c - 如何减少从 sem_wait 唤醒后的上下文切换时间

Question

我发现 sem_post 和 sem_wait 的时间间隔至少有 50 微秒。

sem_t gSema;
struct timeval gTv;

void *run(void *arg) {
    int result;
    struct timeval tv;
    struct timespec nano = {0, 1};

    while( 1 ) {
        result = sem_trywait(&gSema);
        if(result < 0) {
            nanosleep(&nano, NULL);
            continue;
        }
        if(result == 0) {
            gettimeofday(&tv, NULL);
            printf("waken up...elapsed time = %ld\n", (tv.tv_sec - gTv.tv_sec) * 1000000 + (tv.tv_usec - gTv.tv_usec));
        }
    }

    return NULL;
}
void *run2(void *arg) {
    int result;
    struct timeval tv;

    while( 1 ) {
        result = sem_wait(&gSema);
        if( result == 0 ) {
            gettimeofday(&tv, NULL);
            printf("waken up...elapsed time = %ld\n", (tv.tv_sec - gTv.tv_sec) * 1000000 + (tv.tv_usec - gTv.tv_usec));
        }
    }
    return NULL;
}
int main(int argc, char **argv) {
    pthread_t thr;
    sem_init(&gSema, 0, 0);
    pthread_create(&thr, NULL, run, NULL);

    while( 1 ) {
        sleep(1);
        gettimeofday(&gTv, NULL);
        sem_post(&gSema);
    }
    return 1;
}

首先，当我使用 run2() 运行代码时，运行时间为 50 微秒。然后 CPU 使用率很低。其次，当我在没有 nanosleep 的情况下使用 run() 运行代码时，运行时间为 0 或 1。但 CPU 使用率预计为 100%。使用 nanosleep，耗时为 1~50 微秒。

有没有更好的低 CPU 使用率和低延迟的方法？

Answer 1

使用 pthread_mutex 和 pthread_cond 函数，我发现每个开关大约需要 4 微秒。下面的示例代码在大约 8 秒内循环了 100 万次(200 万个信号)。

#include <stdio.h>
#include <pthread.h>

#define LOOPS 1000000

pthread_mutex_t mutex1 = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cond1 = PTHREAD_COND_INITIALIZER;
pthread_mutex_t mutex2 = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cond2 = PTHREAD_COND_INITIALIZER;
int done = 0;

void *test( void *arg )
{
    // lock the second mutex before signaling the parent, 
    // so that the parent cannot signal before the child is ready
    pthread_mutex_lock( &mutex2 );

    for (;;)
    {
        // signal the parent
        pthread_mutex_lock( &mutex1 );
        pthread_cond_signal( &cond1 );
        pthread_mutex_unlock( &mutex1 );

        // wait for the parent to signal back
        pthread_cond_wait( &cond2, &mutex2 );

        // check the done indicator, note that the done indicator is 
        // protected by the second mutex, which the child has locked
        if ( done )
        {
            printf( "done\n" );
            break;
        }
    }

    // release the second mutex and exit
    pthread_mutex_unlock( &mutex2 );
    return NULL;
}

int main( void )
{
    // lock the first mutex before launching the child, 
    // so that the child cannot signal before the parent is ready
    pthread_mutex_lock( &mutex1 );

    // launch the child thread
    pthread_t child;
    pthread_create( &child, NULL, test, NULL);

    for ( int i = 0; i < LOOPS; i++ )
    {
        // give the user some feedback
        if ( i % 100000 == 0 )
        {
            printf( "%7d\r", i );
            fflush( stdout );
        }

        // wait for a signal from the child
        pthread_cond_wait( &cond1, &mutex1 );

        // lock the second mutex and signal the child, 
        // setting the done indicator if needed
        pthread_mutex_lock( &mutex2 );
        if ( i == LOOPS - 1 )
        {
            printf( "%7d\n", i + 1 );
            done = 1;
        }
        pthread_cond_signal( &cond2 );
        pthread_mutex_unlock( &mutex2 );
    }

    // release the first mutex, and wait for the child to finish
    pthread_mutex_unlock( &mutex1 );
    pthread_join( child, NULL );
}