In the book "Musical Applications of Microprocessors," the author gives the following algorithm to do a 4 quadrant multiplication of two 8 bit signed integers with a 16 bit signed result:
Do an unsigned multiply on the raw operands. Then to correct the result, if the multiplicand sign is negative, unsigned single precision subtract the multiplier from the top 8 bits of the raw 16 bit result. If the multiplier sign is also negative, unsigned single precision subtract the multiplicand from the top 8 bits of the raw 16 bit result.
I tried implementing this in assembler and can't seem to get it to work. For example, if I unsigned multiply -2 times -2 the raw result in binary is B11111100.00000100. When I subtract B1111110 twice from the top 8 bits according to the algorithm, I get B11111110.00000100, not B00000000.00000100 as one would want. Thanks for any insight into where I might be going wrong!
Edit - code:
#define smultfix(a,b) \
({ \
int16_t sproduct; \
int8_t smultiplier = a, smultiplicand = b; \
uint16_t uproduct = umultfix(smultiplier,smultiplicand);\
asm volatile ( \
"add %2, r1 \n\t" \
"brpl smult_"QUOTE(__LINE__)"\n\t" \
"sec \n\t" \
"sbc %B3, %1 \n\t" \
"smult_"QUOTE(__LINE__)": add %1, r1 \n\t" \
"brpl send_"QUOTE(__LINE__)" \n\t" \
"sec \n\t" \
"sbc %B3, %2 \n\t" \
"send_"QUOTE(__LINE__)": movw %A0,%A3 \n\t" \
:"=&r" (sproduct):"a" (smultiplier), "a" (smultiplicand), "a" (uproduct)\
); \
sproduct; \
})
Edit: You got the subtraction wrong.
1111'1110b * 1111'1110b == 1111'1100'0000'0100b
-1111'1110'0000'0000b
-1111'1110'0000'0000b
---------------------
100b
Otherwise your algorithm is correct: In the fourth quadrant, you need to subtract 100h multiplied with the sum (a+b). Writing the two-complement bytes as (100h-x) I get:
(100h-a)(100h-b) = 10000h - 100h*(a+b) + ab = 100h*(100h-a) + 100h*(100h-b) + ab mod 10000h
(100h-a)(100h-b) - 100h*(100h-a) - 100*(100h-b) = ab mod 10000h