Lukasz Kosewski wrote:
+ if (plugdata) {
+ writeb(plugdata, mmio_base + hotplug_offset);
+ for (i = 0; i < host_set->n_ports; ++i) {
+ ap = host_set->ports[i];
+ if (!(ap->flags & ATA_FLAG_SATA))
+ continue; //No PATA support here... yet
+ // Check unplug flag
+ if (plugdata & 0x1) {
+ /* Do stuff related to unplugging a device */
+ ata_hotplug_unplug(ap);
+ handled = 1;
+ } else if ((plugdata >> 4) & 0x1) { //Check plug flag
+ /* Do stuff related to plugging in a device */
+ ata_hotplug_plug(ap);
+ handled = 1;
What happens if both bits are set? Seems like that could happen, if a
plug+unplug (cable blip?) occurs in rapid succession.
What IF both bits are set? This is why we have a debounce timer to
take care of the problem :P
The way this is set up, unplugging will win out (plugging will come
first, unplugging will come and destroy 'plug's timer, and then the
unplug action will be performed on timer expiry). Personally, I like
it this way, but I can reverse the order of these two to make plugging
the default action. Which do you prefer?
The above logic
* acks multiple events
* handles only a single event
so either way you lose an event. In the code as it is written above, if
both 'plug' and 'unplug' events are noted, then only the unplug get
handled, and the newly-plugged device is never noticed.
Jeff
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