how to always round up to the next integer

Question

i am trying to find total pages in building a pager on a website  so i want the result to be an integer   i get a list of records and i want to split into 10 per page  the page count   when i do this   list Count     10   or  list Count      decimal 10   and the list Count    12  i get a result of 1    How would I code it so i get 2 in this case  the remainder should always add 1

User · Answer

list Count     9    10  Everything else here is either overkill or simply wrong  except for bestsss  answer  which is awesome    We do not want the overhead of a function call  Math Truncate    Math Ceiling    etc   when simple math is enough     OP s question generalizes  pigeonhole principle  to      How many boxes do I need to store x objects if only y objects fit into each box    The solution    derives from the realization that the last box might be partially empty  and is  x   y - 1     y using integer division    You ll recall from 3rd grade math that integer division is what we re doing when we say 5    2   2   Floating-point division is when we say 5    2   2 5  but we don t want that here   Many programming languages support integer division   In languages derived from C  you get it automatically when you divide int types  short  int  long  etc     The remainder fractional part of any division operation is simply dropped  thus   5   2    2  Replacing our original question with x   5 and y   2 we have      How many boxes do I need to store 5 objects if only 2 objects fit into each box    The answer should now be obvious  3 boxes -- the first two boxes hold two objects each and the last box holds one    x   y - 1     y    5   2 - 1     2   6    2   3   So for the original question  x   list Count    y   10  which gives the solution using no additional function calls    list Count     9    10

User · Answer

Math Ceiling  double list Count     10

User · Answer

Check by using mod - if there is a remainder  simply increment the value by one

User · Answer

I think the easiest way is to divide two integers and increase by one    int r   list Count     10  r     list Count     10    0   0   1     No need of libraries or functions   edited with the right code

User · Answer

This will also work   c    count - 1    10   1

User · Answer

You can use Math Ceiling  http   msdn microsoft com en-us library system math ceiling 28v VS 100 29 aspx

User · Answer

Xform to double  and back  for a simple ceil   list Count   10    list Count   10  gt 0 1 0  - this bad  div   mod  edit 1st  on a 2n thought that s probably faster  depends on the optimization   div   mul  mul is faster than div and mod   int c list Count   10  if  c 10 lt list Count    c      edit2 scarpe all  forgot the most natural  adding 9 ensures rounding up for integers    list Count   9  10

User · Answer

A proper benchmark or how the number may lie  Following the argument about Math ceil value 10d  and   value 9  10 I ended up coding a proper non-dead code  non-interpret mode benchmark  I ve been telling that writing micro benchmark is not an easy task  The code below illustrates this   00 21 40 109 starting up     00 21 40 140 doubleCeil  19444599 00 21 40 140 integerCeil  19444599 00 21 40 140 warming up    00 21 44 375 warmup doubleCeil  194445990000 00 21 44 625 warmup integerCeil  194445990000 00 22 27 437 exec doubleCeil  1944459900000  elapsed  42 806s 00 22 29 796 exec integerCeil  1944459900000  elapsed  2 363s   The benchmark is in Java since I know well how Hotspot optimizes and ensures it s a fair result  With such results  no statistics  noise or anything can taint it    Integer ceil is insanely much faster   The code  package t1   import java math BigDecimal   import java util Random   public class Div       static int   vals       static long doubleCeil            int   v  vals          long sum   0          for  int i 0 i lt v length i                 int value   v i               sum  Math ceil value 10d                     return sum             static long integerCeil                  int   v  vals          long sum   0          for  int i 0 i lt v length i                 int value   v i               sum   value 9  10                    return sum                  public static void main String   args            vals   new  int 7000           Random r  new Random 77           for  int i   0  i  lt  vals length  i                  vals i    r nextInt 55555                     log  starting up                 log  doubleCeil   d   doubleCeil             log  integerCeil   d   integerCeil             log  warming up                       final int warmupCount    int  1e4          log  warmup doubleCeil   d   execDoubleCeil warmupCount            log  warmup integerCeil   d   execIntegerCeil warmupCount             final int execCount    int  1e5                            long time   System nanoTime            long s   execDoubleCeil execCount           long elapsed   System nanoTime   - time          log  exec doubleCeil   d  elapsed    3fs    s  BigDecimal valueOf elapsed  9                                 long time   System nanoTime            long s   execIntegerCeil execCount           long elapsed   System nanoTime   - time          log  exec integerCeil   d  elapsed    3fs    s  BigDecimal valueOf elapsed  9                                     static long execDoubleCeil int count           long sum   0          for int i 0 i lt count i                 sum  doubleCeil                      return sum              static long execIntegerCeil int count           long sum   0          for int i 0 i lt count i                 sum  integerCeil                      return sum             static void log String msg  Object    params           String s   params length gt 0 String format msg  params  msg          System out printf   tH   lt tM   lt tS   lt tL  s n   new Long System currentTimeMillis     s

[c#] how to always round up to the next integer

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