> I understand 'r & 3' to be logically equivalent to '(r & 2) && (r & 1)', > right?
For r == 2, r & 3 == 2, whereas (r & 2) && (r & 1) == 0, so no? Aldy
Aldy Hernandez via Gcc-patches Thu, 20 Oct 2022 05:23:54 -0700
> I understand 'r & 3' to be logically equivalent to '(r & 2) && (r & 1)', > right?
For r == 2, r & 3 == 2, whereas (r & 2) && (r & 1) == 0, so no? Aldy