On Tue, Aug 26, 2014 at 09:06:17AM -0700, Valentina Manea wrote:
> On Tue, Aug 26, 2014 at 6:12 AM, Dan Carpenter <dan.carpenter@xxxxxxxxxx> wrote:
> > 707 /*
> > 708 * loop over all packets from last to first (to prevent overwritting
> > 709 * memory when padding) and move them into the proper place
> > 710 */
> > 711 for (i = np-1; i > 0; i--) {
> >
> > I don't understand this loop. If we really intended to skip the first
> > packet then why isn't the "if (np == 0" condition "if (np <= 1"? That
> > would be more clear and the comment would say "if 1 packet then nothing
> > to unpack."
> >
> > 712 actualoffset -= urb->iso_frame_desc[i].actual_length;
> > 713 memmove(urb->transfer_buffer + urb->iso_frame_desc[i].offset,
> > 714 urb->transfer_buffer + actualoffset,
> > 715 urb->iso_frame_desc[i].actual_length);
> > 716 }
> > 717 }
> >
>
> urb->actual_length is the sum of individual packets' lengths,
> urb->iso_frame_desc[i].actual_length. Since actualoffset in the loop
> keeps decreasing, it will reach 0 when the first packet is processed.
> The first packet's offset is 0 as well and this just does memmove to
> the same buffer location.
It actually won't do the memove() if the i = 0, because the for loop
condition says i > 0.
> The (np == 0) condition is *not* the same as (np <= 1) because we want
> to return only is there are any packets.
Read the code again. For np == 1 then it doesn't do anything because it
doesn't enter the for loop. In other words, it returns without doing
anything either way.
I'm not necessarily saying the code is wrong... I just think that it
would be more clear if we said explicitly that we should return without
doing anything if np == 1.
regards,
dan carpenter
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