Hello,
So as I recall, the following can be an infinite loop now with optimizations, right?
for (int i(1); i!=0; ++i) { ... }
What about:
unsigned int x = 0xFFFFFFFFU;
x = x+1;
if (x) { ... can we get here because "positive x + 1 must still positive"? ... }
If not, given the first, why not?
Thanks,
Jason Mancini
