Messages in this thread | | | Date | Wed, 22 Jan 2003 11:32:06 -0800 (PST) | From | "Randy.Dunlap" <> | Subject | Re: {sys_,/dev/}epoll waiting timeout |
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On Wed, 22 Jan 2003, Jamie Lokier wrote:
| 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.
Like I said, I didn't get it. Now I do. Thanks.
-- ~Randy
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