[PATCH tip/core/rcu 9/9] Documentation/memory-barriers: Fix typos

paulmck
2020-02-14T23:39:46Z

From: SeongJae Park <sjpark@amazon.de>

Signed-off-by: SeongJae Park <sjpark@amazon.de>
Signed-off-by: Paul E. McKenney <paulmck@kernel.org>
---
Documentation/memory-barriers.txt | 8 ++++----
1 file changed, 4 insertions(+), 4 deletions(-)

diff --git a/Documentation/memory-barriers.txt b/Documentation/memory-barriers.txt
index 7146da0..e1c355e 100644
--- a/Documentation/memory-barriers.txt
+++ b/Documentation/memory-barriers.txt
@@ -185,7 +185,7 @@ As a further example, consider this sequence of events:
=============== ===============
{ A == 1, B == 2, C == 3, P == &A, Q == &C }
B = 4; Q = P;
- P = &B D = *Q;
+ P = &B; D = *Q;

There is an obvious data dependency here, as the value loaded into D depends on
the address retrieved from P by CPU 2. At the end of the sequence, any of the
@@ -569,7 +569,7 @@ following sequence of events:
{ A == 1, B == 2, C == 3, P == &A, Q == &C }
B = 4;
<write barrier>
- WRITE_ONCE(P, &B)
+ WRITE_ONCE(P, &B);
Q = READ_ONCE(P);
D = *Q;

@@ -1721,7 +1721,7 @@ of optimizations:
and WRITE_ONCE() are more selective: With READ_ONCE() and
WRITE_ONCE(), the compiler need only forget the contents of the
indicated memory locations, while with barrier() the compiler must
- discard the value of all memory locations that it has currented
+ discard the value of all memory locations that it has currently
cached in any machine registers. Of course, the compiler must also
respect the order in which the READ_ONCE()s and WRITE_ONCE()s occur,
though the CPU of course need not do so.
@@ -1833,7 +1833,7 @@ Aside: In the case of data dependencies, the compiler would be expected
to issue the loads in the correct order (eg. `a[b]` would have to load
the value of b before loading a[b]), however there is no guarantee in
the C specification that the compiler may not speculate the value of b
-(eg. is equal to 1) and load a before b (eg. tmp = a[1]; if (b != 1)
+(eg. is equal to 1) and load a[b] before b (eg. tmp = a[1]; if (b != 1)
tmp = a[b]; ). There is also the problem of a compiler reloading b after
having loaded a[b], thus having a newer copy of b than a[b]. A consensus
has not yet been reached about these problems, however the READ_ONCE()
--
2.9.5