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

[Jamie Lokier]

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