在linux了sched_setaffinity CPU亲和力 [英] sched_setaffinity cpu affinity in linux

问题描述

我已经做在Linux中了sched_setaffinity测试中的服务器与1插槽，4核心，
下面的/ proc内/ cpuinfo showes CPU的信息：

I have done a sched_setaffinity test in Linux in a server with 1 socket ,4 cores , the following /proc/cpuinfo showes the cpu information :

processor       : 0
model name      : Intel(R) Core(TM)2 Quad CPU    Q8400  @ 2.66GHz
cache size      : 2048 KB
physical id     : 0
siblings        : 4
cpu cores       : 4

processor       : 1
model name      : Intel(R) Core(TM)2 Quad CPU    Q8400  @ 2.66GHz
cache size      : 2048 KB
physical id     : 0
siblings        : 4
cpu cores       : 4

processor       : 2
model name      : Intel(R) Core(TM)2 Quad CPU    Q8400  @ 2.66GHz
cache size      : 2048 KB
physical id     : 0
siblings        : 4
cpu cores       : 4

processor       : 3
model name      : Intel(R) Core(TM)2 Quad CPU    Q8400  @ 2.66GHz
cache size      : 2048 KB
physical id     : 0
siblings        : 4
cpu cores       : 4

我有一个简单的测试应用程序：

I have a simple test application :

struct foo {
    int x;
    int y;
}  ;

//globar var
volatile struct foo fvar ;

pid_t gettid( void )
{
    return syscall( __NR_gettid );
}

void *test_func0(void *arg)
{
    int proc_num = (int)(long)arg;
    cpu_set_t set;

    CPU_ZERO( &set );
    CPU_SET( proc_num, &set );
    printf("proc_num=(%d)\n",proc_num) ;
    if (sched_setaffinity( gettid(), sizeof( cpu_set_t ), &set ))
    {
        perror( "sched_setaffinity" );
        return NULL;
    }


    int i=0;
    for(i=0;i<1000000000;++i){
        __sync_fetch_and_add(&fvar.x,1);
    }
    return NULL;
} //test_func0

编译：
    GCC testsync.c -D_GNU_SOURCE -lpthread -o testsync.exe
以下是测试结果：

compiled : gcc testsync.c -D_GNU_SOURCE -lpthread -o testsync.exe The following is the test results :

2 threads running test_func0 in core 0,1  take 35 secs ;
2 threads running test_func0 in core 0,2  take 55 secs ;
2 threads running test_func0 in core 0,3  take 55 secs ;
2 threads running test_func0 in core 1,2  take 55 secs ;
2 threads running test_func0 in core 1,3  take 55 secs ;
2 threads running test_func0 in core 2,3  take 35 secs ;

我不知道为什么2个线程核心（0,1）运行或核心（2,3）是多少
在别人快？如果我运行2个线程在同一核心，核心一样（1,1），
核心（2,2），核心（3,3），这将是采取28秒，也困惑，为什么出现这种情况？

I wonder why 2 threads running in core (0,1) or in core(2,3) would be much faster in others ? if I running 2 threads at the same core , like core(1,1) , core(2,2),core(3,3) , that would be take 28 secs , also confused why this happen ?

推荐答案

核心0和1共用一个二级缓存，所以做芯2和3。运行在共享缓存使得共享变量留在L2双核高速缓存，这使得事情更快。

Cores 0 and 1 share an L2 cache, and so do cores 2 and 3. Running on two cores that share the cache makes the shared variable stay in the L2 cache, which makes things faster.

这是不是在当今的英特尔处理器true，其中​​L 2是每个核心。但是，你正在使用的CPU上，这是它如何工作的（它实际上是用胶粘在一起的两个双核CPU做了四核CPU）。

This is not true in today's Intel processors, where L2 is per core. But on the CPU you're using, this is how it works (it's actually a quad-core CPU made by gluing together two dual-core CPUs).

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