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SubjectRe: {sys_,/dev/}epoll waiting timeout
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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