Keep scanning the tree for overlapping IO otherwise Q2G and process
traces will be incorrect.
Let assume we have 2 IOs:
A A+a
|---------------------------------------|
B B+b
|-----------------|
In the red/black tree we have:
o -> [A,A+a]
/ \
left right
/ \
[...]o o -> [B, B+b]
In the current code, if we would not be able to find [B+b] in the tree:
B is greater than A, so we won't go left
B+b is smaller than A+a, so we are not going right either.
When we have a [X, X+x] IO to look for:
We need to check for right when either:
X+x >= A+a (for merged IO)
and
X > A (for overlapping IO)
TEST=Check with a trace with overlapping IO: Q2C and Q2G are expected.
Signed-off-by: Gwendal Grignou <gwendal@xxxxxxxxxxxx>
---
btt/dip_rb.c | 4 ++--
1 file changed, 2 insertions(+), 2 deletions(-)
diff --git a/btt/dip_rb.c b/btt/dip_rb.c
index 2aa7ffc..6efef6c 100644
--- a/btt/dip_rb.c
+++ b/btt/dip_rb.c
@@ -57,7 +57,7 @@ struct io *rb_find_sec(struct rb_root *root, __u64 sec)
__iop = rb_entry(n, struct io, rb_node);
if (sec < BIT_START(__iop))
n = n->rb_left;
- else if (sec >= BIT_END(__iop))
+ else if (sec > BIT_START(__iop))
n = n->rb_right;
else
return __iop;
@@ -82,7 +82,7 @@ void rb_foreach(struct rb_node *n, struct io *iop,
}
if (iop_s < this_s)
rb_foreach(n->rb_left, iop, fnc, head);
- if (this_e < iop_e)
+ if ((this_e < iop_e) || (this_s < iop_s))
rb_foreach(n->rb_right, iop, fnc, head);
}
}
--
2.14.1.480.gb18f417b89-goog
--
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