1 files changed, 39 insertions, 29 deletions

diff --git a/src/mem/bus.cc b/src/mem/bus.cc
index 7584ffffd..66cd581e7 100644
--- a/src/mem/bus.cc
+++ b/src/mem/bus.cc
@@ -82,33 +82,6 @@ const char * Bus::BusFreeEvent::description()
     return "bus became available";
 }
 
-void
-Bus::occupyBus(int numCycles)
-{
-    //Move up when the bus will next be free
-    //We avoid the use of divide by adding repeatedly
-    //This should be faster if the value is updated frequently, but should
-    //be may be slower otherwise.
-
-    //Bring tickNextIdle up to the present tick
-    //There is some potential ambiguity where a cycle starts, which might make
-    //a difference when devices are acting right around a cycle boundary. Using
-    //a < allows things which happen exactly on a cycle boundary to take up only
-    //the following cycle. Anthing that happens later will have to "wait" for the
-    //end of that cycle, and then start using the bus after that.
-    while (tickNextIdle < curTick)
-        tickNextIdle += clock;
-    //Advance it numCycles bus cycles.
-    //XXX Should this use the repeating add trick as well?
-    tickNextIdle += (numCycles * clock);
-    if (!busIdle.scheduled()) {
-        busIdle.schedule(tickNextIdle);
-    } else {
-        busIdle.reschedule(tickNextIdle);
-    }
-    DPRINTF(Bus, "The bus is now occupied from tick %d to %d\n", curTick, tickNextIdle);
-}
-
 /** Function called by the port when the bus is receiving a Timing
  * transaction.*/
 bool
@@ -120,6 +93,14 @@ Bus::recvTiming(Packet *pkt)
 
     Port *pktPort = interfaces[pkt->getSrc()];
 
+    // If the bus is busy, or other devices are in line ahead of the current
+    // one, put this device on the retry list.
+    if (tickNextIdle > curTick ||
+            (retryList.size() && pktPort != retryingPort)) {
+        addToRetryList(pktPort);
+        return false;
+    }
+
     short dest = pkt->getDest();
     if (dest == Packet::Broadcast) {
         if (timingSnoop(pkt)) {
@@ -146,7 +127,17 @@ Bus::recvTiming(Packet *pkt)
         port = interfaces[dest];
     }
 
-    // The packet will be sent. Figure out how long it occupies the bus.
+    //Bring tickNextIdle up to the present tick
+    //There is some potential ambiguity where a cycle starts, which might make
+    //a difference when devices are acting right around a cycle boundary. Using
+    //a < allows things which happen exactly on a cycle boundary to take up only
+    //the following cycle. Anthing that happens later will have to "wait" for
+    //the end of that cycle, and then start using the bus after that.
+    while (tickNextIdle < curTick)
+        tickNextIdle += clock;
+
+    // The packet will be sent. Figure out how long it occupies the bus, and
+    // how much of that time is for the first "word", aka bus width.
     int numCycles = 0;
     // Requests need one cycle to send an address
     if (pkt->isRequest())
@@ -167,7 +158,26 @@ Bus::recvTiming(Packet *pkt)
         }
     }
 
-    occupyBus(numCycles);
+    // The first word will be delivered after the current tick, the delivery
+    // of the address if any, and one bus cycle to deliver the data
+    pkt->firstWordTime =
+        tickNextIdle +
+        pkt->isRequest() ? clock : 0 +
+        clock;
+
+    //Advance it numCycles bus cycles.
+    //XXX Should this use the repeated addition trick as well?
+    tickNextIdle += (numCycles * clock);
+    if (!busIdle.scheduled()) {
+        busIdle.schedule(tickNextIdle);
+    } else {
+        busIdle.reschedule(tickNextIdle);
+    }
+    DPRINTF(Bus, "The bus is now occupied from tick %d to %d\n",
+            curTick, tickNextIdle);
+
+    // The bus will become idle once the current packet is delivered.
+    pkt->finishTime = tickNextIdle;
 
     if (port->sendTiming(pkt))  {
         // Packet was successfully sent. Return true.