Messages in this thread | ![/](/images/icornerl.gif) | | Date | Wed, 22 Jan 2003 19:34:46 +0000 | From | Jamie Lokier <> | Subject | Re: {sys_,/dev/}epoll waiting timeout |
| |
Randy.Dunlap wrote: > | > Why assume HZ=1000? Would not: > | > > | > timeout = (unsigned long)(timeout*HZ+(HZ-1))/HZ+1; > | > > | > make more sense? > | > | No, that's silly. Why do you want to multiply by HZ and then divide by HZ? > > OK, I don't get it. All Ed did was replace 1000 with HZ and > 999 with (HZ-1). What's bad about that? Seems to me like > the right thing to do. Much more portable. > > What if HZ changes? Who's going to audit the kernel for changes?
You're being dense. The input timeout is measured in milliseconds; see poll(2). The calculated timeout is measured in jiffies. Hence multiply by jiffies and divide by milliseconds.
-- Jamie - To unsubscribe from this list: send the line "unsubscribe linux-kernel" in the body of a message to majordomo@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html Please read the FAQ at http://www.tux.org/lkml/
| ![\](/images/icornerr.gif) |