Forum Discussion
Altera_Forum
Honored Contributor
15 years agoMy implementation requires me to multiply a 14-bit number with a 18-bit number, and take 10 MSB bits as my output. So I simply let all the 32-bits be outputted and then discarded the 22 LSBs.
So after reading that link I posted earlier, I thought why should I keep two sign bits in my 10-bit output and thought of discarding the the second MSB. Leaving all the 10 MSB as my output should be fine I suppose?