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 minu

Answer 1

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.

Answer 2

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. */

Answer 3

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).

Answer 4

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

Answer 5

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

Answer 6

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.