C Best way to get integer division and remainder

Question

I am just wondering  if I want to divide a by b  and am interested both in the result c and the remainder  e g  say I have number of seconds and want to split that into minutes and seconds   what is the best way to go about it   Would it be  int c    int a   b  int d   a   b    or  int c    int a   b  int d   a - b   c    or  double tmp   a   b  int c    int tmp  int d    int  0 5  tmp-c  b     or  maybe there is a magical function that gives one both at once

User · Answer

On x86 at least  g   4 6 1 just uses IDIVL and gets both from that single instruction   C   code   void foo int a  int b  int  c  int  d       c   a   b     d   a   b      x86 code     Z3fooiiPiS   LFB4      movq     rdx   r8     movl     edi   edx     movl     edi   eax     sarl     31   edx     idivl    esi     movl     eax    r8      movl     edx    rcx      ret

User · Answer

All else being equal  the best solution is one that clearly expresses your intent  So   int totalSeconds   453  int minutes   totalSeconds   60  int remainingSeconds   totalSeconds   60    is probably the best of the three options you presented  As noted in other answers however  the div method will calculate both values for you at once

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You can use a modulus to get the remainder  Though  cnicutar s answer seems cleaner more direct

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On x86 the remainder is a by-product of the division itself so any half-decent compiler should be able to just use it  and not perform a div again   This is probably done on other architectures too      Instruction  DIV src      Note  Unsigned division  Divides accumulator  AX  by  src   If divisor   is a byte value  result is put to AL and remainder to AH  If divisor   is a word value  then DX AX is divided by  src  and result is stored   in AX and remainder is stored in DX    int c    int a   b  int d   a   b     Likely uses the result of the division

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In addition to the aforementioned std  div family of functions  there is also the std  remquo family of functions  return the rem-ainder and getting the quo-tient via a passed-in pointer    Edit   It looks like std  remquo doesn t really return the quotient after all

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std  div returns a structure with both result and remainder

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You cannot trust g   4 6 3 here with 64 bit integers on a 32 bit intel platform  a b is computed by a call to divdi3 and a b is computed by a call to moddi3  I can even come up with an example that computes a b and a-b  a b  with these calls  So I use c a b and a-b c    The div method gives a call to a function which computes the div structure  but a function call seems inefficient on platforms which have hardware support for the integral type  i e  64 bit integers on 64 bit intel amd platforms

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Sample code testing div   and combined division  amp  mod  I compiled these with gcc -O3  I had to add the call to doNothing to stop the compiler from optimising everything out  output would be 0 for the division   mod solution    Take it with a grain of salt    include  lt stdio h gt   include  lt sys time h gt   include  lt stdlib h gt   extern doNothing int int      Empty function in another compilation unit  int main         int i      struct timeval timeval      struct timeval timeval2      div t result      gettimeofday  amp timeval NULL       for  i   0  i  lt  1000    i            result   div i 3           doNothing result quot result rem             gettimeofday  amp timeval2 NULL       printf   d  timeval2 tv usec - timeval tv usec       Outputs  150   include  lt stdio h gt   include  lt sys time h gt   include  lt stdlib h gt   extern doNothing int int      Empty function in another compilation unit  int main         int i      struct timeval timeval      struct timeval timeval2      int dividend      int rem      gettimeofday  amp timeval NULL       for  i   0  i  lt  1000    i            dividend   i   3          rem   i   3          doNothing dividend rem             gettimeofday  amp timeval2 NULL       printf   d  timeval2 tv usec - timeval tv usec       Outputs  25

[c++] C++ Best way to get integer division and remainder

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