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问题说明

在下面的代码中，我尝试使用线程顺序计算 x1 和 x2 的值。（在实际情况下，x1 和 x2 将是大计算）但等待两个线程使用 while 循环计算各个变量的值正在成为处理器成本高。问题是，我希望两个线程并行运行，但是两个线程的循环应该同样序列化（意味着应该在一次调用中运行一次）。因此，有什么方法可以删除这些 while 循环并串行获取结果。我对使用信号量和互斥量感到很困惑，因为 x1 和 x2 是相互独立的？请帮忙。提前致谢。

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

pthread_t pth1,pth2;
//Values to calculate
int x1 = 0, x2 = 0;
//Values for condition
int cond1 = 0,cond2 = 0;


void *threadfunc1(void *parm)
{
    for (;;) {
        // Is this while loop is very costly for the processor?
        while(!cond1) {}
        x1++;
        cond1 = 0;
    }
    return NULL ;
}
void *threadfunc2(void *parm)
{
    for (;;) {
        // Is this while loop is very costly for the processor?
        while(!cond2) {}
        x2++;
        cond2 = 0;
    }
    return NULL ;
}



int main () {
    pthread_create(&pth1, NULL, threadfunc1, "foo");
    pthread_create(&pth2, NULL, threadfunc2, "foo");
    int loop = 0;
    while (loop < 10) {
        // iterated as a step
        loop++;
        printf("Initial : x1 = %d, x2 = %d\n", x1, x2);
        cond1 = 1;
        cond2 = 1;
        // Is this while loop is very costly for the processor?
        while(cond1) {}
        while(cond2) {}
        printf("Final   : x1 = %d, x2 = %d\n", x1, x2);
    }

    pthread_cancel(pth1);
    pthread_cancel(pth2);
    return 1;
}
4

1 回答 1

2

如果您删除 threadfunc1 和 threadfunc2 中的 while 循环，则线程将返回并且不会继续。因此，您肯定需要一种方法让线程保持活动状态，直到它完成对序列的计算（在本例中为 10）。

这是一个示例代码，显示了尝试在主线程和两个线程之间进行协调。使用 pthread_cond_wait()，等待从简单的“while”循环转移到 pthread_cond_wait()。

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

//Values to calculate
int x1 = 0, x2 = 0;
//Values for condition
int cond1 = 0, cond2 = 0;

static pthread_mutex_t m1 = PTHREAD_MUTEX_INITIALIZER;
static pthread_mutex_t m2 = PTHREAD_MUTEX_INITIALIZER;
static pthread_cond_t c = PTHREAD_COND_INITIALIZER;

void *threadfunc1(void *parm) {
    int i = 0;
    while (i < 10) {
        printf("\t\tThread1 with x1 = %d\n", x1);
        pthread_mutex_lock(&m1);
        if (cond1 == 0) {
            pthread_cond_wait(&c, &m1);
        }
        x1++;
        cond1 = 0;
        pthread_mutex_unlock(&m1);
        i++;
    }
    printf("\t\tThread1 returns with x1 = %d\n", x1);
    return NULL ;
}

void *threadfunc2(void *parm) {
    int i = 0;
    while (i < 10) {
        printf("\t\tThread2 with x2 = %d\n", x2);
        pthread_mutex_lock(&m2);
        if (cond2 == 0) {
            pthread_cond_wait(&c, &m2);
        }
        x2++;
        cond2 = 0;
        pthread_mutex_unlock(&m2);
        i++;
    }
    printf("\t\tThread2 returns with x2 = %d\n", x2);
    return NULL ;
}


int main () {
    pthread_t pth1, pth2;
    pthread_create(&pth1, NULL, threadfunc1, "foo");
    pthread_create(&pth2, NULL, threadfunc2, "foo");
    int retVal;
    int loop = 0;
    while (loop <= 10) {
        loop++;
        cond1 = 1;
        pthread_cond_signal(&c);

        cond2 = 1;
        pthread_cond_signal(&c);
        printf("Loop [%d]: x1 = %d, x2 = %d\n", loop, x1, x2);
    }

    retVal = pthread_join(pth1, NULL);
    retVal = pthread_join(pth2, NULL);
    printf("Final : x1 = %d, x2 = %d\n", x1, x2);
    return 0;
}
于 2013-08-22T07:41:13.503 回答