Understanding Factorialize solution





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function factorialize(num) {

    var count = 1;

    for (var i = 1; i < num; i++) {

        count = count + (count * i);

    }

    return count;

}



factorialize(1);


Why does this return 1 when you use 1? Shouldn't it return 2 because count = 1 + (1 * 1)?






javascript algorithm







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  • 2





    It is not entering the loop as i=1 < 1 is false. With <= what you expect for 1 would happen

    – juvian
    May 9 '16 at 17:23






  • 1





    In the for loop, var i = 1; sets i to 1. i < num is the condition for this loop to run. Since num is 1, the loop does not run, and the function simply returns count.

    – blex
    May 9 '16 at 17:24


















1















function factorialize(num) {

    var count = 1;

    for (var i = 1; i < num; i++) {

        count = count + (count * i);

    }

    return count;

}



factorialize(1);


Why does this return 1 when you use 1? Shouldn't it return 2 because count = 1 + (1 * 1)?






javascript algorithm







share|improve this question




















  • 2





    It is not entering the loop as i=1 < 1 is false. With <= what you expect for 1 would happen

    – juvian
    May 9 '16 at 17:23






  • 1





    In the for loop, var i = 1; sets i to 1. i < num is the condition for this loop to run. Since num is 1, the loop does not run, and the function simply returns count.

    – blex
    May 9 '16 at 17:24














1












1








1








function factorialize(num) {

    var count = 1;

    for (var i = 1; i < num; i++) {

        count = count + (count * i);

    }

    return count;

}



factorialize(1);


Why does this return 1 when you use 1? Shouldn't it return 2 because count = 1 + (1 * 1)?






javascript algorithm







share|improve this question
















function factorialize(num) {

    var count = 1;

    for (var i = 1; i < num; i++) {

        count = count + (count * i);

    }

    return count;

}



factorialize(1);


Why does this return 1 when you use 1? Shouldn't it return 2 because count = 1 + (1 * 1)?






javascript algorithm




javascript algorithm






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited May 9 '16 at 17:35









Alex Pan

2,73452332




2,73452332










asked May 9 '16 at 17:21









wquwqu

80117




80117








  • 2





    It is not entering the loop as i=1 < 1 is false. With <= what you expect for 1 would happen

    – juvian
    May 9 '16 at 17:23






  • 1





    In the for loop, var i = 1; sets i to 1. i < num is the condition for this loop to run. Since num is 1, the loop does not run, and the function simply returns count.

    – blex
    May 9 '16 at 17:24














  • 2





    It is not entering the loop as i=1 < 1 is false. With <= what you expect for 1 would happen

    – juvian
    May 9 '16 at 17:23






  • 1





    In the for loop, var i = 1; sets i to 1. i < num is the condition for this loop to run. Since num is 1, the loop does not run, and the function simply returns count.

    – blex
    May 9 '16 at 17:24








2




2





It is not entering the loop as i=1 < 1 is false. With <= what you expect for 1 would happen

– juvian
May 9 '16 at 17:23





It is not entering the loop as i=1 < 1 is false. With <= what you expect for 1 would happen

– juvian
May 9 '16 at 17:23




1




1





In the for loop, var i = 1; sets i to 1. i < num is the condition for this loop to run. Since num is 1, the loop does not run, and the function simply returns count.

– blex
May 9 '16 at 17:24





In the for loop, var i = 1; sets i to 1. i < num is the condition for this loop to run. Since num is 1, the loop does not run, and the function simply returns count.

– blex
May 9 '16 at 17:24












1 Answer
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For num = 1, when we enter the loop, the check i < num returned false, so we do not get to the count = count + (count * i); and go to return with count = 1.






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    1 Answer
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    active

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    0














    For num = 1, when we enter the loop, the check i < num returned false, so we do not get to the count = count + (count * i); and go to return with count = 1.






    share|improve this answer






























      0














      For num = 1, when we enter the loop, the check i < num returned false, so we do not get to the count = count + (count * i); and go to return with count = 1.






      share|improve this answer




























        0












        0








        0







        For num = 1, when we enter the loop, the check i < num returned false, so we do not get to the count = count + (count * i); and go to return with count = 1.






        share|improve this answer















        For num = 1, when we enter the loop, the check i < num returned false, so we do not get to the count = count + (count * i); and go to return with count = 1.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 16 '18 at 11:43

























        answered Nov 1 '18 at 11:24









        Eugene MihaylinEugene Mihaylin

        1,0101725




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