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
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This archive was generated by hypermail 2b29 : Thu Jan 23 2003 - 22:00:30 EST