if/else statements only logs if, not else












1















I am stuck on a problem for my coding homework.
Here it is:




  • Write a loop that logs "Marco!" when i is even,

  • "Polo!" when i is odd.

  • Do not edit the existing code.

  • Hint 1: Use an if/else statement

  • Hint 2: Google the mod operator (%)


My attempt



let x=11;

let y=4;

let i=x%y;



  if (i) {

 console.log("Marco!")

 }



  else  {

 console.log("Polo")

 }


This logs Marco when I need it to log polo. So while I continue to try and solve this I wanted to see how experts would do it.






javascript if-statement math modulo







share|improve this question




















  • 5





    Use % 2 to check if a number's even (to be more precise, % 2 === 0)

    – CertainPerformance
    Jan 2 at 0:52


















1















I am stuck on a problem for my coding homework.
Here it is:




  • Write a loop that logs "Marco!" when i is even,

  • "Polo!" when i is odd.

  • Do not edit the existing code.

  • Hint 1: Use an if/else statement

  • Hint 2: Google the mod operator (%)


My attempt



let x=11;

let y=4;

let i=x%y;



  if (i) {

 console.log("Marco!")

 }



  else  {

 console.log("Polo")

 }


This logs Marco when I need it to log polo. So while I continue to try and solve this I wanted to see how experts would do it.






javascript if-statement math modulo







share|improve this question




















  • 5





    Use % 2 to check if a number's even (to be more precise, % 2 === 0)

    – CertainPerformance
    Jan 2 at 0:52
















1












1








1








I am stuck on a problem for my coding homework.
Here it is:




  • Write a loop that logs "Marco!" when i is even,

  • "Polo!" when i is odd.

  • Do not edit the existing code.

  • Hint 1: Use an if/else statement

  • Hint 2: Google the mod operator (%)


My attempt



let x=11;

let y=4;

let i=x%y;



  if (i) {

 console.log("Marco!")

 }



  else  {

 console.log("Polo")

 }


This logs Marco when I need it to log polo. So while I continue to try and solve this I wanted to see how experts would do it.






javascript if-statement math modulo







share|improve this question
















I am stuck on a problem for my coding homework.
Here it is:




  • Write a loop that logs "Marco!" when i is even,

  • "Polo!" when i is odd.

  • Do not edit the existing code.

  • Hint 1: Use an if/else statement

  • Hint 2: Google the mod operator (%)


My attempt



let x=11;

let y=4;

let i=x%y;



  if (i) {

 console.log("Marco!")

 }



  else  {

 console.log("Polo")

 }


This logs Marco when I need it to log polo. So while I continue to try and solve this I wanted to see how experts would do it.






javascript if-statement math modulo




javascript if-statement math modulo






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Jan 2 at 2:13









Dale Burrell

3,36042655




3,36042655










asked Jan 2 at 0:50









Marco RMarco R

61




61








  • 5





    Use % 2 to check if a number's even (to be more precise, % 2 === 0)

    – CertainPerformance
    Jan 2 at 0:52
















  • 5





    Use % 2 to check if a number's even (to be more precise, % 2 === 0)

    – CertainPerformance
    Jan 2 at 0:52










5




5





Use % 2 to check if a number's even (to be more precise, % 2 === 0)

– CertainPerformance
Jan 2 at 0:52







Use % 2 to check if a number's even (to be more precise, % 2 === 0)

– CertainPerformance
Jan 2 at 0:52














1 Answer
1






active

oldest

votes


















1














If you want to check if a number is even or odd, use the modulo operator (%), which returns the remainder of dividing one number by the other. You should reverse your logic:






let x = 11;

let y = 4;

let i = x % y;



if (i % 2) {

  console.log("Polo!");

} else {

  console.log("Marco!");

}



console.log(i); //So you can see if the above works or not





Here's how this works:



let i = x % y;


What this does is it divides x by y (divides 11 by 4), and takes away the remainder - in this case the remainder would be 3, so i = 3.



Now, here comes the tricky bit. If you want to find out if a number is even, you can use % 2, which is what we're doing in the if statement. If the number is even, it will return 0 as there will be no remainder from dividing by two. It's tricky, but I'll show you as best I can:



If we have 6 (which we know is even), and we test if it is even by dividing it by 2, it should return 0 as there is no remainder:






console.log(6 % 2);





And this is how our logic in the first snippet works, only it uses Boolean truthy and falsy values. Falsy values are:



false

0

''

""

``

null

undefined

NaN


So if i is even, the modulo will return 0, meaning that the first if statement will not run because i % 2 will return 0 which evaluates to false, therefore the code will run console.log("Marco!") if i is even, but console.log("Polo!") if i is odd.



Further reading:




  • https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Arithmetic_Operators

  • How can I use modulo operator (%) in JavaScript?

  • https://developer.mozilla.org/en-US/docs/Glossary/Falsy

  • https://developer.mozilla.org/en-US/docs/Glossary/Truthy

  • https://www.sitepoint.com/javascript-truthy-falsy/






share|improve this answer























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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    If you want to check if a number is even or odd, use the modulo operator (%), which returns the remainder of dividing one number by the other. You should reverse your logic:






    let x = 11;
    
let y = 4;
    
let i = x % y;
    

    
if (i % 2) {
    
  console.log("Polo!");
    
} else {
    
  console.log("Marco!");
    
}
    

    
console.log(i); //So you can see if the above works or not





    Here's how this works:



    let i = x % y;


    What this does is it divides x by y (divides 11 by 4), and takes away the remainder - in this case the remainder would be 3, so i = 3.



    Now, here comes the tricky bit. If you want to find out if a number is even, you can use % 2, which is what we're doing in the if statement. If the number is even, it will return 0 as there will be no remainder from dividing by two. It's tricky, but I'll show you as best I can:



    If we have 6 (which we know is even), and we test if it is even by dividing it by 2, it should return 0 as there is no remainder:






    console.log(6 % 2);





    And this is how our logic in the first snippet works, only it uses Boolean truthy and falsy values. Falsy values are:



    false
    
0
    
''
    
""
    
``
    
null
    
undefined
    
NaN


    So if i is even, the modulo will return 0, meaning that the first if statement will not run because i % 2 will return 0 which evaluates to false, therefore the code will run console.log("Marco!") if i is even, but console.log("Polo!") if i is odd.



    Further reading:




    • https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Arithmetic_Operators

    • How can I use modulo operator (%) in JavaScript?

    • https://developer.mozilla.org/en-US/docs/Glossary/Falsy

    • https://developer.mozilla.org/en-US/docs/Glossary/Truthy

    • https://www.sitepoint.com/javascript-truthy-falsy/






    share|improve this answer




























      1














      If you want to check if a number is even or odd, use the modulo operator (%), which returns the remainder of dividing one number by the other. You should reverse your logic:






      let x = 11;
      
let y = 4;
      
let i = x % y;
      

      
if (i % 2) {
      
  console.log("Polo!");
      
} else {
      
  console.log("Marco!");
      
}
      

      
console.log(i); //So you can see if the above works or not





      Here's how this works:



      let i = x % y;


      What this does is it divides x by y (divides 11 by 4), and takes away the remainder - in this case the remainder would be 3, so i = 3.



      Now, here comes the tricky bit. If you want to find out if a number is even, you can use % 2, which is what we're doing in the if statement. If the number is even, it will return 0 as there will be no remainder from dividing by two. It's tricky, but I'll show you as best I can:



      If we have 6 (which we know is even), and we test if it is even by dividing it by 2, it should return 0 as there is no remainder:






      console.log(6 % 2);





      And this is how our logic in the first snippet works, only it uses Boolean truthy and falsy values. Falsy values are:



      false
      
0
      
''
      
""
      
``
      
null
      
undefined
      
NaN


      So if i is even, the modulo will return 0, meaning that the first if statement will not run because i % 2 will return 0 which evaluates to false, therefore the code will run console.log("Marco!") if i is even, but console.log("Polo!") if i is odd.



      Further reading:




      • https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Arithmetic_Operators

      • How can I use modulo operator (%) in JavaScript?

      • https://developer.mozilla.org/en-US/docs/Glossary/Falsy

      • https://developer.mozilla.org/en-US/docs/Glossary/Truthy

      • https://www.sitepoint.com/javascript-truthy-falsy/






      share|improve this answer


























        1












        1








        1







        If you want to check if a number is even or odd, use the modulo operator (%), which returns the remainder of dividing one number by the other. You should reverse your logic:






        let x = 11;
        
let y = 4;
        
let i = x % y;
        

        
if (i % 2) {
        
  console.log("Polo!");
        
} else {
        
  console.log("Marco!");
        
}
        

        
console.log(i); //So you can see if the above works or not





        Here's how this works:



        let i = x % y;


        What this does is it divides x by y (divides 11 by 4), and takes away the remainder - in this case the remainder would be 3, so i = 3.



        Now, here comes the tricky bit. If you want to find out if a number is even, you can use % 2, which is what we're doing in the if statement. If the number is even, it will return 0 as there will be no remainder from dividing by two. It's tricky, but I'll show you as best I can:



        If we have 6 (which we know is even), and we test if it is even by dividing it by 2, it should return 0 as there is no remainder:






        console.log(6 % 2);





        And this is how our logic in the first snippet works, only it uses Boolean truthy and falsy values. Falsy values are:



        false
        
0
        
''
        
""
        
``
        
null
        
undefined
        
NaN


        So if i is even, the modulo will return 0, meaning that the first if statement will not run because i % 2 will return 0 which evaluates to false, therefore the code will run console.log("Marco!") if i is even, but console.log("Polo!") if i is odd.



        Further reading:




        • https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Arithmetic_Operators

        • How can I use modulo operator (%) in JavaScript?

        • https://developer.mozilla.org/en-US/docs/Glossary/Falsy

        • https://developer.mozilla.org/en-US/docs/Glossary/Truthy

        • https://www.sitepoint.com/javascript-truthy-falsy/






        share|improve this answer













        If you want to check if a number is even or odd, use the modulo operator (%), which returns the remainder of dividing one number by the other. You should reverse your logic:






        let x = 11;
        
let y = 4;
        
let i = x % y;
        

        
if (i % 2) {
        
  console.log("Polo!");
        
} else {
        
  console.log("Marco!");
        
}
        

        
console.log(i); //So you can see if the above works or not





        Here's how this works:



        let i = x % y;


        What this does is it divides x by y (divides 11 by 4), and takes away the remainder - in this case the remainder would be 3, so i = 3.



        Now, here comes the tricky bit. If you want to find out if a number is even, you can use % 2, which is what we're doing in the if statement. If the number is even, it will return 0 as there will be no remainder from dividing by two. It's tricky, but I'll show you as best I can:



        If we have 6 (which we know is even), and we test if it is even by dividing it by 2, it should return 0 as there is no remainder:






        console.log(6 % 2);





        And this is how our logic in the first snippet works, only it uses Boolean truthy and falsy values. Falsy values are:



        false
        
0
        
''
        
""
        
``
        
null
        
undefined
        
NaN


        So if i is even, the modulo will return 0, meaning that the first if statement will not run because i % 2 will return 0 which evaluates to false, therefore the code will run console.log("Marco!") if i is even, but console.log("Polo!") if i is odd.



        Further reading:




        • https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Arithmetic_Operators

        • How can I use modulo operator (%) in JavaScript?

        • https://developer.mozilla.org/en-US/docs/Glossary/Falsy

        • https://developer.mozilla.org/en-US/docs/Glossary/Truthy

        • https://www.sitepoint.com/javascript-truthy-falsy/






        let x = 11;
        
let y = 4;
        
let i = x % y;
        

        
if (i % 2) {
        
  console.log("Polo!");
        
} else {
        
  console.log("Marco!");
        
}
        

        
console.log(i); //So you can see if the above works or not





        let x = 11;
        
let y = 4;
        
let i = x % y;
        

        
if (i % 2) {
        
  console.log("Polo!");
        
} else {
        
  console.log("Marco!");
        
}
        

        
console.log(i); //So you can see if the above works or not





        console.log(6 % 2);





        console.log(6 % 2);






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Jan 2 at 1:07









        Jack BashfordJack Bashford

        11.7k31846




        11.7k31846
































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