Hi, On Thu, Jan 28, 2010 at 09:11:41AM -0700, Curtis Gedak wrote: > Petr Uzel wrote: > >Don't you mean something like 'if (retry_count % 3 == 0)' instead? > >That would pause on the 3rd, 6th and 9th (last) iteration. > > My original patch did try to do something like you suggest. After > analyzing the patch, I realized that it only paused once in all of > the 10 iterations. This single pause was sufficient to catch all of > the "failed to inform kernel of partition changes" problems that > occurred during my testing. > > Hence this time I did intend that the code only pause once for 1 > second. :-) > > My thoughts are that 1 second is a very long time for a computer. If > more than an single 1 second pause is used, then in my opinion this > would unnecessarily increase the amount of time required until the > call would finally fail. For example this function call would still > fail when used on a device with at least one partition mounted. I > did not want this to creep up to 2 or 3 seconds without significant > benefits to the user.
Yes, this make sense. But what confused me (and made me assume your intention was to sleep 3 times) is why it should sleep on 7th iteration out of 10. I don't think that after 'unsuccessful' one second sleep in 7th iteration the two other non-sleeping iterations would make any difference. Wouldn't be better if it sleeps on the last iteration instead? Or even get rid of the 9 non-sleeping iterations and just retry the ioctl after one second sleep and then give up? I know, this is nitpicking, but it simply isn't clear to me why the '3' is there. Regards, Petr -- Petr Uzel, openSUSE Boosters Team IRC: ptr_uzl @ freenode
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