keon/awesome-bits

m & (n - 1) doesn't seem to work as m % (pow(2, n))

calebsander opened this issue · 2 comments

calebsander commented

I assume "Modulo 2n against m" means m % (pow(2, n)). In that case, for m=11 and n = 3, m & (n - 1) gives 2, but 11 % 8 gives 3. Perhaps you mean m & ((1 << n) - 1)?

xmikus01 commented

You are right, closed by #18.

keon commented

Thanks guys.
I am closing this issue since it is solved.