speedup-example.cpp
/* Copyright (c) 2013, Regents of the Columbia University
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#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
#include <errno.h>
#include <assert.h>
//#include "tern/user.h"
#define N 4
//#define M 30000
#define M 30000*10
//int nwait = 0;
volatile long long sum;
volatile long long sum_2;
long loops = 6e3;
pthread_mutex_t mutex;
pthread_cond_t cond;
pthread_barrier_t bar;
void set_affinity(int core_id) {
cpu_set_t cpuset;
CPU_ZERO(&cpuset);
CPU_SET(core_id, &cpuset);
assert(pthread_setaffinity_np(pthread_self(), sizeof(cpu_set_t), &cpuset) == 0);
}
void* thread_func(void *arg) {
set_affinity((int)(long)arg);
for (int j = 0; j < M; j++) {
//pthread_mutex_lock(&mutex);
//nwait++;
for (long i = 0; i < loops; i++) // This is the key of speedup for parrot: the mutex needs to be a little bit congested.
{
sum += i;
sum_2 += i;
}
//pthread_cond_wait(&cond, &mutex);
//printf("being in the lock is: %lu\n", pthread_self());
//pthread_mutex_unlock(&mutex);
//soba_wait(0);
//pthread_barrier_wait(&bar);
for (long i = 0; i < loops; i++)
{
sum += i*i*i*i*i*i;
}
//fprintf(stderr, "compute thread %u %d\n", (unsigned)thread, sched_getcpu());
}
}
int main(int argc, char *argv[]) {
set_affinity(23);
//soba_init(0, N, 20);
pthread_t th[N];
int ret;
//pthread_cond_init(&cond, NULL);
//pthread_barrier_init(&bar, NULL, N);
for(unsigned i=0; i<N; ++i) {
ret = pthread_create(&th[i], NULL, thread_func, (void*)i);
assert(!ret && "pthread_create() failed!");
}
/*for (int j = 0; j < M; j++) {
while (nwait < N) {
sched_yield();
}
pthread_mutex_lock(&mutex);
nwait = 0;
//fprintf(stderr, "broadcast %u %d\n", (unsigned)pthread_self(), sched_getcpu());
pthread_cond_broadcast(&cond);
pthread_mutex_unlock(&mutex);
}
*/
for(unsigned i=0; i<N; ++i)
pthread_join(th[i], NULL);
// test the validity of results
printf("sum_2 = %lld\n", sum_2);
//printf("sum = %lld\n", sum);
exit(0);
}上面是我们整个旅程的路线，下面是我们的始发地。
xtern/lib/runtime/record-scheduler.cpp
void RRScheduler::getTurn()
{
int tid = self();
assert(tid>=0 && tid < Scheduler::nthread);
waits[tid].wait();
dprintf("RRScheduler: %d gets turn\n", self());
SELFCHECK;
}
/// tern tid for current thread
static int self() { return self_tid; }RRScheduler::getTurn是RRScheduler::wait_t::wait的唯一入口。
getTurn的作用是让一个线程运行，而让其他线程等待，实现串行化。
实际上getTurn最早见于struct Serializer，所以这里引出一些重要类的继承关系。
接下来要弄明白是哪里调用了getTurn，以及它在RR调度中起的作用。
在34次直接调用getTurn()的共有五处：
而调用SCHED_TIMER_START的函数有（它们都在xtern/lib/runtime/record-runtime.cpp）：
（B）利用tern_pthread_create创建四个子线程。
在第三回中我们来探索一下tern_pthread_create都干了些什么。
第三回
上回说到我们遇到了第一个强敌“两种顺序”，在首回交锋中，他使出了大招“用tern_pthread_create来创建子线程”。接下来看我们如何破。
这就是大招的原理分解图。在分析RecorderRT<_S>::pthreadCreate时，发现有提示：
/// The pthread_create wrapper solves three problems. /// /// Problem 1. We must assign a logical tern tid to the new thread while /// holding turn, or multiple newly created thread could get their logical /// tids nondeterministically. To do so, we assign a logical tid to a new /// thread in the thread that creates the new thread. /// /// If we were to assign this logical id in the new thread itself, we may /// run into nondeterministic runs, as illustrated by the following /// example /// /// t0 t1 t2 t3 /// getTurn(); /// create t2 /// putTurn(); /// getTurn(); /// create t3 /// putTurn(); /// getTurn(); /// get turn tid 2 /// putTurn(); /// getTurn(); /// get turn tid 3 /// putTurn(); /// /// in a different run, t1 may run first and get turn tid 2. /// /// Problem 2. When a child thread is created, the child thread may run /// into a getTurn() before the parent thread has assigned a logical tid /// to the child thread. This causes getTurn to refer to self_tid, which /// is undefined. To solve this problem, we use @thread_begin_sem to /// create a thread in suspended mode, until the parent thread has /// assigned a logical tid for the child thread. /// /// Problem 3. We can use one semaphore to create a child thread /// suspended. However, when there are two pthread_create() calls, we may /// still run into cases where a child thread tries to get its /// thread-local tid but gets -1. consider /// /// t0 t1 t2 t3 /// getTurn(); /// create t2 /// sem_post(&thread_begin_sem) /// putTurn(); /// getTurn(); /// create t3 /// sem_wait(&thread_begin_sem); /// self_tid = TidMap[pthread_self()] /// sem_post(&thread_begin_sem) /// putTurn(); /// getTurn(); /// get turn tid 2 /// putTurn(); /// sem_wait(&thread_begin_sem); /// self_tid = TidMap[pthread_self()] /// getTurn(); /// get turn tid 3 /// putTurn(); /// /// The crux of the problem is that multiple sem_post can pair up with /// multiple sem_down in different ways. We solve this problem using /// another semaphore, thread_begin_done_sem. ///
看来是上乘剑法。
有了招数分解图和一部分剑谱还是费解，那么先来看一下我们的伙伴厨师top探得的消息：
情景一：./speedup-example（有mutex，四个子线程，M=30000*10，没使用Parrot）
情景二：./run_example_tern（有mutex，四个子线程，M=30000*10，使用Parrot）
情景三：./run_example_tern（没有mutex，四个子线程，M=30000*10，使用Parrot）
通过top，我们发现——
这让我们想探究一下在情景二和情景三下四个子线程的创建顺序是什么样的。先设想一下：情景二是四个子线程近乎同时创建，而情景二是一开始先近乎同时创建两个子线程，此后每当一个子线程完成自己的任务结束时再创建一个新的子线程。
情景四：./run_example_tern（有mutex，八个子线程，M=30000*10，使用Parrot）
情景五：./run_example_tern（没有mutex，八个子线程，M=30000*10，使用Parrot）
原文地址：http://blog.csdn.net/bluecloudmatrix/article/details/37955147
