fixed point multiplication without 64 bit temporary

问题

Hi I'm implementing some fixed point math stuff for embedded systems and I'm trying to do the multiplication of two 16.16 fixed point numbers without creating a 64bit temporary. So far here is the code I came up with that generates the least instructions.

int multiply(int x, int y){
    int result;
    long long temp = x;
    temp *= y;
    temp >>= 16;
    result = temp;
    return result;
}

the problem with this code is that it uses a temporary 64 bit integer which seem to generate bad assembly code. I'm trying to make a system that uses two 32 bit integers instead of a 64 bit one. Anyone know how to do this?

回答1:

Think of your numbers as each composed of two large "digits."

  A.B
x C.D

The "base" of the digits is the 2^bit_width, i.e., 2^16, or 65536.

So, the product is

D*B       + D*A*65536 + C*B*65536 + C*A*65536*65536

However, to get the product shifted right by 16, you need to divide all these terms by 65536, so

D*B/65536 + D*A       + C*B        + C*A*65536

In C:

uint16_t a = x >> 16;
uint16_t b = x & 0xffff;
uint16_t c = y >> 16;
uint16_t d = y & 0xffff;

return ((d * b) >> 16) + (d * a) + (c * b) + ((c * a) << 16);

The signed version is a bit more complicated; it is often easiest to perform the arithmetic on the absolute values of x and y and then fix the sign (unless you overflow, which you can check for rather tediously).

来源：https://stackoverflow.com/questions/15123638/fixed-point-multiplication-without-64-bit-temporary