Why modulus operation work different in ruby than other languages?












1















In Ruby, I get:



-5 % 3 # => 1


whereas other languages like PHP, Javascript, C++, and Java all produce the result -2. I don't understand this concept. I hope someone can explain this ruby's calculation method. It would be better if you could use an example of how it works.










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  • 4





    You want -5.remainder(3) which returns -2

    – Stefan
    Nov 19 '18 at 16:51






  • 1





    AFAIK -5 % 3 is implementation defined in C and C++, that can be either -2 or 1 depending on how the CPU implements modulus with negatives and what it thinks "smallest" means: -2 is smaller than 1 because -2 < 1 but -2 is bigger than 1 because |-2| > |1|.

    – mu is too short
    Nov 19 '18 at 23:55











  • It is a matter of what you want as the set of representatives of the elements derived by modulo. It is whether you think {-2, -1, 0} more natural or {0, 1, 2} more natural. Ruby takes the latter, and so do I.

    – sawa
    Nov 21 '18 at 5:56


















1















In Ruby, I get:



-5 % 3 # => 1


whereas other languages like PHP, Javascript, C++, and Java all produce the result -2. I don't understand this concept. I hope someone can explain this ruby's calculation method. It would be better if you could use an example of how it works.










share|improve this question




















  • 4





    You want -5.remainder(3) which returns -2

    – Stefan
    Nov 19 '18 at 16:51






  • 1





    AFAIK -5 % 3 is implementation defined in C and C++, that can be either -2 or 1 depending on how the CPU implements modulus with negatives and what it thinks "smallest" means: -2 is smaller than 1 because -2 < 1 but -2 is bigger than 1 because |-2| > |1|.

    – mu is too short
    Nov 19 '18 at 23:55











  • It is a matter of what you want as the set of representatives of the elements derived by modulo. It is whether you think {-2, -1, 0} more natural or {0, 1, 2} more natural. Ruby takes the latter, and so do I.

    – sawa
    Nov 21 '18 at 5:56
















1












1








1








In Ruby, I get:



-5 % 3 # => 1


whereas other languages like PHP, Javascript, C++, and Java all produce the result -2. I don't understand this concept. I hope someone can explain this ruby's calculation method. It would be better if you could use an example of how it works.










share|improve this question
















In Ruby, I get:



-5 % 3 # => 1


whereas other languages like PHP, Javascript, C++, and Java all produce the result -2. I don't understand this concept. I hope someone can explain this ruby's calculation method. It would be better if you could use an example of how it works.







ruby






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edited Nov 21 '18 at 5:50









sawa

130k29202301




130k29202301










asked Nov 19 '18 at 16:06









Pradhumn SharmaPradhumn Sharma

1116




1116








  • 4





    You want -5.remainder(3) which returns -2

    – Stefan
    Nov 19 '18 at 16:51






  • 1





    AFAIK -5 % 3 is implementation defined in C and C++, that can be either -2 or 1 depending on how the CPU implements modulus with negatives and what it thinks "smallest" means: -2 is smaller than 1 because -2 < 1 but -2 is bigger than 1 because |-2| > |1|.

    – mu is too short
    Nov 19 '18 at 23:55











  • It is a matter of what you want as the set of representatives of the elements derived by modulo. It is whether you think {-2, -1, 0} more natural or {0, 1, 2} more natural. Ruby takes the latter, and so do I.

    – sawa
    Nov 21 '18 at 5:56
















  • 4





    You want -5.remainder(3) which returns -2

    – Stefan
    Nov 19 '18 at 16:51






  • 1





    AFAIK -5 % 3 is implementation defined in C and C++, that can be either -2 or 1 depending on how the CPU implements modulus with negatives and what it thinks "smallest" means: -2 is smaller than 1 because -2 < 1 but -2 is bigger than 1 because |-2| > |1|.

    – mu is too short
    Nov 19 '18 at 23:55











  • It is a matter of what you want as the set of representatives of the elements derived by modulo. It is whether you think {-2, -1, 0} more natural or {0, 1, 2} more natural. Ruby takes the latter, and so do I.

    – sawa
    Nov 21 '18 at 5:56










4




4





You want -5.remainder(3) which returns -2

– Stefan
Nov 19 '18 at 16:51





You want -5.remainder(3) which returns -2

– Stefan
Nov 19 '18 at 16:51




1




1





AFAIK -5 % 3 is implementation defined in C and C++, that can be either -2 or 1 depending on how the CPU implements modulus with negatives and what it thinks "smallest" means: -2 is smaller than 1 because -2 < 1 but -2 is bigger than 1 because |-2| > |1|.

– mu is too short
Nov 19 '18 at 23:55





AFAIK -5 % 3 is implementation defined in C and C++, that can be either -2 or 1 depending on how the CPU implements modulus with negatives and what it thinks "smallest" means: -2 is smaller than 1 because -2 < 1 but -2 is bigger than 1 because |-2| > |1|.

– mu is too short
Nov 19 '18 at 23:55













It is a matter of what you want as the set of representatives of the elements derived by modulo. It is whether you think {-2, -1, 0} more natural or {0, 1, 2} more natural. Ruby takes the latter, and so do I.

– sawa
Nov 21 '18 at 5:56







It is a matter of what you want as the set of representatives of the elements derived by modulo. It is whether you think {-2, -1, 0} more natural or {0, 1, 2} more natural. Ruby takes the latter, and so do I.

– sawa
Nov 21 '18 at 5:56














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It's in the docs: https://ruby-doc.org/core-2.5.0/Numeric.html#method-i-divmod




If q, r = x.divmod(y), then



q = floor(x/y)
x = q*y + r


The quotient is rounded toward negative infinity




So q is -3 (-5 / 2 and round down, as per usual integer division rules). And r = x - q * y = -5 - -3 * 2 = 1






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    1 Answer
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    2














    It's in the docs: https://ruby-doc.org/core-2.5.0/Numeric.html#method-i-divmod




    If q, r = x.divmod(y), then



    q = floor(x/y)
    x = q*y + r


    The quotient is rounded toward negative infinity




    So q is -3 (-5 / 2 and round down, as per usual integer division rules). And r = x - q * y = -5 - -3 * 2 = 1






    share|improve this answer




























      2














      It's in the docs: https://ruby-doc.org/core-2.5.0/Numeric.html#method-i-divmod




      If q, r = x.divmod(y), then



      q = floor(x/y)
      x = q*y + r


      The quotient is rounded toward negative infinity




      So q is -3 (-5 / 2 and round down, as per usual integer division rules). And r = x - q * y = -5 - -3 * 2 = 1






      share|improve this answer


























        2












        2








        2







        It's in the docs: https://ruby-doc.org/core-2.5.0/Numeric.html#method-i-divmod




        If q, r = x.divmod(y), then



        q = floor(x/y)
        x = q*y + r


        The quotient is rounded toward negative infinity




        So q is -3 (-5 / 2 and round down, as per usual integer division rules). And r = x - q * y = -5 - -3 * 2 = 1






        share|improve this answer













        It's in the docs: https://ruby-doc.org/core-2.5.0/Numeric.html#method-i-divmod




        If q, r = x.divmod(y), then



        q = floor(x/y)
        x = q*y + r


        The quotient is rounded toward negative infinity




        So q is -3 (-5 / 2 and round down, as per usual integer division rules). And r = x - q * y = -5 - -3 * 2 = 1







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 19 '18 at 16:18









        Sergio TulentsevSergio Tulentsev

        180k30289304




        180k30289304






























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