>> 1) Look again; it's O(1) work per entry, or O(n) work for an n-entry >> directory. And O(1) space. With very small constant factors, > Yes. I was thinking about it this morning (after coffee). Thank you for the second look. > One variant on those algorithms that might make sense here is to save > the current cookie each time we see that the result of a cookie search > is a filp->f_pos offset < the current filp->f_pos offset. That means we > will in general only detect the loop after going through an entire > cycle, but that should be sufficient... All of these low-overhead algorithms can take a couple of loop iterations before they detect it; their job is to achieve a reasonably low constant factor in time using O(1) space. The worst case for the power-of-two algorithm is when the loop is n = 2^k+1 items long. When you get to item 2^(k+1), you'll be comparing to item 2^k, which is a mismatch. Then you'll save the cookie from 2^(k+1) and have to go to 2^(k+1) + 2^k + 1, or about 3*n, before detecting it. I don't consider this a problem, because it wastes a few seconds of computer time, to be followed by wasting a few hours trying to pass a bug report upstream about the broken NFS server... I don't quite follow how your proposed variant works. Pardon my ignorance of NFS, but is the f->pos something that comes from the server, or something that is synthesized locally? Obviously, if you keep a record of all the server cookies, you can detect loops quite easily. If it comes from the server, there's a risk that there might be two backward jumps in the cycle, and thus you'll never notice it. -- To unsubscribe from this list: send the line "unsubscribe linux-nfs" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
