如何在不使用long long的情况下在C上将64位无符号整数与32位无符号整数相乘？

Question

How can I multiply a 64-bit unsigned int with a 32-bit unsigned int without using long long . The 64-bit number is stored as an array of two 32-bit numbers.

In other words, how I can I obtain e and f from a , b and c given the following:

( a * 2^32 + b ) * c = e * 2^32 + f

a , b , c , e and f are all 32-bit unsigned integers, and only 32-bit math operations are available.

Answer 1

The first thing to realize is that we'll obtain a 96-bit number, as the following shows:

(2 ⁶⁴ - 1) × (2 ³² - 1)
= 2 ⁶⁴ × 2 ³² - 2 ⁶⁴ - 2 ³² + 1
= 2 ⁹⁶ - 2 ⁶⁴ - 2 ³² + 1
< 2 ⁹⁶

The next thing to realize is that we can't use 32-bit multiplication. Even without a carry, the results are too large.

(2 ³² - 1) × (2 ³² - 1)
= 2 ³² × 2 ³² - 2 ³² - 2 ³² + 1
= 2 ⁶⁴ - 2 ³³ + 1
≥ 2 ³²

We'll need to perform 16-bit multiplications instead. 16-bit multiplication can be done using 32-bit numbers without the possibility of overflow, even with a 16-bit carry.

FFFF ₁₆ × FFFF ₁₆ + FFFF ₁₆
= FFFF_0000 ₁₆
< 2 ³²

If we view the numbers as 16-bit values, the multiplication looks like:

                  +-----+-----+-----+-----+
                  | ahi | alo | bhi | blo |
                  +-----+-----+-----+-----+

                              +-----+-----+
                           ×  | chi | clo |
                              +-----+-----+

===========================================

      +-----+-----+-----+-----+-----+-----+
      | dhi | dlo | ehi | elo | fhi | flo |
      +-----+-----+-----+-----+-----+-----+

We all know how to do the above multiplication from elementary school, right?

      +-----+-----+-----+-----+     +-----+
      | ahi | alo | bhi | blo |  ×  | clo |
      +-----+-----+-----+-----+     +-----+

      +-----+-----+-----+-----+-----+     +-----+
   +  | ahi | alo | bhi | blo |  0  |  ×  | chi |
      +-----+-----+-----+-----+-----+     +-----+

The means the following are all the tools we need:

void mul16c(
   uint16_t i, uint16_t j, uint16_t cin,
   uint16_t *prod_ptr, uint16_t *cout_ptr
) {
   uint32_t prod = i * j + cin;
   *prod_ptr = prod & 0xFFFF:
   *cout_ptr = prod >> 16;
}

void add16c(
   uint16_t i, uint16_t j, uint16_t cin,
   uint16_t *sum_ptr, uint16_t* cout_ptr
) {
   uint32_t sum = i + j + cin;
   *sum_ptr = sum & 0xFFFF:
   *cout_ptr = sum >> 16;
}

And we can use them as follows:

void mul64x32(
   uint32_t i[2], uint32_t j,
   uint32_t (*prod_ptr)[2], uint32_t *cout_ptr
) {
   uint16_t i16[4];
   i16[3] = i[1] >> 16;
   i16[2] = i[1] & 0xFFFF;
   i16[1] = i[0] >> 16;
   i16[0] = i[0] & 0xFFFF;

   uint16_t j16[2];
   j16[1] = j[0] >> 16;
   j16[0] = j[0] & 0xFFFF;

   uint16_t p[6];
   {
      uint16_t carry = 0;
      mul16c(i16[0], j16[0], carry, &(p[0]), &carry);
      mul16c(i16[1], j16[0], carry, &(p[1]), &carry);
      mul16c(i16[2], j16[0], carry, &(p[2]), &carry);
      mul16c(i16[3], j16[0], carry, &(p[3]), &carry);
      p[4] = carry;
      p[5] = 0;
   }

   uint16_t q[6];
   {
      uint16_t carry = 0;
      q[0] = 0;
      mul16c(i16[0], j16[1], carry, &(q[1]), &carry);
      mul16c(i16[1], j16[1], carry, &(q[2]), &carry);
      mul16c(i16[2], j16[1], carry, &(q[3]), &carry);
      mul16c(i16[3], j16[1], carry, &(q[4]), &carry);
      q[5] = carry;
   }

   uint16_t product[6];
   {
      uint16_t carry = 0;
      add16c(p[0], q[0], carry, &(product[0]), &carry);
      add16c(p[1], q[1], carry, &(product[1]), &carry);
      add16c(p[2], q[2], carry, &(product[2]), &carry);
      add16c(p[3], q[3], carry, &(product[3]), &carry);
      add16c(p[4], q[4], carry, &(product[4]), &carry);
      add16c(p[5], q[5], carry, &(product[5]), &carry);
   }

   (*prod_ptr)[0] = ( product[1] << 16 ) | product[0];
   (*prod_ptr)[1] = ( product[3] << 16 ) | product[2];
   *cout_ptr      = ( product[5] << 16 ) | product[4];
}

Element zero of each array is expected to be the least significant portion of the number.