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Test whether all array elements are factors of a number


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7















I have the following question:




Write a function that returns true if all integers in an array are factors of a number, and false otherwise.




I tried the code below:






function checkFactors(factors, num) {

for (let i=0; i<factors.length; i++){
let element = factors[i];
console.log(element)

if (num % element !== 0){
return false
}
else {
return true
}
}
}

console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false





My solution returns true which is wrong. I know it's the else statement that's messing it up. But I want to understand why the else statement can't go there.










share|improve this question




















  • 2





    You should get the return true out of the loop ;) and leave the return false inside the loop :) The reason for it is that otherwise your loop really stops after the first checking, returning either true or false, and you want it to return false as soon as something is not a factor. The return true outside of the loop would then indicate that all numbers supplied where factors of the number argument

    – Icepickle
    13 hours ago








  • 2





    FYI, it's not just you, this is one of the most common errors people make when they first start out with programming and loops. :-)

    – T.J. Crowder
    13 hours ago






  • 1





    @t.j.crowder yet we don't have a good dupetarget for it.

    – Jonas Wilms
    12 hours ago






  • 2





    @JonasWilms - Yeah, I bookmarked this one on the basis that there isn't a lot of extraneous stuff, it really is just the earlier return thing. :-)

    – T.J. Crowder
    12 hours ago











  • @T.J.Crowder stackoverflow.com/questions/42913798/… is the general dupe I wrote for this class of problems.

    – Barmar
    7 hours ago




















7















I have the following question:




Write a function that returns true if all integers in an array are factors of a number, and false otherwise.




I tried the code below:






function checkFactors(factors, num) {

for (let i=0; i<factors.length; i++){
let element = factors[i];
console.log(element)

if (num % element !== 0){
return false
}
else {
return true
}
}
}

console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false





My solution returns true which is wrong. I know it's the else statement that's messing it up. But I want to understand why the else statement can't go there.










share|improve this question




















  • 2





    You should get the return true out of the loop ;) and leave the return false inside the loop :) The reason for it is that otherwise your loop really stops after the first checking, returning either true or false, and you want it to return false as soon as something is not a factor. The return true outside of the loop would then indicate that all numbers supplied where factors of the number argument

    – Icepickle
    13 hours ago








  • 2





    FYI, it's not just you, this is one of the most common errors people make when they first start out with programming and loops. :-)

    – T.J. Crowder
    13 hours ago






  • 1





    @t.j.crowder yet we don't have a good dupetarget for it.

    – Jonas Wilms
    12 hours ago






  • 2





    @JonasWilms - Yeah, I bookmarked this one on the basis that there isn't a lot of extraneous stuff, it really is just the earlier return thing. :-)

    – T.J. Crowder
    12 hours ago











  • @T.J.Crowder stackoverflow.com/questions/42913798/… is the general dupe I wrote for this class of problems.

    – Barmar
    7 hours ago
















7












7








7


1






I have the following question:




Write a function that returns true if all integers in an array are factors of a number, and false otherwise.




I tried the code below:






function checkFactors(factors, num) {

for (let i=0; i<factors.length; i++){
let element = factors[i];
console.log(element)

if (num % element !== 0){
return false
}
else {
return true
}
}
}

console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false





My solution returns true which is wrong. I know it's the else statement that's messing it up. But I want to understand why the else statement can't go there.










share|improve this question
















I have the following question:




Write a function that returns true if all integers in an array are factors of a number, and false otherwise.




I tried the code below:






function checkFactors(factors, num) {

for (let i=0; i<factors.length; i++){
let element = factors[i];
console.log(element)

if (num % element !== 0){
return false
}
else {
return true
}
}
}

console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false





My solution returns true which is wrong. I know it's the else statement that's messing it up. But I want to understand why the else statement can't go there.






function checkFactors(factors, num) {

for (let i=0; i<factors.length; i++){
let element = factors[i];
console.log(element)

if (num % element !== 0){
return false
}
else {
return true
}
}
}

console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false





function checkFactors(factors, num) {

for (let i=0; i<factors.length; i++){
let element = factors[i];
console.log(element)

if (num % element !== 0){
return false
}
else {
return true
}
}
}

console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false






javascript arrays loops if-statement






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 9 hours ago









isanae

2,53711437




2,53711437










asked 13 hours ago









PineNuts0PineNuts0

87431432




87431432








  • 2





    You should get the return true out of the loop ;) and leave the return false inside the loop :) The reason for it is that otherwise your loop really stops after the first checking, returning either true or false, and you want it to return false as soon as something is not a factor. The return true outside of the loop would then indicate that all numbers supplied where factors of the number argument

    – Icepickle
    13 hours ago








  • 2





    FYI, it's not just you, this is one of the most common errors people make when they first start out with programming and loops. :-)

    – T.J. Crowder
    13 hours ago






  • 1





    @t.j.crowder yet we don't have a good dupetarget for it.

    – Jonas Wilms
    12 hours ago






  • 2





    @JonasWilms - Yeah, I bookmarked this one on the basis that there isn't a lot of extraneous stuff, it really is just the earlier return thing. :-)

    – T.J. Crowder
    12 hours ago











  • @T.J.Crowder stackoverflow.com/questions/42913798/… is the general dupe I wrote for this class of problems.

    – Barmar
    7 hours ago
















  • 2





    You should get the return true out of the loop ;) and leave the return false inside the loop :) The reason for it is that otherwise your loop really stops after the first checking, returning either true or false, and you want it to return false as soon as something is not a factor. The return true outside of the loop would then indicate that all numbers supplied where factors of the number argument

    – Icepickle
    13 hours ago








  • 2





    FYI, it's not just you, this is one of the most common errors people make when they first start out with programming and loops. :-)

    – T.J. Crowder
    13 hours ago






  • 1





    @t.j.crowder yet we don't have a good dupetarget for it.

    – Jonas Wilms
    12 hours ago






  • 2





    @JonasWilms - Yeah, I bookmarked this one on the basis that there isn't a lot of extraneous stuff, it really is just the earlier return thing. :-)

    – T.J. Crowder
    12 hours ago











  • @T.J.Crowder stackoverflow.com/questions/42913798/… is the general dupe I wrote for this class of problems.

    – Barmar
    7 hours ago










2




2





You should get the return true out of the loop ;) and leave the return false inside the loop :) The reason for it is that otherwise your loop really stops after the first checking, returning either true or false, and you want it to return false as soon as something is not a factor. The return true outside of the loop would then indicate that all numbers supplied where factors of the number argument

– Icepickle
13 hours ago







You should get the return true out of the loop ;) and leave the return false inside the loop :) The reason for it is that otherwise your loop really stops after the first checking, returning either true or false, and you want it to return false as soon as something is not a factor. The return true outside of the loop would then indicate that all numbers supplied where factors of the number argument

– Icepickle
13 hours ago






2




2





FYI, it's not just you, this is one of the most common errors people make when they first start out with programming and loops. :-)

– T.J. Crowder
13 hours ago





FYI, it's not just you, this is one of the most common errors people make when they first start out with programming and loops. :-)

– T.J. Crowder
13 hours ago




1




1





@t.j.crowder yet we don't have a good dupetarget for it.

– Jonas Wilms
12 hours ago





@t.j.crowder yet we don't have a good dupetarget for it.

– Jonas Wilms
12 hours ago




2




2





@JonasWilms - Yeah, I bookmarked this one on the basis that there isn't a lot of extraneous stuff, it really is just the earlier return thing. :-)

– T.J. Crowder
12 hours ago





@JonasWilms - Yeah, I bookmarked this one on the basis that there isn't a lot of extraneous stuff, it really is just the earlier return thing. :-)

– T.J. Crowder
12 hours ago













@T.J.Crowder stackoverflow.com/questions/42913798/… is the general dupe I wrote for this class of problems.

– Barmar
7 hours ago







@T.J.Crowder stackoverflow.com/questions/42913798/… is the general dupe I wrote for this class of problems.

– Barmar
7 hours ago














5 Answers
5






active

oldest

votes


















5














Just place return true out of for loop,



If you keep return true in else part as soon as any of value which does not satisfies num % element !== 0 your code will return true which should not happen in this case as you're checking for all the values in array should be factor of given number




Let's understand by 1st example





  • On first element in array 1 it will check if condition num % element !== 0 which turns out false, so it will go to else condition and return true from function and will not check for rest of values.

  • So you need to keep return true at the end so if any of the value in loop doesn't satisfy the if condition than only control will go to return true





function checkFactors(factors, num) {

for (let i=0; i<factors.length; i++){
let element = factors[i];
if (num % element !== 0){
return false
}
}
return true
}



console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false
console.log(checkFactors([1, 2], 2))






In short - In such case where you want all of them must match a condition as a thumb rule you can consider it like





  1. keep the failing case return value inside for loop

  2. keep the passing case return value at the end of function


JS have a inbuilt method Array.every for such cases






function checkFactors(factors, num) {
return factors.every(element => num % element === 0);
}
console.log(checkFactors([1, 2, 3, 8], 12));
console.log(checkFactors([1, 2], 2));








share|improve this answer

































    7














    You are working in a chocolate store, and your boss tells you to check wether all chocolates (there are chili chocolate, caramel chocolate and coffee chocolate) are delicious. He tells you the following:



    Go through all chocolates, and for each chocolate, taste it, if it is fine, tell me that everything is fine, otherwise tell me that something is wrong¹



    You start with the first chocolate, which is chili chocolate, it tastes delucious, you go to your boss and tell him that everything is fine. Your boss yells at you because you haven't tasted the caramel chocolate and the coffee chocolate yet.



    You realize that your boss actually wanted you to do:



    Go through the chocolates, for each chocolate, taste it, if it doesnt taste well tell, tell me immeadiately, otherwise continue until you tasted them all, then return to me and tell me that everything is fine.²



    Or in code:



     // ¹
    function checkChocolates(chocolates) {
    for(const chocolate of chocolates) {
    if(isTasty(chocolate)) {
    return true;
    } else {
    return false;
    }
    }
    }

    // ²
    function checkChocolates(chocolates) {
    for(const chocolate of chocolates) {
    if(isTasty(chocolate)) {
    continue; // this could be omitted, as a loop keeps looping nevertheless
    } else {
    return false;
    }
    }
    return true;
    }


    As this is a very common task in programming, there is already a shorter way to express this:



     if(chocolates.every(isTasty)) {
    alert("all chocolates are fine");
    } else {
    alert("Oh, that doesnt taste good");
    }


    whereas isTasty is a function taking a chocolate and returning either true or false.





    If you didn't grasp it yet, just try it out! Buy some chocolate, and taste it! If someone tells you "eating choclate isn't learning", respond with "I'm doing rubber duck debugging" and no one can complain :)






    share|improve this answer





















    • 1





      A tiny quibble - it's "chocolate", not "choclate".

      – Wai Ha Lee
      12 hours ago






    • 1





      @waiHaLee oh, pronounciation tricked me ...

      – Jonas Wilms
      12 hours ago



















    1














    Inside the loop the input num was was tested for divisibility, if the num was divisible the control was going in the else block from where the function returned true.



    The loop was not checking for all numbers of the input array it was returning true when the first number was divisible.



    Just use a flag variable to see if all of the elements are divisible by the input number num, if any one is not divisible the flag will set to false and then we can break out of the loop and return it as there is no point to check the other numbers.






    function checkFactors(factors, num) {
    let flag = true;
    for (let i=0; i<factors.length; i++){
    let element = factors[i];
    if (num % element !== 0){
    flag = false;
    break;
    }
    }
    return flag;
    }

    console.log(checkFactors([1, 2, 3, 8], 12));
    console.log(checkFactors([1, 2], 2));
    console.log(checkFactors([2, 4, 3, 6, 9], 12));
    console.log(checkFactors([3, 5, 2, 6, 9], 15));
    console.log(checkFactors([4, 2, 8, 1], 16));





    You can also use Array.every to check the same in a concise way:






    function checkFactors(factors, num) {
    return factors.every(element => num % element === 0);
    }
    console.log(checkFactors([1, 2, 3, 8], 12));
    console.log(checkFactors([1, 2], 2));
    console.log(checkFactors([2, 4, 3, 6, 9], 12));
    console.log(checkFactors([3, 5, 2, 6, 9], 15));
    console.log(checkFactors([4, 2, 8, 1], 16));








    share|improve this answer

































      0














      Yes, "else" causing the issue. I removed it and added "return true" outside of for loop.



      function checkFactors(factors, num) {

      for (let i=0; i<factors.length; i++){
      let element = factors[i];
      console.log(element)

      if (num % element !== 0){
      return false
      }
      }
      return true;
      }





      share|improve this answer



















      • 1





        He is looking for an explanation as well though

        – Icepickle
        12 hours ago











      • I see Icepickle has explained in the question comments nicely! Jonas Wilms's answer too.

        – javapedia.net
        12 hours ago





















      0














      Your code's logic is wrong. You should check all the array's Elements, if all elements satisfy the condition, return true, but if one of them not satisfy the condition, return false immediately. The else means one item satisfy the condition, but not all elements. That's where the problem is.






      share|improve this answer








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        5 Answers
        5






        active

        oldest

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        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        5














        Just place return true out of for loop,



        If you keep return true in else part as soon as any of value which does not satisfies num % element !== 0 your code will return true which should not happen in this case as you're checking for all the values in array should be factor of given number




        Let's understand by 1st example





        • On first element in array 1 it will check if condition num % element !== 0 which turns out false, so it will go to else condition and return true from function and will not check for rest of values.

        • So you need to keep return true at the end so if any of the value in loop doesn't satisfy the if condition than only control will go to return true





        function checkFactors(factors, num) {

        for (let i=0; i<factors.length; i++){
        let element = factors[i];
        if (num % element !== 0){
        return false
        }
        }
        return true
        }



        console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false
        console.log(checkFactors([1, 2], 2))






        In short - In such case where you want all of them must match a condition as a thumb rule you can consider it like





        1. keep the failing case return value inside for loop

        2. keep the passing case return value at the end of function


        JS have a inbuilt method Array.every for such cases






        function checkFactors(factors, num) {
        return factors.every(element => num % element === 0);
        }
        console.log(checkFactors([1, 2, 3, 8], 12));
        console.log(checkFactors([1, 2], 2));








        share|improve this answer






























          5














          Just place return true out of for loop,



          If you keep return true in else part as soon as any of value which does not satisfies num % element !== 0 your code will return true which should not happen in this case as you're checking for all the values in array should be factor of given number




          Let's understand by 1st example





          • On first element in array 1 it will check if condition num % element !== 0 which turns out false, so it will go to else condition and return true from function and will not check for rest of values.

          • So you need to keep return true at the end so if any of the value in loop doesn't satisfy the if condition than only control will go to return true





          function checkFactors(factors, num) {

          for (let i=0; i<factors.length; i++){
          let element = factors[i];
          if (num % element !== 0){
          return false
          }
          }
          return true
          }



          console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false
          console.log(checkFactors([1, 2], 2))






          In short - In such case where you want all of them must match a condition as a thumb rule you can consider it like





          1. keep the failing case return value inside for loop

          2. keep the passing case return value at the end of function


          JS have a inbuilt method Array.every for such cases






          function checkFactors(factors, num) {
          return factors.every(element => num % element === 0);
          }
          console.log(checkFactors([1, 2, 3, 8], 12));
          console.log(checkFactors([1, 2], 2));








          share|improve this answer




























            5












            5








            5







            Just place return true out of for loop,



            If you keep return true in else part as soon as any of value which does not satisfies num % element !== 0 your code will return true which should not happen in this case as you're checking for all the values in array should be factor of given number




            Let's understand by 1st example





            • On first element in array 1 it will check if condition num % element !== 0 which turns out false, so it will go to else condition and return true from function and will not check for rest of values.

            • So you need to keep return true at the end so if any of the value in loop doesn't satisfy the if condition than only control will go to return true





            function checkFactors(factors, num) {

            for (let i=0; i<factors.length; i++){
            let element = factors[i];
            if (num % element !== 0){
            return false
            }
            }
            return true
            }



            console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false
            console.log(checkFactors([1, 2], 2))






            In short - In such case where you want all of them must match a condition as a thumb rule you can consider it like





            1. keep the failing case return value inside for loop

            2. keep the passing case return value at the end of function


            JS have a inbuilt method Array.every for such cases






            function checkFactors(factors, num) {
            return factors.every(element => num % element === 0);
            }
            console.log(checkFactors([1, 2, 3, 8], 12));
            console.log(checkFactors([1, 2], 2));








            share|improve this answer















            Just place return true out of for loop,



            If you keep return true in else part as soon as any of value which does not satisfies num % element !== 0 your code will return true which should not happen in this case as you're checking for all the values in array should be factor of given number




            Let's understand by 1st example





            • On first element in array 1 it will check if condition num % element !== 0 which turns out false, so it will go to else condition and return true from function and will not check for rest of values.

            • So you need to keep return true at the end so if any of the value in loop doesn't satisfy the if condition than only control will go to return true





            function checkFactors(factors, num) {

            for (let i=0; i<factors.length; i++){
            let element = factors[i];
            if (num % element !== 0){
            return false
            }
            }
            return true
            }



            console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false
            console.log(checkFactors([1, 2], 2))






            In short - In such case where you want all of them must match a condition as a thumb rule you can consider it like





            1. keep the failing case return value inside for loop

            2. keep the passing case return value at the end of function


            JS have a inbuilt method Array.every for such cases






            function checkFactors(factors, num) {
            return factors.every(element => num % element === 0);
            }
            console.log(checkFactors([1, 2, 3, 8], 12));
            console.log(checkFactors([1, 2], 2));








            function checkFactors(factors, num) {

            for (let i=0; i<factors.length; i++){
            let element = factors[i];
            if (num % element !== 0){
            return false
            }
            }
            return true
            }



            console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false
            console.log(checkFactors([1, 2], 2))





            function checkFactors(factors, num) {

            for (let i=0; i<factors.length; i++){
            let element = factors[i];
            if (num % element !== 0){
            return false
            }
            }
            return true
            }



            console.log(checkFactors([1, 2, 3, 8], 12)) //➞ false
            console.log(checkFactors([1, 2], 2))





            function checkFactors(factors, num) {
            return factors.every(element => num % element === 0);
            }
            console.log(checkFactors([1, 2, 3, 8], 12));
            console.log(checkFactors([1, 2], 2));





            function checkFactors(factors, num) {
            return factors.every(element => num % element === 0);
            }
            console.log(checkFactors([1, 2, 3, 8], 12));
            console.log(checkFactors([1, 2], 2));






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 9 hours ago

























            answered 13 hours ago









            Code ManiacCode Maniac

            11.5k2833




            11.5k2833

























                7














                You are working in a chocolate store, and your boss tells you to check wether all chocolates (there are chili chocolate, caramel chocolate and coffee chocolate) are delicious. He tells you the following:



                Go through all chocolates, and for each chocolate, taste it, if it is fine, tell me that everything is fine, otherwise tell me that something is wrong¹



                You start with the first chocolate, which is chili chocolate, it tastes delucious, you go to your boss and tell him that everything is fine. Your boss yells at you because you haven't tasted the caramel chocolate and the coffee chocolate yet.



                You realize that your boss actually wanted you to do:



                Go through the chocolates, for each chocolate, taste it, if it doesnt taste well tell, tell me immeadiately, otherwise continue until you tasted them all, then return to me and tell me that everything is fine.²



                Or in code:



                 // ¹
                function checkChocolates(chocolates) {
                for(const chocolate of chocolates) {
                if(isTasty(chocolate)) {
                return true;
                } else {
                return false;
                }
                }
                }

                // ²
                function checkChocolates(chocolates) {
                for(const chocolate of chocolates) {
                if(isTasty(chocolate)) {
                continue; // this could be omitted, as a loop keeps looping nevertheless
                } else {
                return false;
                }
                }
                return true;
                }


                As this is a very common task in programming, there is already a shorter way to express this:



                 if(chocolates.every(isTasty)) {
                alert("all chocolates are fine");
                } else {
                alert("Oh, that doesnt taste good");
                }


                whereas isTasty is a function taking a chocolate and returning either true or false.





                If you didn't grasp it yet, just try it out! Buy some chocolate, and taste it! If someone tells you "eating choclate isn't learning", respond with "I'm doing rubber duck debugging" and no one can complain :)






                share|improve this answer





















                • 1





                  A tiny quibble - it's "chocolate", not "choclate".

                  – Wai Ha Lee
                  12 hours ago






                • 1





                  @waiHaLee oh, pronounciation tricked me ...

                  – Jonas Wilms
                  12 hours ago
















                7














                You are working in a chocolate store, and your boss tells you to check wether all chocolates (there are chili chocolate, caramel chocolate and coffee chocolate) are delicious. He tells you the following:



                Go through all chocolates, and for each chocolate, taste it, if it is fine, tell me that everything is fine, otherwise tell me that something is wrong¹



                You start with the first chocolate, which is chili chocolate, it tastes delucious, you go to your boss and tell him that everything is fine. Your boss yells at you because you haven't tasted the caramel chocolate and the coffee chocolate yet.



                You realize that your boss actually wanted you to do:



                Go through the chocolates, for each chocolate, taste it, if it doesnt taste well tell, tell me immeadiately, otherwise continue until you tasted them all, then return to me and tell me that everything is fine.²



                Or in code:



                 // ¹
                function checkChocolates(chocolates) {
                for(const chocolate of chocolates) {
                if(isTasty(chocolate)) {
                return true;
                } else {
                return false;
                }
                }
                }

                // ²
                function checkChocolates(chocolates) {
                for(const chocolate of chocolates) {
                if(isTasty(chocolate)) {
                continue; // this could be omitted, as a loop keeps looping nevertheless
                } else {
                return false;
                }
                }
                return true;
                }


                As this is a very common task in programming, there is already a shorter way to express this:



                 if(chocolates.every(isTasty)) {
                alert("all chocolates are fine");
                } else {
                alert("Oh, that doesnt taste good");
                }


                whereas isTasty is a function taking a chocolate and returning either true or false.





                If you didn't grasp it yet, just try it out! Buy some chocolate, and taste it! If someone tells you "eating choclate isn't learning", respond with "I'm doing rubber duck debugging" and no one can complain :)






                share|improve this answer





















                • 1





                  A tiny quibble - it's "chocolate", not "choclate".

                  – Wai Ha Lee
                  12 hours ago






                • 1





                  @waiHaLee oh, pronounciation tricked me ...

                  – Jonas Wilms
                  12 hours ago














                7












                7








                7







                You are working in a chocolate store, and your boss tells you to check wether all chocolates (there are chili chocolate, caramel chocolate and coffee chocolate) are delicious. He tells you the following:



                Go through all chocolates, and for each chocolate, taste it, if it is fine, tell me that everything is fine, otherwise tell me that something is wrong¹



                You start with the first chocolate, which is chili chocolate, it tastes delucious, you go to your boss and tell him that everything is fine. Your boss yells at you because you haven't tasted the caramel chocolate and the coffee chocolate yet.



                You realize that your boss actually wanted you to do:



                Go through the chocolates, for each chocolate, taste it, if it doesnt taste well tell, tell me immeadiately, otherwise continue until you tasted them all, then return to me and tell me that everything is fine.²



                Or in code:



                 // ¹
                function checkChocolates(chocolates) {
                for(const chocolate of chocolates) {
                if(isTasty(chocolate)) {
                return true;
                } else {
                return false;
                }
                }
                }

                // ²
                function checkChocolates(chocolates) {
                for(const chocolate of chocolates) {
                if(isTasty(chocolate)) {
                continue; // this could be omitted, as a loop keeps looping nevertheless
                } else {
                return false;
                }
                }
                return true;
                }


                As this is a very common task in programming, there is already a shorter way to express this:



                 if(chocolates.every(isTasty)) {
                alert("all chocolates are fine");
                } else {
                alert("Oh, that doesnt taste good");
                }


                whereas isTasty is a function taking a chocolate and returning either true or false.





                If you didn't grasp it yet, just try it out! Buy some chocolate, and taste it! If someone tells you "eating choclate isn't learning", respond with "I'm doing rubber duck debugging" and no one can complain :)






                share|improve this answer















                You are working in a chocolate store, and your boss tells you to check wether all chocolates (there are chili chocolate, caramel chocolate and coffee chocolate) are delicious. He tells you the following:



                Go through all chocolates, and for each chocolate, taste it, if it is fine, tell me that everything is fine, otherwise tell me that something is wrong¹



                You start with the first chocolate, which is chili chocolate, it tastes delucious, you go to your boss and tell him that everything is fine. Your boss yells at you because you haven't tasted the caramel chocolate and the coffee chocolate yet.



                You realize that your boss actually wanted you to do:



                Go through the chocolates, for each chocolate, taste it, if it doesnt taste well tell, tell me immeadiately, otherwise continue until you tasted them all, then return to me and tell me that everything is fine.²



                Or in code:



                 // ¹
                function checkChocolates(chocolates) {
                for(const chocolate of chocolates) {
                if(isTasty(chocolate)) {
                return true;
                } else {
                return false;
                }
                }
                }

                // ²
                function checkChocolates(chocolates) {
                for(const chocolate of chocolates) {
                if(isTasty(chocolate)) {
                continue; // this could be omitted, as a loop keeps looping nevertheless
                } else {
                return false;
                }
                }
                return true;
                }


                As this is a very common task in programming, there is already a shorter way to express this:



                 if(chocolates.every(isTasty)) {
                alert("all chocolates are fine");
                } else {
                alert("Oh, that doesnt taste good");
                }


                whereas isTasty is a function taking a chocolate and returning either true or false.





                If you didn't grasp it yet, just try it out! Buy some chocolate, and taste it! If someone tells you "eating choclate isn't learning", respond with "I'm doing rubber duck debugging" and no one can complain :)







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 12 hours ago

























                answered 12 hours ago









                Jonas WilmsJonas Wilms

                65.4k53660




                65.4k53660








                • 1





                  A tiny quibble - it's "chocolate", not "choclate".

                  – Wai Ha Lee
                  12 hours ago






                • 1





                  @waiHaLee oh, pronounciation tricked me ...

                  – Jonas Wilms
                  12 hours ago














                • 1





                  A tiny quibble - it's "chocolate", not "choclate".

                  – Wai Ha Lee
                  12 hours ago






                • 1





                  @waiHaLee oh, pronounciation tricked me ...

                  – Jonas Wilms
                  12 hours ago








                1




                1





                A tiny quibble - it's "chocolate", not "choclate".

                – Wai Ha Lee
                12 hours ago





                A tiny quibble - it's "chocolate", not "choclate".

                – Wai Ha Lee
                12 hours ago




                1




                1





                @waiHaLee oh, pronounciation tricked me ...

                – Jonas Wilms
                12 hours ago





                @waiHaLee oh, pronounciation tricked me ...

                – Jonas Wilms
                12 hours ago











                1














                Inside the loop the input num was was tested for divisibility, if the num was divisible the control was going in the else block from where the function returned true.



                The loop was not checking for all numbers of the input array it was returning true when the first number was divisible.



                Just use a flag variable to see if all of the elements are divisible by the input number num, if any one is not divisible the flag will set to false and then we can break out of the loop and return it as there is no point to check the other numbers.






                function checkFactors(factors, num) {
                let flag = true;
                for (let i=0; i<factors.length; i++){
                let element = factors[i];
                if (num % element !== 0){
                flag = false;
                break;
                }
                }
                return flag;
                }

                console.log(checkFactors([1, 2, 3, 8], 12));
                console.log(checkFactors([1, 2], 2));
                console.log(checkFactors([2, 4, 3, 6, 9], 12));
                console.log(checkFactors([3, 5, 2, 6, 9], 15));
                console.log(checkFactors([4, 2, 8, 1], 16));





                You can also use Array.every to check the same in a concise way:






                function checkFactors(factors, num) {
                return factors.every(element => num % element === 0);
                }
                console.log(checkFactors([1, 2, 3, 8], 12));
                console.log(checkFactors([1, 2], 2));
                console.log(checkFactors([2, 4, 3, 6, 9], 12));
                console.log(checkFactors([3, 5, 2, 6, 9], 15));
                console.log(checkFactors([4, 2, 8, 1], 16));








                share|improve this answer






























                  1














                  Inside the loop the input num was was tested for divisibility, if the num was divisible the control was going in the else block from where the function returned true.



                  The loop was not checking for all numbers of the input array it was returning true when the first number was divisible.



                  Just use a flag variable to see if all of the elements are divisible by the input number num, if any one is not divisible the flag will set to false and then we can break out of the loop and return it as there is no point to check the other numbers.






                  function checkFactors(factors, num) {
                  let flag = true;
                  for (let i=0; i<factors.length; i++){
                  let element = factors[i];
                  if (num % element !== 0){
                  flag = false;
                  break;
                  }
                  }
                  return flag;
                  }

                  console.log(checkFactors([1, 2, 3, 8], 12));
                  console.log(checkFactors([1, 2], 2));
                  console.log(checkFactors([2, 4, 3, 6, 9], 12));
                  console.log(checkFactors([3, 5, 2, 6, 9], 15));
                  console.log(checkFactors([4, 2, 8, 1], 16));





                  You can also use Array.every to check the same in a concise way:






                  function checkFactors(factors, num) {
                  return factors.every(element => num % element === 0);
                  }
                  console.log(checkFactors([1, 2, 3, 8], 12));
                  console.log(checkFactors([1, 2], 2));
                  console.log(checkFactors([2, 4, 3, 6, 9], 12));
                  console.log(checkFactors([3, 5, 2, 6, 9], 15));
                  console.log(checkFactors([4, 2, 8, 1], 16));








                  share|improve this answer




























                    1












                    1








                    1







                    Inside the loop the input num was was tested for divisibility, if the num was divisible the control was going in the else block from where the function returned true.



                    The loop was not checking for all numbers of the input array it was returning true when the first number was divisible.



                    Just use a flag variable to see if all of the elements are divisible by the input number num, if any one is not divisible the flag will set to false and then we can break out of the loop and return it as there is no point to check the other numbers.






                    function checkFactors(factors, num) {
                    let flag = true;
                    for (let i=0; i<factors.length; i++){
                    let element = factors[i];
                    if (num % element !== 0){
                    flag = false;
                    break;
                    }
                    }
                    return flag;
                    }

                    console.log(checkFactors([1, 2, 3, 8], 12));
                    console.log(checkFactors([1, 2], 2));
                    console.log(checkFactors([2, 4, 3, 6, 9], 12));
                    console.log(checkFactors([3, 5, 2, 6, 9], 15));
                    console.log(checkFactors([4, 2, 8, 1], 16));





                    You can also use Array.every to check the same in a concise way:






                    function checkFactors(factors, num) {
                    return factors.every(element => num % element === 0);
                    }
                    console.log(checkFactors([1, 2, 3, 8], 12));
                    console.log(checkFactors([1, 2], 2));
                    console.log(checkFactors([2, 4, 3, 6, 9], 12));
                    console.log(checkFactors([3, 5, 2, 6, 9], 15));
                    console.log(checkFactors([4, 2, 8, 1], 16));








                    share|improve this answer















                    Inside the loop the input num was was tested for divisibility, if the num was divisible the control was going in the else block from where the function returned true.



                    The loop was not checking for all numbers of the input array it was returning true when the first number was divisible.



                    Just use a flag variable to see if all of the elements are divisible by the input number num, if any one is not divisible the flag will set to false and then we can break out of the loop and return it as there is no point to check the other numbers.






                    function checkFactors(factors, num) {
                    let flag = true;
                    for (let i=0; i<factors.length; i++){
                    let element = factors[i];
                    if (num % element !== 0){
                    flag = false;
                    break;
                    }
                    }
                    return flag;
                    }

                    console.log(checkFactors([1, 2, 3, 8], 12));
                    console.log(checkFactors([1, 2], 2));
                    console.log(checkFactors([2, 4, 3, 6, 9], 12));
                    console.log(checkFactors([3, 5, 2, 6, 9], 15));
                    console.log(checkFactors([4, 2, 8, 1], 16));





                    You can also use Array.every to check the same in a concise way:






                    function checkFactors(factors, num) {
                    return factors.every(element => num % element === 0);
                    }
                    console.log(checkFactors([1, 2, 3, 8], 12));
                    console.log(checkFactors([1, 2], 2));
                    console.log(checkFactors([2, 4, 3, 6, 9], 12));
                    console.log(checkFactors([3, 5, 2, 6, 9], 15));
                    console.log(checkFactors([4, 2, 8, 1], 16));








                    function checkFactors(factors, num) {
                    let flag = true;
                    for (let i=0; i<factors.length; i++){
                    let element = factors[i];
                    if (num % element !== 0){
                    flag = false;
                    break;
                    }
                    }
                    return flag;
                    }

                    console.log(checkFactors([1, 2, 3, 8], 12));
                    console.log(checkFactors([1, 2], 2));
                    console.log(checkFactors([2, 4, 3, 6, 9], 12));
                    console.log(checkFactors([3, 5, 2, 6, 9], 15));
                    console.log(checkFactors([4, 2, 8, 1], 16));





                    function checkFactors(factors, num) {
                    let flag = true;
                    for (let i=0; i<factors.length; i++){
                    let element = factors[i];
                    if (num % element !== 0){
                    flag = false;
                    break;
                    }
                    }
                    return flag;
                    }

                    console.log(checkFactors([1, 2, 3, 8], 12));
                    console.log(checkFactors([1, 2], 2));
                    console.log(checkFactors([2, 4, 3, 6, 9], 12));
                    console.log(checkFactors([3, 5, 2, 6, 9], 15));
                    console.log(checkFactors([4, 2, 8, 1], 16));





                    function checkFactors(factors, num) {
                    return factors.every(element => num % element === 0);
                    }
                    console.log(checkFactors([1, 2, 3, 8], 12));
                    console.log(checkFactors([1, 2], 2));
                    console.log(checkFactors([2, 4, 3, 6, 9], 12));
                    console.log(checkFactors([3, 5, 2, 6, 9], 15));
                    console.log(checkFactors([4, 2, 8, 1], 16));





                    function checkFactors(factors, num) {
                    return factors.every(element => num % element === 0);
                    }
                    console.log(checkFactors([1, 2, 3, 8], 12));
                    console.log(checkFactors([1, 2], 2));
                    console.log(checkFactors([2, 4, 3, 6, 9], 12));
                    console.log(checkFactors([3, 5, 2, 6, 9], 15));
                    console.log(checkFactors([4, 2, 8, 1], 16));






                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited 12 hours ago

























                    answered 12 hours ago









                    Amardeep BhowmickAmardeep Bhowmick

                    5,38321128




                    5,38321128























                        0














                        Yes, "else" causing the issue. I removed it and added "return true" outside of for loop.



                        function checkFactors(factors, num) {

                        for (let i=0; i<factors.length; i++){
                        let element = factors[i];
                        console.log(element)

                        if (num % element !== 0){
                        return false
                        }
                        }
                        return true;
                        }





                        share|improve this answer



















                        • 1





                          He is looking for an explanation as well though

                          – Icepickle
                          12 hours ago











                        • I see Icepickle has explained in the question comments nicely! Jonas Wilms's answer too.

                          – javapedia.net
                          12 hours ago


















                        0














                        Yes, "else" causing the issue. I removed it and added "return true" outside of for loop.



                        function checkFactors(factors, num) {

                        for (let i=0; i<factors.length; i++){
                        let element = factors[i];
                        console.log(element)

                        if (num % element !== 0){
                        return false
                        }
                        }
                        return true;
                        }





                        share|improve this answer



















                        • 1





                          He is looking for an explanation as well though

                          – Icepickle
                          12 hours ago











                        • I see Icepickle has explained in the question comments nicely! Jonas Wilms's answer too.

                          – javapedia.net
                          12 hours ago
















                        0












                        0








                        0







                        Yes, "else" causing the issue. I removed it and added "return true" outside of for loop.



                        function checkFactors(factors, num) {

                        for (let i=0; i<factors.length; i++){
                        let element = factors[i];
                        console.log(element)

                        if (num % element !== 0){
                        return false
                        }
                        }
                        return true;
                        }





                        share|improve this answer













                        Yes, "else" causing the issue. I removed it and added "return true" outside of for loop.



                        function checkFactors(factors, num) {

                        for (let i=0; i<factors.length; i++){
                        let element = factors[i];
                        console.log(element)

                        if (num % element !== 0){
                        return false
                        }
                        }
                        return true;
                        }






                        share|improve this answer












                        share|improve this answer



                        share|improve this answer










                        answered 13 hours ago









                        javapedia.netjavapedia.net

                        582311




                        582311








                        • 1





                          He is looking for an explanation as well though

                          – Icepickle
                          12 hours ago











                        • I see Icepickle has explained in the question comments nicely! Jonas Wilms's answer too.

                          – javapedia.net
                          12 hours ago
















                        • 1





                          He is looking for an explanation as well though

                          – Icepickle
                          12 hours ago











                        • I see Icepickle has explained in the question comments nicely! Jonas Wilms's answer too.

                          – javapedia.net
                          12 hours ago










                        1




                        1





                        He is looking for an explanation as well though

                        – Icepickle
                        12 hours ago





                        He is looking for an explanation as well though

                        – Icepickle
                        12 hours ago













                        I see Icepickle has explained in the question comments nicely! Jonas Wilms's answer too.

                        – javapedia.net
                        12 hours ago







                        I see Icepickle has explained in the question comments nicely! Jonas Wilms's answer too.

                        – javapedia.net
                        12 hours ago













                        0














                        Your code's logic is wrong. You should check all the array's Elements, if all elements satisfy the condition, return true, but if one of them not satisfy the condition, return false immediately. The else means one item satisfy the condition, but not all elements. That's where the problem is.






                        share|improve this answer








                        New contributor




                        W.Bright is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.

























                          0














                          Your code's logic is wrong. You should check all the array's Elements, if all elements satisfy the condition, return true, but if one of them not satisfy the condition, return false immediately. The else means one item satisfy the condition, but not all elements. That's where the problem is.






                          share|improve this answer








                          New contributor




                          W.Bright is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.























                            0












                            0








                            0







                            Your code's logic is wrong. You should check all the array's Elements, if all elements satisfy the condition, return true, but if one of them not satisfy the condition, return false immediately. The else means one item satisfy the condition, but not all elements. That's where the problem is.






                            share|improve this answer








                            New contributor




                            W.Bright is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.










                            Your code's logic is wrong. You should check all the array's Elements, if all elements satisfy the condition, return true, but if one of them not satisfy the condition, return false immediately. The else means one item satisfy the condition, but not all elements. That's where the problem is.







                            share|improve this answer








                            New contributor




                            W.Bright is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            share|improve this answer



                            share|improve this answer






                            New contributor




                            W.Bright is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            answered 12 hours ago









                            W.BrightW.Bright

                            133




                            133




                            New contributor




                            W.Bright is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.





                            New contributor





                            W.Bright is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






                            W.Bright is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






























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