implement 64-bit arithmetic on a 32-bit machine

Question

The following code computes the product of x and y and stores the result in memory. Data type ll_t is defined to be equivalent to long long.

typedef long lon         


        

Answer 1

I've converted it to intel syntax.

mov esi, y_low
mov eax, x
mov edx, eax
sar edx, 31
mov ecx, y_high

imul ecx, eax ; ecx = y_high *{signed} x

mov ebx, edx

imul ebx, esi ; ebx = sign_extension(x) *{signed} y_low

add ecx, ebx ; ecx = y_high *{signed} x_low + x_high *{signed} y_low

mul esi ; edx:eax = x_low *{unsigned} y_low

lea edx, [ecx + edx] ; edx = high(x_low *{unsigned} y_low + y_high *{signed} x_low + x_high *{signed} y_low)

mov ecx, dest
mov [ecx], eax
mov [ecx + 4], edx

What the above code does is multiplication of 2 64-bit signed integers that keeps the least-significant 64 bits of the product.

Where does the other 64-bit multiplicand come from? It's x sign-extended from 32 bits to 64. The sar instruction is used to replicate x's sign bit into all bits of edx. I call this value consisting only of the x's sign x_high. x_low is the value of x actually passed into the routine.

y_low and y_high are the least and most significant parts of y, just like x's x_low and x_high are.

From here it's pretty easy:

product = y *{signed} x =
(y_high * 2³² + y_low) *{signed} (x_high * 2³² + x_low) =
y_high *{signed} x_high * 2⁶⁴ +
y_high *{signed} x_low * 2³² +
y_low *{signed} x_high * 2³² +
y_low *{signed} x_low

y_high *{signed} x_high * 2⁶⁴ isn't calculated because it doesn't contribute to the least significant 64 bits of the product. We'd calculate it if we were interested in the full 128-bit product (full 96-bit product for the picky).

y_low *{signed} x_low is calculated using unsigned multiplication. It's legal to do so because 2's complement signed multiplication gives the same least significant bits as unsigned multiplication. Example:
-1 *{signed} -1 = 1
0xFFFFFFFFFFFFFFFF *{unsigned} 0xFFFFFFFFFFFFFFFF = 0xFFFFFFFFFFFFFFFE0000000000000001 (64 least significant bits are equivalent to 1)