C Rounding up to the nearest multiple of a number

Question

OK - I m almost embarrassed posting this here  and I will delete if anyone votes to close  as it seems like a basic question   Is this the correct way to round up to a multiple of a number in C      I know there are other questions related to this but I am specficially interested to know what is the best way to do this in C     int roundUp int numToRound  int multiple     if multiple    0       return numToRound       int roundDown      int   numToRound    multiple    multiple   int roundUp   roundDown   multiple    int roundCalc   roundUp   return  roundCalc       Update  Sorry I probably didn t make intention clear   Here are some examples   roundUp 7  100    return 100  roundUp 117  100    return 200  roundUp 477  100    return 500  roundUp 1077  100    return 1100  roundUp 52  20    return 60  roundUp 74  30    return 90

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To always round up  int alwaysRoundUp int n  int multiple        if  n   multiple    0            n     n   multiple    multiple    multiple              Another way           n   n - n   multiple   multiple             return n      alwaysRoundUp 1  10  -  10  alwaysRoundUp 5  10  -  10  alwaysRoundUp 10  10  -  10    To always round down  int alwaysRoundDown int n  int multiple        n    n   multiple    multiple       return n      alwaysRoundDown 1  10  -  0  alwaysRoundDown 5  10  -  0  alwaysRoundDown 10  10  -  10    To round the normal way  int normalRound int n  int multiple        n     n   multiple 2  multiple    multiple       return n      normalRound 1  10  -  0  normalRound 5  10  -  10  normalRound 10  10  -  10

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Round to Power of Two   Just in case anyone needs a solution for positive numbers rounded to the nearest multiple of a power of two  because that s how I ended up here       number  the number to be rounded  ex  5  123  98345  etc      pow2    the power to be rounded to  ex  to round to 16  use  4   int roundPow2  int number  int pow2        pow2--                         because  2 exp x      1  lt  lt   x -1       pow2   0x01  lt  lt  pow2       pow2--                         because for any                                                                        x   2 exp x                                                                        subtracting one will                                    yield a field of ones                                    which we can use in a                                    bitwise OR      number--                       yield a similar field for                                    bitwise OR     number   number   pow2      number                         restore value by adding one back      return number      The input number will stay the same if it is already a multiple   Here is the x86 64 output that GCC gives with -O2 or -Os  9Sep2013 Build - godbolt GCC online    roundPow2 int  int       lea ecx   rsi-1      mov eax  1     sub edi  1     sal eax  cl     sub eax  1     or  eax  edi     add eax  1     ret   Each C line of code corresponds perfectly with its line in the assembly  http   goo gl DZigfX  Each of those instructions are extremely fast  so the function is extremely fast too  Since the code is so small and quick  it might be useful to inline the function when using it     Credit    Algorithm  Hagen von Eitzen   Math SE Godbolt Interactive Compiler   mattgodbolt gcc-explorer on GitHub

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int roundUp  int numToRound  int multiple      return multiple     numToRound   multiple - 1    multiple       although    won t work for negative numbers won t work if numRound   multiple overflows   would suggest using unsigned integers instead  which has defined overflow behaviour   You ll get an exception is multiple    0  but it isn t a well-defined problem in that case anyway

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Endless possibilities  for signed integers only  n     r - n    r

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I think this should help you  I have written the below program in C     include  lt stdio h gt  int main       int i  j    printf   nEnter Two Integers i and j         scanf   d  d    amp i   amp j     int Round Off i j-i j    printf  The Rounded Off Integer Is    d n   Round Off     return 0

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c   int roundUp int numToRound  int multiple      return  multiple      numToRound multiple-1    multiple    multiple    numToRound       and for your    bashrc   roundup       echo       2        1    2 -1    2     2      1

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This is a generalization of the problem of  how do I find out how many bytes n bits will take   A   n bits   7    8    int RoundUp int n  int roundTo           fails on negative   What does that mean      if  roundTo    0  return 0      return   n   roundTo - 1    roundTo    roundTo     edit - fixed error

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First off  your error condition  multiple    0  should probably have a return value  What  I don t know  Maybe you want to throw an exception  that s up to you  But  returning nothing is dangerous   Second  you should check that numToRound isn t already a multiple  Otherwise  when you add multiple to roundDown  you ll get the wrong answer   Thirdly  your casts are wrong  You cast numToRound to an integer  but it s already an integer  You need to cast to to double before the division  and back to int after the multiplication   Lastly  what do you want for negative numbers  Rounding  up  can mean rounding to zero  rounding in the same direction as positive numbers   or away from zero  a  larger  negative number   Or  maybe you don t care   Here s a version with the first three fixes  but I don t deal with the negative issue   int roundUp int numToRound  int multiple     if multiple    0       return 0      else if numToRound   multiple    0       return numToRound      int roundDown    int      double  numToRound   multiple     multiple    int roundUp   roundDown   multiple    int roundCalc   roundUp   return  roundCalc

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may be this can help   int RoundUpToNearestMultOfNumber int val  int num      assert 0    num     return  floor  val   num    num    num

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This works for positive numbers  not sure about negative  It only uses integer math   int roundUp int numToRound  int multiple        if  multiple    0          return numToRound       int remainder   numToRound   multiple      if  remainder    0          return numToRound       return numToRound   multiple - remainder      Edit  Here s a version that works with negative numbers  if by  up  you mean a result that s always    the input   int roundUp int numToRound  int multiple        if  multiple    0          return numToRound       int remainder   abs numToRound    multiple      if  remainder    0          return numToRound       if  numToRound  lt  0          return - abs numToRound  - remainder       else         return numToRound   multiple - remainder

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This is the modern c   approach using a template function which is working for float  double  long  int and short  but not for long long  and long double because of the used double values     include  lt cmath gt   include  lt iostream gt   template lt typename T gt  T roundMultiple  T value  T multiple         if  multiple    0  return value      return static cast lt T gt  std  round static cast lt double gt  value  static cast lt double gt  multiple   static cast lt double gt  multiple       int main         std  cout  lt  lt  roundMultiple 39298 0  100 0   lt  lt  std  endl      std  cout  lt  lt  roundMultiple 20930 0f  1000 0f   lt  lt  std  endl      std  cout  lt  lt  roundMultiple 287399  10   lt  lt  std  endl      But you can easily add support for long long and long double with template specialisation as shown below   template lt  gt  long double roundMultiple lt long double gt   long double value  long double multiple        if  multiple    0 0l  return value      return std  round value multiple  multiple     template lt  gt  long long roundMultiple lt long long gt   long long value  long long multiple        if  multiple    0 0l  return value      return static cast lt long long gt  std  round static cast lt long double gt  value  static cast lt long double gt  multiple   static cast lt long double gt  multiple        To create functions to round up  use std  ceil and to always round down use std  floor  My  example from above is rounding using std  round   Create the  round up  or better known as  round ceiling  template function as shown below   template lt typename T gt  T roundCeilMultiple  T value  T multiple         if  multiple    0  return value      return static cast lt T gt  std  ceil static cast lt double gt  value  static cast lt double gt  multiple   static cast lt double gt  multiple        Create the  round down  or better known as  round floor  template function as shown below   template lt typename T gt  T roundFloorMultiple  T value  T multiple         if  multiple    0  return value      return static cast lt T gt  std  floor static cast lt double gt  value  static cast lt double gt  multiple   static cast lt double gt  multiple

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For negative numToRound   It should be really easy to do this but the standard modulo   operator doesn t handle negative numbers like one might expect  For instance -14   12   -2 and not 10  First thing to do is to get modulo operator that never returns negative numbers  Then roundUp is really simple   public static int mod int x  int n         return   x   n    n    n     public static int roundUp int numToRound  int multiple         return numRound   mod -numToRound  multiple

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I m using     template  lt class  Ty gt  inline  Ty n Align Up  Ty n x   Ty n alignment        assert n alignment  gt  0         n x     n x  gt   0   n alignment - 1   1 - n alignment     causes to round away from zero  greatest absolute value      n x     n x  gt   0   n alignment - 1   -1     causes to round up  towards positive infinity        n x      Ty - n x  gt   0    amp  n alignment  - 1     the same as above  avoids branch and integer multiplication       n x    n alignment - 1     only works for positive numbers  fastest      return n x - n x   n alignment     rounds negative towards zero     and for powers of two   template  lt class  Ty gt  bool b Is POT  Ty n x        return   n x  amp   n x - 1       template  lt class  Ty gt  inline  Ty n Align Up POT  Ty n x   Ty n pot alignment        assert n pot alignment  gt  0       assert b Is POT n pot alignment       alignment must be power of two     -- n pot alignment      return  n x   n pot alignment   amp   n pot alignment     rounds towards positive infinity  i e  negative towards zero      Note that both of those round negative values towards zero  that means round to positive infinity for all values   neither of them relies on signed overflow  which is undefined in C C      This gives   n Align Up 10  100    100 n Align Up 110  100    200 n Align Up 0  100    0 n Align Up -10  100    0 n Align Up -110  100    -100 n Align Up -210  100    -200 n Align Up POT 10  128    128 n Align Up POT 130  128    256 n Align Up POT 0  128    0 n Align Up POT -10  128    0 n Align Up POT -130  128    -128 n Align Up POT -260  128    -256

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I use a combination of modulus to nullify the addition of the remainder if x is already a multiple   int round up int x  int div        return x    div - x   div    div      We find the inverse of the remainder then modulus that with the divisor again to nullify it if it is the divisor itself then add x   round up 19  3    21

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Round to nearest multiple that happens to be a power of 2  unsigned int round unsigned int value  unsigned int multiple       return   value-1u   amp    multiple-1u     multiple      This can be useful for when allocating along cachelines  where the rounding increment you want is a power of two  but the resulting value only needs to be a multiple of it   On gcc the body of this function generates 8 assembly instructions with no division or branches   round   0   16  - gt    0 round   1   16  - gt   16 round  16   16  - gt   16 round 257  128  - gt  384  128   3  round 333    2  - gt  334

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Here is a super simple solution to show the concept of elegance  It s basically for grid snaps   pseudo code  nearestPos   Math Ceil  numberToRound   multiple     multiple

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Without conditions   int roundUp int numToRound  int multiple         assert multiple       return   numToRound   multiple - 1    multiple    multiple      This works like rounding away from zero for negative numbers  EDIT  Version that works also for negative numbers  int roundUp int numToRound  int multiple         assert multiple       int isPositive    int  numToRound  gt   0       return   numToRound   isPositive    multiple - 1     multiple    multiple      Tests    If multiple is a power of 2  faster in  3 7 times http   quick-bench com sgPEZV9AUDqtx2uujRSa3-eTE80   int roundUp int numToRound  int multiple         assert multiple  amp  amp    multiple  amp   multiple - 1      0        return  numToRound   multiple - 1   amp  -multiple      Tests

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Rounding up  n  to the nearest multiple of number  b       - Not tested for negative numbers       see http   stackoverflow com questions 3407012   define roundUp n b     b   0    n        n   b -1  -    n -1   b             c test- gt roundUp    void test roundUp               yes roundUp n b     b   0    n       n   b   0   n    n   b - n   b             yes roundUp n b     b   0    n        n   b - 1    b    b             no roundUp n b     n   b   0   n    b     n   b     b           no roundUp n b     n   b  -  n   b     if  true     couldn t make it work without             test  roundUp       unsigned m                    m   roundUp 17 8       m      assertTrue  24    roundUp 17 8                       m   roundUp 24 8         assertTrue  24    roundUp 24 8          assertTrue  24    roundUp 24 4         assertTrue  24    roundUp 23 4                       m   roundUp 23 4         assertTrue  24    roundUp 21 4          assertTrue  20    roundUp 20 4         assertTrue  20    roundUp 19 4         assertTrue  20    roundUp 18 4         assertTrue  20    roundUp 17 4          assertTrue  17    roundUp 17 0         assertTrue  20    roundUp 20 0

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I found an algorithm which is somewhat similar to one posted above   int   x  n-1  n    nx   x    where x is a user-input value and n is the multiple being used   It works for all values x  where x is an integer  positive or negative  including zero   I wrote it specifically for a C   program  but this can basically be implemented in any language

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This is getting the results you are seeking for positive integers    include  lt iostream gt  using namespace std   int roundUp int numToRound  int multiple    int main         cout  lt  lt   answer is     lt  lt  roundUp 7  100   lt  lt  endl      cout  lt  lt   answer is     lt  lt  roundUp 117  100   lt  lt  endl      cout  lt  lt   answer is     lt  lt  roundUp 477  100   lt  lt  endl      cout  lt  lt   answer is     lt  lt  roundUp 1077  100   lt  lt  endl      cout  lt  lt   answer is     lt  lt  roundUp 52 20   lt  lt  endl      cout  lt  lt   answer is     lt  lt  roundUp 74 30   lt  lt  endl      return 0     int roundUp int numToRound  int multiple        if  multiple    0            return 0            int result    int   numToRound   multiple    multiple      if  numToRound   multiple            result    multiple             return result      And here are the outputs   answer is  100 answer is  200 answer is  500 answer is  1100 answer is  60 answer is  90

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well for one thing  since i dont really understand what you want to do   the lines   int roundUp   roundDown   multiple  int roundCalc   roundUp  return  roundCalc      could definitely be shortened to   int roundUp   roundDown   multiple  return roundUp

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This works when factor will always be positive  int round up int num  int factor        return num   factor - 1 -  num   factor - 1    factor     Edit  This returns round up 0 100  100  Please see Paul s comment below for a solution that returns round up 0 100  0

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int roundUp int numToRound  int multiple     if multiple    0       return 0      return   numToRound - 1    multiple   1    multiple        And no need to mess around with conditions

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This is what I would do    include  lt cmath gt   int roundUp int numToRound  int multiple           if our number is zero  return immediately    if  numToRound    0          return multiple          if multiplier is zero  return immediately     if  multiple    0          return numToRound          how many times are number greater than multiple     float rounds   static cast lt float gt  numToRound    static cast lt float gt  multiple           determine  whether if number is multiplier of multiple     int floorRounds   static cast lt int gt  floor rounds         if  rounds - floorRounds  gt  0             multiple is not multiplier of number - gt  advance to the next multiplier         return  floorRounds 1    multiple      else            multiple is multiplier of number - gt  return actual multiplier         return  floorRounds    multiple      The code might not be optimal  but I prefer clean code than dry performance

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For anyone looking for a short and sweet answer  This is what I used  No accounting for negatives   n -  n   r    That will return the previous factor    n   r  -  n   r    Will return the next  Hope this helps someone

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I think this works  int roundUp int numToRound  int multiple        return multiple    numToRound multiple   numToRound     numToRound multiple  1  multiple  numToRound

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Here s my solution based on the OP s suggestion  and the examples given by everyone else  Since most everyone was looking for it to handle negative numbers  this solution does just that  without the use of any special functions  i e  abs  and the like   By avoiding the modulus and using division instead  the negative number is a natural result  although it s rounded down  After the rounded down version is calculated  then it does the required math to round up  either in the negative or positive direction   Also note that no special functions are used to calculate anything  so there is a small speed boost there   int RoundUp int n  int multiple           prevent divide by 0 by returning n     if  multiple    0  return n          calculate the rounded down version     int roundedDown   n   multiple   multiple          if the rounded version and original are the same  then return the original     if  roundedDown    n  return n          handle negative number and round up according to the sign        NOTE  if n is  lt  0 then subtract the multiple  otherwise add it     return  n  lt  0    roundedDown - multiple   roundedDown   multiple

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This works for me but did not try to handle negatives  public static int roundUp int numToRound  int multiple        if  multiple    0            return 0        else if  numToRound   multiple    0        return numToRound             int mod   numToRound   multiple      int diff   multiple - mod      return numToRound   diff

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Probably safer to cast to floats and use ceil   - unless you know that the int division is going to produce the correct result

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int noOfMultiples   int  numToRound   multiple  0 5   return noOfMultiples multiple   C   rounds each number down so if you add 0 5  if its 1 5 it will be 2  but 1 49 will be 1 99 therefore 1   EDIT - Sorry didn t see you wanted to round up  i would suggest using a ceil   method instead of the  0 5

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float roundUp float number  float fixedBase        if  fixedBase    0  amp  amp  number    0            float sign   number  gt  0   1   -1          number    sign          number    fixedBase          int fixedPoint    int  ceil number           number   fixedPoint   fixedBase          number    sign            return number      This works for any float number or base  e g  you can round -4 to the nearest 6 75    In essence it is converting to fixed point  rounding there  then converting back   It handles negatives by rounding AWAY from 0   It also handles a negative round to value by essentially turning the function into roundDown   An int specific version looks like   int roundUp int number  int fixedBase        if  fixedBase    0  amp  amp  number    0            int sign   number  gt  0   1   -1          int baseSign   fixedBase  gt  0   1   0          number    sign          int fixedPoint    number   baseSign    fixedBase - 1     fixedBase          number   fixedPoint   fixedBase          number    sign            return number      Which is more or less plinth s answer  with the added negative input support

[c++] C++: Rounding up to the nearest multiple of a number

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