Commit c8241f85 authored by Paul E. McKenney's avatar Paul E. McKenney

doc: Update control-dependencies section of memory-barriers.txt

This commit adds consistency to examples, formatting, and a couple of
additional warnings.
Signed-off-by: default avatarPaul E. McKenney <paulmck@linux.vnet.ibm.com>
Reviewed-by: default avatarJosh Triplett <josh@joshtriplett.org>
parent 526914a0
......@@ -640,6 +640,10 @@ See also the subsection on "Cache Coherency" for a more thorough example.
CONTROL DEPENDENCIES
--------------------
Control dependencies can be a bit tricky because current compilers do
not understand them. The purpose of this section is to help you prevent
the compiler's ignorance from breaking your code.
A load-load control dependency requires a full read memory barrier, not
simply a data dependency barrier to make it work correctly. Consider the
following bit of code:
......@@ -667,14 +671,15 @@ for load-store control dependencies, as in the following example:
q = READ_ONCE(a);
if (q) {
WRITE_ONCE(b, p);
WRITE_ONCE(b, 1);
}
Control dependencies pair normally with other types of barriers. That
said, please note that READ_ONCE() is not optional! Without the
READ_ONCE(), the compiler might combine the load from 'a' with other
loads from 'a', and the store to 'b' with other stores to 'b', with
possible highly counterintuitive effects on ordering.
Control dependencies pair normally with other types of barriers.
That said, please note that neither READ_ONCE() nor WRITE_ONCE()
are optional! Without the READ_ONCE(), the compiler might combine the
load from 'a' with other loads from 'a'. Without the WRITE_ONCE(),
the compiler might combine the store to 'b' with other stores to 'b'.
Either can result in highly counterintuitive effects on ordering.
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
......@@ -682,7 +687,7 @@ to optimize the original example by eliminating the "if" statement
as follows:
q = a;
b = p; /* BUG: Compiler and CPU can both reorder!!! */
b = 1; /* BUG: Compiler and CPU can both reorder!!! */
So don't leave out the READ_ONCE().
......@@ -692,11 +697,11 @@ branches of the "if" statement as follows:
q = READ_ONCE(a);
if (q) {
barrier();
WRITE_ONCE(b, p);
WRITE_ONCE(b, 1);
do_something();
} else {
barrier();
WRITE_ONCE(b, p);
WRITE_ONCE(b, 1);
do_something_else();
}
......@@ -705,12 +710,12 @@ optimization levels:
q = READ_ONCE(a);
barrier();
WRITE_ONCE(b, p); /* BUG: No ordering vs. load from a!!! */
WRITE_ONCE(b, 1); /* BUG: No ordering vs. load from a!!! */
if (q) {
/* WRITE_ONCE(b, p); -- moved up, BUG!!! */
/* WRITE_ONCE(b, 1); -- moved up, BUG!!! */
do_something();
} else {
/* WRITE_ONCE(b, p); -- moved up, BUG!!! */
/* WRITE_ONCE(b, 1); -- moved up, BUG!!! */
do_something_else();
}
......@@ -723,10 +728,10 @@ memory barriers, for example, smp_store_release():
q = READ_ONCE(a);
if (q) {
smp_store_release(&b, p);
smp_store_release(&b, 1);
do_something();
} else {
smp_store_release(&b, p);
smp_store_release(&b, 1);
do_something_else();
}
......@@ -735,10 +740,10 @@ ordering is guaranteed only when the stores differ, for example:
q = READ_ONCE(a);
if (q) {
WRITE_ONCE(b, p);
WRITE_ONCE(b, 1);
do_something();
} else {
WRITE_ONCE(b, r);
WRITE_ONCE(b, 2);
do_something_else();
}
......@@ -751,10 +756,10 @@ the needed conditional. For example:
q = READ_ONCE(a);
if (q % MAX) {
WRITE_ONCE(b, p);
WRITE_ONCE(b, 1);
do_something();
} else {
WRITE_ONCE(b, r);
WRITE_ONCE(b, 2);
do_something_else();
}
......@@ -763,7 +768,7 @@ equal to zero, in which case the compiler is within its rights to
transform the above code into the following:
q = READ_ONCE(a);
WRITE_ONCE(b, p);
WRITE_ONCE(b, 1);
do_something_else();
Given this transformation, the CPU is not required to respect the ordering
......@@ -776,10 +781,10 @@ one, perhaps as follows:
q = READ_ONCE(a);
BUILD_BUG_ON(MAX <= 1); /* Order load from a with store to b. */
if (q % MAX) {
WRITE_ONCE(b, p);
WRITE_ONCE(b, 1);
do_something();
} else {
WRITE_ONCE(b, r);
WRITE_ONCE(b, 2);
do_something_else();
}
......@@ -812,30 +817,28 @@ not necessarily apply to code following the if-statement:
q = READ_ONCE(a);
if (q) {
WRITE_ONCE(b, p);
WRITE_ONCE(b, 1);
} else {
WRITE_ONCE(b, r);
WRITE_ONCE(b, 2);
}
WRITE_ONCE(c, 1); /* BUG: No ordering against the read from "a". */
WRITE_ONCE(c, 1); /* BUG: No ordering against the read from 'a'. */
It is tempting to argue that there in fact is ordering because the
compiler cannot reorder volatile accesses and also cannot reorder
the writes to "b" with the condition. Unfortunately for this line
of reasoning, the compiler might compile the two writes to "b" as
the writes to 'b' with the condition. Unfortunately for this line
of reasoning, the compiler might compile the two writes to 'b' as
conditional-move instructions, as in this fanciful pseudo-assembly
language:
ld r1,a
ld r2,p
ld r3,r
cmp r1,$0
cmov,ne r4,r2
cmov,eq r4,r3
cmov,ne r4,$1
cmov,eq r4,$2
st r4,b
st $1,c
A weakly ordered CPU would have no dependency of any sort between the load
from "a" and the store to "c". The control dependencies would extend
from 'a' and the store to 'c'. The control dependencies would extend
only to the pair of cmov instructions and the store depending on them.
In short, control dependencies apply only to the stores in the then-clause
and else-clause of the if-statement in question (including functions
......@@ -843,7 +846,7 @@ invoked by those two clauses), not to code following that if-statement.
Finally, control dependencies do -not- provide transitivity. This is
demonstrated by two related examples, with the initial values of
x and y both being zero:
'x' and 'y' both being zero:
CPU 0 CPU 1
======================= =======================
......@@ -915,6 +918,9 @@ In summary:
(*) Control dependencies do -not- provide transitivity. If you
need transitivity, use smp_mb().
(*) Compilers do not understand control dependencies. It is therefore
your job to ensure that they do not break your code.
SMP BARRIER PAIRING
-------------------
......
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