qemu-targetstep: Attempt to track guest process gfn when running in userspace

Seems like single-stepping the guest with LAPIC influences the guest scheduler behaviour, since just a single step inside the target gfn (to determine if its running in userspace), is enough to for us to never reach the GUEST_STOP_TRACK event. FWICT the single-stepping is not frequent and does not take long enough to justify never reaching the stop event.

1 files changed, 0 insertions, 45 deletions

diff --git a/test/qemu-eviction_guest.c b/test/qemu-eviction_guest.c
deleted file mode 100644
index 086fee5..0000000
--- a/test/qemu-eviction_guest.c
+++ /dev/null
@@ -1,45 +0,0 @@
-#include "cachepc/uapi.h"
-
-#include <sys/time.h>
-#include <sys/resource.h>
-#include <err.h>
-#include <unistd.h>
-#include <stdint.h>
-#include <string.h>
-#include <stdio.h>
-#include <stdlib.h>
-
-int
-main(int argc, const char **argv)
-{
-	void *buf;
-
-	buf = NULL;
-	if (posix_memalign(&buf, L1_LINESIZE * L1_SETS, L1_LINESIZE * L1_SETS))
-		err(1, "memalign");
-	memset(buf, 0, L1_LINESIZE * L1_SETS);
-
-	setpriority(PRIO_PROCESS, 0, -20);
-
-	while (1) {
-		printf("LOOP\n");
-		CPC_DO_VMMCALL(KVM_HC_CPC_VMMCALL_SIGNAL,
-			CPC_GUEST_START_TRACK, 0);
-		*(uint8_t *)(buf + L1_LINESIZE * 9) = 1;
-		*(uint8_t *)(buf + L1_LINESIZE * 10) = 1;
-		*(uint8_t *)(buf + L1_LINESIZE * 11) = 1;
-		*(uint8_t *)(buf + L1_LINESIZE * 12) = 1;
-		*(uint8_t *)(buf + L1_LINESIZE * 13) = 1;
-		*(uint8_t *)(buf + L1_LINESIZE * 14) = 1;
-		*(uint8_t *)(buf + L1_LINESIZE * 15) = 1;
-		*(uint8_t *)(buf + L1_LINESIZE * 9) = 1;
-		*(uint8_t *)(buf + L1_LINESIZE * 10) = 1;
-		*(uint8_t *)(buf + L1_LINESIZE * 11) = 1;
-		*(uint8_t *)(buf + L1_LINESIZE * 12) = 1;
-		*(uint8_t *)(buf + L1_LINESIZE * 13) = 1;
-		*(uint8_t *)(buf + L1_LINESIZE * 14) = 1;
-		*(uint8_t *)(buf + L1_LINESIZE * 15) = 1;
-		CPC_DO_VMMCALL(KVM_HC_CPC_VMMCALL_SIGNAL,
-			CPC_GUEST_STOP_TRACK, 0);
-	}
-}