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?
Sample code testing div() and combined division & 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 <stdio.h>
#include <sys/time.h>
#include <stdlib.h>
extern doNothing(int,int); // Empty function in another compilation unit
int main() {
int i;
struct timeval timeval;
struct timeval timeval2;
div_t result;
gettimeofday(&timeval,NULL);
for (i = 0; i < 1000; ++i) {
result = div(i,3);
doNothing(result.quot,result.rem);
}
gettimeofday(&timeval2,NULL);
printf("%d",timeval2.tv_usec - timeval.tv_usec);
}
Outputs: 150
#include <stdio.h>
#include <sys/time.h>
#include <stdlib.h>
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(&timeval,NULL);
for (i = 0; i < 1000; ++i) {
dividend = i / 3;
rem = i % 3;
doNothing(dividend,rem);
}
gettimeofday(&timeval2,NULL);
printf("%d",timeval2.tv_usec - timeval.tv_usec);
}
Outputs: 25
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.
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).
You can use a modulus to get the remainder. Though @cnicutar's answer seems cleaner/more direct.
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
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.
std::div
returns a structure with both result and remainder.
Source: Stackoverflow.com