C++ Best way to get integer division and remainder

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;

int c = (int)a / b;
int d = a - b * c;

double tmp = a / b;
int c = (int)tmp;
int d = (int)(0.5+(tmp-c)*b);

maybe there is a magical function that gives one both at once?

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. */

std::div returns a structure with both result and remainder.

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

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

Jamin Grey

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.

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.

来源：https://stackoverflow.com/questions/7070346/c-best-way-to-get-integer-division-and-remainder

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