4 ms·
> under/overflow Assuming the 2 variables are treated as unsigned ints, even with under/overflows the algorithm works, since if a=a+b overflows, then b=a-b is g
by setuids 5y ago
> under/overflow
Assuming the 2 variables are treated as unsigned ints, even with under/overflows the algorithm works, since if a=a+b overflows, then b=a-b is guaranteed to underflow, thus returning the original a.
The overflowed bit is irrelevant here
- wolpoli 5y agoI never considered that this algorithm would work so well with unsigned int. Looks like I learn something new today.