From: Andi Kleen <a...@linux.intel.com>

Add a simple heuristic to detect overlap of LBR entries and the call
stack when in lbr-as-callgraph mode. The return address in the
normal callstack is one off compared to the from entry in the
branch stack. Handle this with a simple "assume call instruction
is not longer than 8 bytes" heuristic. With that we can remove
any redundant call in the callstack that is already in the branch
stack.

Signed-off-by: Andi Kleen <a...@linux.intel.com>
---
 tools/perf/util/machine.c | 21 +++++++++++++++------
 1 file changed, 15 insertions(+), 6 deletions(-)

diff --git a/tools/perf/util/machine.c b/tools/perf/util/machine.c
index 853639c..1f167fe 100644
--- a/tools/perf/util/machine.c
+++ b/tools/perf/util/machine.c
@@ -1349,6 +1349,7 @@ static int machine__resolve_callchain_sample(struct 
machine *machine,
        int chain_nr = min(max_stack, (int)chain->nr);
        int i;
        int err;
+       int first_call = 0;
 
        callchain_cursor_reset(&callchain_cursor);
 
@@ -1362,8 +1363,6 @@ static int machine__resolve_callchain_sample(struct 
machine *machine,
         * Limitations for now:
         * - No extra filters
         * - No annotations (should annotate somehow)
-        * - When the sample is near the beginning of the function
-        *   we may overlap with the real callstack. 
         */
 
        if (branch->nr > PERF_MAX_BRANCH_DEPTH) {
@@ -1372,13 +1371,23 @@ static int machine__resolve_callchain_sample(struct 
machine *machine,
        }
 
        if (callchain_param.branch_callstack) {
-               int nr = min(max_stack, branch->nr);
+               int nr = min(max_stack, (int)branch->nr);
                struct branch_entry be[nr];
 
                for (i = 0; i < nr; i++) { 
-                       if (callchain_param.order == ORDER_CALLEE)
+                       if (callchain_param.order == ORDER_CALLEE) {
                                be[i] = branch->entries[i];
-                       else
+                               /* 
+                                * Check for overlap into the callchain.
+                                * The return address is one off compared to
+                                * the branch entry. To adjust for this 
+                                * assume the calling instruction is not longer
+                                * than 8 bytes.
+                                */
+                               if (be[i].from < chain->ips[first_call] &&
+                                   be[i].from >= chain->ips[first_call] - 8)
+                                       first_call++;
+                       } else
                                be[i] = branch->entries[branch->nr - i - 1];
                }
 
@@ -1405,7 +1414,7 @@ static int machine__resolve_callchain_sample(struct 
machine *machine,
                return 0;
        }
 
-       for (i = 0; i < chain_nr; i++) {
+       for (i = first_call; i < chain_nr; i++) {
                u64 ip;
 
                if (callchain_param.order == ORDER_CALLEE)
-- 
1.8.3.1

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