here's an ignorant, naive, and very likely wrong attempt:
what happens if you mask off the high and low bytes of the
larger number, do two 8,8->16 multiplies, left shift the result
of the result of the higher one, and add, as a macro?
#define _mul8x16(c,s) ( \
(long int) ((c) * (unsigned
I would like to multiply a 16-bit number by an 8-bit number and
produce a 24-bit result on the AVR. The AVR has a hardware 8-bit *
8-bit -> 16-bit multiplier.
If I multiply a 16-bit number by a 16-bit number, it produces a 16-bit
result, which isn't wide enough to hold the result.
If I cast one
