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無需等待就實現互斥鎖

[英]Implementing Mutex Lock Without Busy Waiting

我接到大學的任務,無需等待就可以實現互斥鎖。 我正在嘗試實施它,但沒有成功。 有時,它有時會一直掛起,從而引發分段溢出,但是當我在gdb中運行它時,它每次都會完美運行。

互斥體

#define _GNU_SOURCE 

#include <stdio.h>
#include <pthread.h>
#include <unistd.h>
#include <signal.h>
#include <stdbool.h>
#include <sys/syscall.h>
#include "stack.h"

#define NUM_THREADS 2

// shared data structure
int sum = 0;

struct Stack waiting_q;

bool mutex_locked = false;

void * got_signal(int x) { return NULL; }

void acquire()
{
    bool first_time = true;
    while (!__sync_bool_compare_and_swap(&mutex_locked, false, true))
    {
        if (first_time)
        {
            push(&waiting_q, pthread_self());
            first_time = false;
        }
        printf("%ld is waiting for mutex\n", syscall(SYS_gettid));
        pause();
    }
    printf("Mutex acquired by %ld\n", syscall(SYS_gettid));
}

void release()
{
    int thread_r = pop(&waiting_q);
    if (waiting_q.size != 0 && thread_r != INT_MIN)
        pthread_kill(thread_r, SIGCONT);

    mutex_locked = false;
    printf("Mutex released by = %ld\n", syscall(SYS_gettid));
}

void * work()
{
    acquire();

    for (int i = 0; i < 10000; i++)
    {
        sum = sum + 1;
    }
    release();
    return NULL;
}

int main()
{
    init_stack(&waiting_q);
    pthread_t threads[NUM_THREADS];
    for (int i = 0; i < NUM_THREADS; i++)
    {
        int rc = pthread_create(&threads[i], NULL, work, NULL);
        if (rc != 0)
            printf("Error creating thread\n");
    }

    for (int i = 0; i < NUM_THREADS; i++)
    {
        pthread_join(threads[i], NULL);
    }

    printf("Value of Sum = %d\n", sum);
    return 0;
}

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

struct Node{
    struct Node * next;
    pthread_t x;
};

struct Stack{
    struct Node * head;
    int size;
};

void push(struct Stack * s, pthread_t n)
{
    struct Node * new_head = malloc(sizeof(struct Node));
    new_head->next = s->head;
    new_head->x = n;
    s->head = new_head;
    s->size++;
}

pthread_t pop(struct Stack * s)
{
    pthread_t rc = INT_MIN;
    if (s->head != NULL)
    {
        rc = s->head->x;
        struct Node * next = s->head->next;
        free(s->head);
        s->head = next;
        return rc;
    }
    s->size--;
    return rc;
}

void init_stack(struct Stack * s)
{
    s->head = 0;
    s->size = 0;
}

從互斥體實現對Stack數據結構的訪問不同步。

多個線程可能試圖同時acquire互斥體,這將導致它們同時訪問waiting_q 同樣,釋放線程對waiting_qpop()訪問不會與獲取線程的push()訪問同步。

Stack上的這種數據爭用很可能是導致段錯誤的原因。

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