how to calculate (a times b) divided by c only using 32-bit integer types even if a times b would not fit such a type

Question

Consider the following as a reference implementation:

/* calculates (a * b) / c */
uint32_t muldiv(uint32_t a, uint32_t b, uint32_t c)
{
    uint64_t x = a;
         


        

Answer 1

I suppose there are reasons you can't do

x = a/c;
x = x*b;

are there? And maybe add

y = b/c;
y = y*a;

if ( x != y )
    return ERROR_VALUE;

Note that, since you're using integer division, a*b/c and a/c*b might lead to different values if c is bigger than a or b. Also, if both a and b are smaller than c it won't work.

Answer 2

The simplest way would be converting the intermediar result to 64 bits, but, depending on value of c, you could use another approach:

((a/c)*b  +  (a%c)*(b/c) + ((a%c)*(b%c))/c

The only problem is that the last term could still overflow for large values of c. still thinking about it..

Answer 3

If b and c are both constants, you can calculate the result very simply using Egyptian fractions.

For example. y = a * 4 / 99 can be written as

y = a / 25 + a / 2475

You can express any fraction as a sum of Egyptian fractions, as explained in answers to Egyptian Fractions in C.

Having b and c fixed in advance might seem like a bit of a restriction, but this method is a lot simpler than the general case answered by others.

Answer 4

If b = 3000000000 => qn = 3000000000, qn*2 will be overflowed. So I edit the code of Sven Marnach.

uint32_t muldiv(uint32_t a, uint32_t b, uint32_t c)
{
uint32_t q = 0;              // the quotient
uint32_t r = 0;              // the remainder
uint32_t qn = b / c;
uint32_t rn = b % c;
while (a)
{
    if (a & 1)
    {
        q += qn;
        if (qn >= UINT32_MAX) {
            cout << "CO CO" << endl;
        }
        r += rn;
        if (r >= c)
        {
            q++;
            r -= c;
        }
    }
    a >>= 1;
    qn <<= 1;
    int temp = rn;
    if (rn > INT32_MAX) {
        // rn times 2: overflow
        rn = UINT32_MAX;// rn 
        temp = (temp - INT32_MAX) * 2; // find the compensator mean: rn * 2  = UINT32_MAX + temp
        qn++;
        rn = rn - c + temp;
    }
    else {
        rn <<= 1;
        if (rn >= c)
        {
            qn++;
            rn -= c;
        }
    }


}

//return r;
return q;

}

Answer 5

Searching on www.google.com/codesearch turns up a number of implementations, including this wonderfuly obvious one. I particularly like the extensive comments and well chosen variable names

INT32 muldiv(INT32 a, INT32 b, INT32 c)
{ INT32 q=0, r=0, qn, rn;
  int qneg=0, rneg=0;
  if (c==0) c=1;
  if (a<0) { qneg=!qneg; rneg=!rneg; a = -a; }
  if (b<0) { qneg=!qneg; rneg=!rneg; b = -b; }
  if (c<0) { qneg=!qneg;             c = -c; }

  qn = b / c;
  rn = b % c;

  while(a)
  { if (a&1) { q += qn;
               r += rn;
               if(r>=c) { q++; r -= c; }
             }
    a  >>= 1;
    qn <<= 1;
    rn <<= 1;
    if (rn>=c) {qn++; rn -= c; }
  }
  result2 = rneg ? -r : r;
  return qneg ? -q : q;
}

http://www.google.com/codesearch/p?hl=en#HTrPUplLEaU/users/mr/MCPL/mcpl.tgz|gIE-sNMlwIs/MCPL/mintcode/sysc/mintsys.c&q=muldiv%20lang:c

Answer 6

I have adapted the algorithm posted by Paul for unsigned ints (by omitting the parts that are dealing with signs). The algorithm is basically Ancient Egyptian multiplication of a with the fraction floor(b/c) + (b%c)/c (with the slash denoting real division here).

uint32_t muldiv(uint32_t a, uint32_t b, uint32_t c)
{
    uint32_t q = 0;              // the quotient
    uint32_t r = 0;              // the remainder
    uint32_t qn = b / c;
    uint32_t rn = b % c;
    while(a)
    {
        if (a & 1)
        {
            q += qn;
            r += rn;
            if (r >= c)
            {
                q++;
                r -= c;
            }
        }
        a  >>= 1;
        qn <<= 1;
        rn <<= 1;
        if (rn >= c)
        {
            qn++; 
            rn -= c;
        }
    }
    return q;
}

This algorithm will yield the exact answer as long as it fits in 32 bits. You can optionally also return the remainder r.