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| -rw-r--r-- | Documentation/memory-barriers.txt | 183 |
1 files changed, 175 insertions, 8 deletions
diff --git a/Documentation/memory-barriers.txt b/Documentation/memory-barriers.txt index 2d22da095a60..deafa36aeea1 100644 --- a/Documentation/memory-barriers.txt +++ b/Documentation/memory-barriers.txt @@ -571,11 +571,10 @@ dependency barrier to make it work correctly. Consider the following bit of code: q = ACCESS_ONCE(a); - if (p) { - <data dependency barrier> - q = ACCESS_ONCE(b); + if (q) { + <data dependency barrier> /* BUG: No data dependency!!! */ + p = ACCESS_ONCE(b); } - x = *q; This will not have the desired effect because there is no actual data dependency, but rather a control dependency that the CPU may short-circuit @@ -584,11 +583,176 @@ the load from b as having happened before the load from a. In such a case what's actually required is: q = ACCESS_ONCE(a); - if (p) { + if (q) { <read barrier> - q = ACCESS_ONCE(b); + p = ACCESS_ONCE(b); } - x = *q; + +However, stores are not speculated. This means that ordering -is- provided +in the following example: + + q = ACCESS_ONCE(a); + if (ACCESS_ONCE(q)) { + ACCESS_ONCE(b) = p; + } + +Please note that ACCESS_ONCE() is not optional! Without the ACCESS_ONCE(), +the compiler is within its rights to transform this example: + + q = a; + if (q) { + b = p; /* BUG: Compiler can reorder!!! */ + do_something(); + } else { + b = p; /* BUG: Compiler can reorder!!! */ + do_something_else(); + } + +into this, which of course defeats the ordering: + + b = p; + q = a; + if (q) + do_something(); + else + do_something_else(); + +Worse yet, if the compiler is able to prove (say) that the value of +variable 'a' is always non-zero, it would be well within its rights +to optimize the original example by eliminating the "if" statement +as follows: + + q = a; + b = p; /* BUG: Compiler can reorder!!! */ + do_something(); + +The solution is again ACCESS_ONCE(), which preserves the ordering between +the load from variable 'a' and the store to variable 'b': + + q = ACCESS_ONCE(a); + if (q) { + ACCESS_ONCE(b) = p; + do_something(); + } else { + ACCESS_ONCE(b) = p; + do_something_else(); + } + +You could also use barrier() to prevent the compiler from moving +the stores to variable 'b', but barrier() would not prevent the +compiler from proving to itself that a==1 always, so ACCESS_ONCE() +is also needed. + +It is important to note that control dependencies absolutely require a +a conditional. For example, the following "optimized" version of +the above example breaks ordering: + + q = ACCESS_ONCE(a); + ACCESS_ONCE(b) = p; /* BUG: No ordering vs. load from a!!! */ + if (q) { + /* ACCESS_ONCE(b) = p; -- moved up, BUG!!! */ + do_something(); + } else { + /* ACCESS_ONCE(b) = p; -- moved up, BUG!!! */ + do_something_else(); + } + +It is of course legal for the prior load to be part of the conditional, +for example, as follows: + + if (ACCESS_ONCE(a) > 0) { + ACCESS_ONCE(b) = q / 2; + do_something(); + } else { + ACCESS_ONCE(b) = q / 3; + do_something_else(); + } + +This will again ensure that the load from variable 'a' is ordered before the +stores to variable 'b'. + +In addition, you need to be careful what you do with the local variable 'q', +otherwise the compiler might be able to guess the value and again remove +the needed conditional. For example: + + q = ACCESS_ONCE(a); + if (q % MAX) { + ACCESS_ONCE(b) = p; + do_something(); + } else { + ACCESS_ONCE(b) = p; + do_something_else(); + } + +If MAX is defined to be 1, then the compiler knows that (q % MAX) is +equal to zero, in which case the compiler is within its rights to +transform the above code into the following: + + q = ACCESS_ONCE(a); + ACCESS_ONCE(b) = p; + do_something_else(); + +This transformation loses the ordering between the load from variable 'a' +and the store to variable 'b'. If you are relying on this ordering, you +should do something like the following: + + q = ACCESS_ONCE(a); + BUILD_BUG_ON(MAX <= 1); /* Order load from a with store to b. */ + if (q % MAX) { + ACCESS_ONCE(b) = p; + do_something(); + } else { + ACCESS_ONCE(b) = p; + do_something_else(); + } + +Finally, control dependencies do -not- provide transitivity. This is +demonstrated by two related examples: + + CPU 0 CPU 1 + ===================== ===================== + r1 = ACCESS_ONCE(x); r2 = ACCESS_ONCE(y); + if (r1 >= 0) if (r2 >= 0) + ACCESS_ONCE(y) = 1; ACCESS_ONCE(x) = 1; + + assert(!(r1 == 1 && r2 == 1)); + +The above two-CPU example will never trigger the assert(). However, +if control dependencies guaranteed transitivity (which they do not), +then adding the following two CPUs would guarantee a related assertion: + + CPU 2 CPU 3 + ===================== ===================== + ACCESS_ONCE(x) = 2; ACCESS_ONCE(y) = 2; + + assert(!(r1 == 2 && r2 == 2 && x == 1 && y == 1)); /* FAILS!!! */ + +But because control dependencies do -not- provide transitivity, the +above assertion can fail after the combined four-CPU example completes. +If you need the four-CPU example to provide ordering, you will need +smp_mb() between the loads and stores in the CPU 0 and CPU 1 code fragments. + +In summary: + + (*) Control dependencies can order prior loads against later stores. + However, they do -not- guarantee any other sort of ordering: + Not prior loads against later loads, nor prior stores against + later anything. If you need these other forms of ordering, + use smb_rmb(), smp_wmb(), or, in the case of prior stores and + later loads, smp_mb(). + + (*) Control dependencies require at least one run-time conditional + between the prior load and the subsequent store. If the compiler + is able to optimize the conditional away, it will have also + optimized away the ordering. Careful use of ACCESS_ONCE() can + help to preserve the needed conditional. + + (*) Control dependencies require that the compiler avoid reordering the + dependency into nonexistence. Careful use of ACCESS_ONCE() or + barrier() can help to preserve your control dependency. + + (*) Control dependencies do -not- provide transitivity. If you + need transitivity, use smp_mb(). SMP BARRIER PAIRING @@ -1083,7 +1247,10 @@ compiler from moving the memory accesses either side of it to the other side: barrier(); -This is a general barrier - lesser varieties of compiler barrier do not exist. +This is a general barrier -- there are no read-read or write-write variants +of barrier(). Howevever, ACCESS_ONCE() can be thought of as a weak form +for barrier() that affects only the specific accesses flagged by the +ACCESS_ONCE(). The compiler barrier has no direct effect on the CPU, which may then reorder things however it wishes. |