually execute any actions. The correct approach is to output the information to the console (or another output stream) depending on the runtime environment of your Javascript.

It's worth noting that numbers divisible by both three and five are multiples of 15, which can simplify the code implementation.

To streamline your code, consider using a structure like this:

for each number in a specified range:
    if the number is divisible by 15:
        print "FizzBuzz"
        continue looping

    if the number is divisible by 3:
        print "Fizz"
        continue looping

    if the number is divisible by 5:
        print "Buzz"
        continue looping

    print the number itself

Some may argue against having multiple exit points in a loop, but as long as you maintain clear control flow within a concise structure, it eliminates the risk of creating spaghetti code.

Here's the corresponding Javascript code embedded in a webpage for testing purposes:

<html><head></head><body><script type="text/javascript">
    var i;
    for (i = 1; i <= 20; i++) {
        if (i % 15 === 0) {
            document.write ("FizzBuzz<br>");
            continue;
        };  

        if (i % 3 === 0) {
            document.write ("Fizz<br>");
            continue;
        };  

        if (i % 5 === 0) {
            document.write ("Buzz<br>");
            continue;
        };  

        document.write (i + "<br>");
    }   
</script></body></html>

When run, the script outputs the expected sequence:

1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz
16
17
Fizz
19
Buzz

Answer 5

Answer №3

for (let x = 1; x <= 30; x++) {
    if (x % 21 === 0) {
        console.log("FizzBuzzPop");
    }
    else if (x % 3 === 0) {
        console.log("FizzPop");
    }
    else if (x % 7 === 0) {
        console.log("Pop");
    }
    else{
        console.log(x);
    };
}

Answer 6

for (let x = 1; x <= 30; x++) {
    if (x % 21 === 0) {
        console.log("FizzBuzzPop");
    }
    else if (x % 3 === 0) {
        console.log("FizzPop");
    }
    else if (x % 7 === 0) {
        console.log("Pop");
    }
    else{
        console.log(x);
    };
}

Answer 7

Answer №4

After reviewing all the insightful responses given here:

It appears that you may be experiencing difficulties with step 1 based on the code provided. I made a similar mistake when I followed your link and checked the instructions. Step 1 does not require solving the Fizzbuzz problem. Instead, it involves a much simpler task. I suggest revisiting the (not very clear) instructions once more ;)

Answer 8

After reviewing all the insightful responses given here:

It appears that you may be experiencing difficulties with step 1 based on the code provided. I made a similar mistake when I followed your link and checked the instructions. Step 1 does not require solving the Fizzbuzz problem. Instead, it involves a much simpler task. I suggest revisiting the (not very clear) instructions once more ;)

Answer 9

Answer №5

When using the statement <code>for ( i = 1; i > 20; i++ )

, it actually means that the program will not execute any action as the condition is incorrect. To make the variable i start at 1 and end at 20, you should modify it to for( i = 1; i <= 20; i++). Additionally, if you intend to test a specific number, you can achieve this by creating a function like:

function TestFizzBuzz(num){
    ...
    ...
}
TestFizzBuzz(1);
TestFizzBuzz(990);
...

Answer 10

When using the statement <code>for ( i = 1; i > 20; i++ )

, it actually means that the program will not execute any action as the condition is incorrect. To make the variable i start at 1 and end at 20, you should modify it to for( i = 1; i <= 20; i++). Additionally, if you intend to test a specific number, you can achieve this by creating a function like:

function TestFizzBuzz(num){
    ...
    ...
}
TestFizzBuzz(1);
TestFizzBuzz(990);
...

Answer 11

Answer №6

What if we increase the level of challenge? 1) Eliminate division and modulo operations; 2) Optimize the loop to avoid unnecessary iterations. Here's a unique solution:

int num3 = 3;
int num5 = 5;
int index = 3;

while (index <= 100)
{
    Console.Write(index.ToString() + " - ");

    if (index == num3)
    {
        Console.Write("fizz");

        num3 += 3;
    }

    if (index == num5)
    {
        Console.Write("buzz");

        num5 += 5;
    }

    Console.WriteLine();

    index = num3 < num5 ? num3 : num5;
}

Answer 12

What if we increase the level of challenge? 1) Eliminate division and modulo operations; 2) Optimize the loop to avoid unnecessary iterations. Here's a unique solution:

int num3 = 3;
int num5 = 5;
int index = 3;

while (index <= 100)
{
    Console.Write(index.ToString() + " - ");

    if (index == num3)
    {
        Console.Write("fizz");

        num3 += 3;
    }

    if (index == num5)
    {
        Console.Write("buzz");

        num5 += 5;
    }

    Console.WriteLine();

    index = num3 < num5 ? num3 : num5;
}

Answer 13

Answer №7

This is my approach:

let numbers = new Array();

for (let j = 0; j < 20; j++){
    numbers[j] = j + 1;
}

for (let k = 0; k < 20; k++){
    if((numbers[k] % 5 == 0) && (numbers[k] % 3 == 0)){
        console.log("FizzBuzz");
    }else if(numbers[k] % 5 == 0){
        console.log("Buzz");
    }else if (numbers[k] % 3 == 0){
        console.log("Fizz");
    }else{
        console.log(numbers[k]);
    }
}

Answer 14

This is my approach:

let numbers = new Array();

for (let j = 0; j < 20; j++){
    numbers[j] = j + 1;
}

for (let k = 0; k < 20; k++){
    if((numbers[k] % 5 == 0) && (numbers[k] % 3 == 0)){
        console.log("FizzBuzz");
    }else if(numbers[k] % 5 == 0){
        console.log("Buzz");
    }else if (numbers[k] % 3 == 0){
        console.log("Fizz");
    }else{
        console.log(numbers[k]);
    }
}

Answer 15

Answer №8

To keep it concise, let's achieve this in a single line of code:

for (i=1;i<21;i++){console.log(i+": "+(i%3?(i%5?i:'Buzz'):(i%5?'Fizz':'FizzBuzz')));};

Answer 16

To keep it concise, let's achieve this in a single line of code:

for (i=1;i<21;i++){console.log(i+": "+(i%3?(i%5?i:'Buzz'):(i%5?'Fizz':'FizzBuzz')));};

Answer 17

Answer №9

  Let's write a simple FizzBuzz program using JavaScript! 

  for(var num = 1; num <= 20; num++){
    if(num % 15 === 0){
        console.log("FizzBuzz");
    }
    else if(num % 5 === 0){
        console.log("Buzz");
    }
    else if(num % 3 === 0){
        console.log("Fizz");
    }
    else {
        console.log(num);
    }
  }

Answer 18

  Let's write a simple FizzBuzz program using JavaScript! 

  for(var num = 1; num <= 20; num++){
    if(num % 15 === 0){
        console.log("FizzBuzz");
    }
    else if(num % 5 === 0){
        console.log("Buzz");
    }
    else if(num % 3 === 0){
        console.log("Fizz");
    }
    else {
        console.log(num);
    }
  }

Answer 19

Answer №10

The objective is to output "Fizz" for numbers that are perfectly divisible by 3 (with no remainder), "Buzz" for numbers divisible by 5, and "FizzBuzz" for numbers divisible by both 3 and 5; otherwise, it should display the number itself.

The modulo (%) operator gives the remainder of a division operation, so when (x % y) equals 0, it means the division is perfect without any remainder
Since modulo can yield 0 as a result, it's important to understand truthy and falsy values - 0 is considered falsy, hence we need to negate the test if we want to check for an actual 0 value

Example: !(3 % 3) => !(0) => !false => true

for (let i = 1; i <= 20; ++i) {
  if (!(i % 3) && !(i % 5)) { // Check if "i" is perfectly divisible by both 3 & 5
    console.log("FizzBuzz");
  } else if (!(i % 3)) {      // Check if "i" is only divisible by 3
    console.log("Fizz");
  } else if (!(i % 5)) {      // Check if "i" is only divisible by 5
    console.log("Buzz");
  } else {
    console.log(i);           // Output the number itself
  }
}

<script src="https://getfirebug.com/firebug-lite-debug.js"></script>

Answer 20

The objective is to output "Fizz" for numbers that are perfectly divisible by 3 (with no remainder), "Buzz" for numbers divisible by 5, and "FizzBuzz" for numbers divisible by both 3 and 5; otherwise, it should display the number itself.

The modulo (%) operator gives the remainder of a division operation, so when (x % y) equals 0, it means the division is perfect without any remainder
Since modulo can yield 0 as a result, it's important to understand truthy and falsy values - 0 is considered falsy, hence we need to negate the test if we want to check for an actual 0 value

Example: !(3 % 3) => !(0) => !false => true

for (let i = 1; i <= 20; ++i) {
  if (!(i % 3) && !(i % 5)) { // Check if "i" is perfectly divisible by both 3 & 5
    console.log("FizzBuzz");
  } else if (!(i % 3)) {      // Check if "i" is only divisible by 3
    console.log("Fizz");
  } else if (!(i % 5)) {      // Check if "i" is only divisible by 5
    console.log("Buzz");
  } else {
    console.log(i);           // Output the number itself
  }
}

<script src="https://getfirebug.com/firebug-lite-debug.js"></script>

Finding myself at a standstill with the first step of the Codecademy FizzBuzz app

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

Answer №8

Answer №9

Answer №10

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