0x0
前些天组里老司机@梁希在jvm的项目榨干机器性能之余，为了检查下gcc编译器和Intel Xoen CPU的正确性，写了一组测试代码测试了下mfence指令的效果
`
mfence Opcode : 0F AE /6
Performs a serializing operation on all load-from-memory and store-to-memory instructions that were issued prior the MFENCE instruction. This serializing operation guarantees that every load and store instruction that precedes in program order the MFENCE instruction is globally visible before any load or store instruction that follows the MFENCE instruction is globally visible. The MFENCE instruction is ordered with respect to all load and store instructions, other MFENCE instructions, any SFENCE and LFENCE instructions, and any serializing instructions (such as the CPUID instruction).
Weakly ordered memory types can be used to achieve higher processor performance through such techniques as out-of-order issue, speculative reads, write-combining, and write-collapsing.
The degree to which a consumer of data recognizes or knows that the data is weakly ordered varies among applications and may be unknown to the producer of this data. The MFENCE instruction provides a performance-efficient way of ensuring load and store ordering between routines that produce weakly-ordered results and routines that consume that data.
It should be noted that processors are free to speculatively fetch and cache data from system memory regions that are assigned a memory-type that permits speculative reads (that is, the WB, WC, and WT memory types). The PREFETCHh instruction is considered a hint to this speculative behavior. Because this speculative fetching can occur at any time and is not tied to instruction execution, the MFENCE instruction is not ordered with respect to PREFETCHh instructions or any other speculative fetching mechanism (that is, data could be speculatively loaded into the cache just before, during, or after the execution of an MFENCE instruction).
`
简单来说就是一个可以在CPU乱序执行中保证真实的load/store顺序的指令
0x1
老司机写了一个小程序(注：有误版)
// file: order.c
#define _GNU_SOURCE
#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
union p64 {
int i;
char padding[64];
long align8;
};
volatile union p64 v1, v2;
int b;
void *
run1(
void *ignore)
{
for (;;) {
while (!b);
if (v1.i || v2.i) {
puts("assert error 1");
exit(-1);
}
v1.i = 1;
asm ("sfence": : :"memory");
v2.i = 1;
asm ("sfence": : :"memory");
b = 0;
}
}
intmain()
{
pthread_t p;
pthread_create(&p, NULL, run1, NULL);
int cnt = 0;
for (;; cnt++) {
v1.i = v2.i = 0;
asm ("sfence": : :"memory");
b = 1;
asm ("sfence": : :"memory");
int icnt = 0;
for (;; icnt++) {
int i1 = v1.i;
asm ("lfence": : :"memory");
int i2 = v2.i;
if (i1 && i2)
break;
if (i1 < i2) {
printf("assert error, cnt = %d, icnt = %d, i1 = %d, i2 = %d\n", cnt, icnt, i1, i2);
exit(-1);
}
}
}
return 0;
}
大概逻辑是： 一共有3个变量，v1.i, v2.i, b ,起了2个线程，一个顺序写入v1和v2，一个读v1和v2，互相通过改变b的值来通讯，然后两个线程不停循环。
这个程序会挂在
printf("assert error, cnt = %d, icnt = %d, i1 = %d, i2 = %d\n", cnt, icnt, i1, i2);
这条断言上，意思是线程1在顺序写入v1和v2，但是主线程却出现读到 v1=0，v2=1的情况。
0x2
然后我帮忙去看了一下，觉得这种写法甚是粗暴，于是原样照搬了一个c++11版:
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <atomic>
#include <thread>
using namespace std;
union p64 {
atomic<
int> i;
char padding[64];
long align8;
};
volatile union p64 v1, v2;
atomic<
int> b;
void *
run1()
{
int rcnt = 0;
for (;; rcnt++) {
while (!b.load());
if (v1.i.load() || v2.i.load()) {
puts("assert error 1");
exit(-1);
}
v1.i.store(1);
v2.i.store(1);
b.store(0);
}
}
intmain()
{
// init
v1.i.store(0);
v2.i.store(0);
thread t(run1);
int cnt = 0;
for (;; cnt++) {
v1.i.store(0);
v2.i.store(0);
b.store(1);
int icnt = 0;
for (;; icnt++) {
int b2 = b.load();
int i1 = v1.i.load();
// *****
int i2 = v2.i.load();
// *****
if (i1 && i2)
break;
if (i1 < i2) {
printf("assert error, cnt = %d, icnt = %d, i1 = %d, i2 = %d\n", cnt, icnt, i1, i2);
exit(-1);
}
if (i1 == 0 && i2 == 0 && b2 == 0)
break;
}
}
return 0;
}
因为是原样照搬，所以肯定还是会挂，但是毕竟语义上更好理解了
我们先来分析一下为什么会挂
- 线程1对于v1，v2的写入顺序一定是一致的
- Memory Barrier也保证了他们写入顺序对其他线程的可见性（很有迷惑性的一点）
- 但是主线程却可以读到 v1=0,v2=1的情况
- 所以情况就是虽然顺序写入了，但是别的线程没有看到正确的顺序？
- Intel: 并不是！
- 原因是搞错了因果关系，他真正保证的顺序是当你读到v2的new value的时候，那么v1也一定被写入了。
- 解决方案就是互换上面代码中我用**星号**标注出的两行
- done
在旧写法中，挂掉的情况是线程1写入v1 = 1，主线程读v1，没有读到，那么主线程认为v1是0，然后线程1继续写入v2，主线程读到了，主线程认为v2是1。 然后挂在了断言上。
两行互换后，主线程首先读取v2，如果v2已经是1了，那么v1也一定是1，反之亦然。
0x3
当然，想让跑通那个例子不需要那么多的atomic<>，精简之后利用c++11的memory_order可以写成如下：
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <atomic>
#include <thread>
using namespace std;
union p64 {
int i;
char padding[64];
long align8;
};
volatile union p64 v1, v2;
atomic<
int> b;
// variable b as a guard
void *
run1()
{
int rcnt = 0;
for (;; rcnt++) {
while (!b.load());
if (v1.i || v2.i) {
puts("assert error 1");
exit(-1);
}
v1.i = 1;
v2.i = 1;
b.store(0, memory_order_release);
}
}
intmain()
{
// init
v1.i = 0;
v2.i = 0;
thread t(run1);
int cnt = 0;
for (;; cnt++) {
v1.i = 0;
v2.i = 0;
b.store(1, memory_order_release);
int icnt = 0;
for (;; icnt++) {
int b2 = b.load(memory_order_acquire);
if (b2 != 0) {
continue;
}
int i1 = v1.i;
int i2 = v2.i;
if (i1 && i2)
break;
if (i1 < i2) {
printf("assert error 2, cnt = %d, icnt = %d, i1 = %d, i2 = %d\n", cnt, icnt, i1, i2);
exit(-1);
}
}
}
return 0;
}
利用变量b在两个线程之间同步，如下图
(Thead 1)
v1.i = 1;
v2.i = 1;
b.store(0, memory_order_release) <---+
|
synchronize with b
(happend before)
|
+-----> b.load(memory_order_acquire)
i1 = v1.i
i2 = v2.i
(Thread 2)
我们查看下生成的代码
g++ -std=c++11 -pthread -g -O2 order.cpp
v1.i = 1;
400be6: c7 05 d0 10 20 00 01 movl $0x1,0x2010d0(%rip) # 601cc0 <v1>
400bed: 00 00 00
v2.i = 1;
400bf0: c7 05 86 10 20 00 01 movl $0x1,0x201086(%rip) # 601c80 <v2>
400bf7: 00 00 00
memory_order __b = __m & __memory_order_mask;
__glibcxx_assert(__b != memory_order_acquire);
__glibcxx_assert(__b != memory_order_acq_rel);
__glibcxx_assert(__b != memory_order_consume);
__atomic_store_n(&_M_i, __i, __m);
400bfa: c7 05 5c 10 20 00 00 movl $0x0,0x20105c(%rip) # 601c60 <b>
400c01: 00 00 00
b.store(0, memory_order_release);
400a58: 8b 05 02 12 20 00 mov 0x201202(%rip),%eax # 601c60 <b>
int b2 = b.load(memory_order_consume);
if (b2 != 0) {
400a5e: 85 c0 test %eax,%eax
400a60: 75 f3 jne 400a55 <main+0x55>
continue;
}
int i1 = v1.i;
400a62: 8b 0d 58 12 20 00 mov 0x201258(%rip),%ecx # 601cc0 <v1>
int i2 = v2.i;
400a68: 44 8b 05 11 12 20 00 mov 0x201211(%rip),%r8d # 601c80 <v2>
看来Intel的Strong Memory Model已经保证了这一点，Memory Barrier都不需要了
（虽然标题里面有MemoryBarrier，但是内容里面根本没涉及的样子。。）
posted on 2016-01-19 16:13
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