如何在沒有競爭條件的情況下等待條件？

如果您按原樣運行代碼，您可能不會遇到任何問題。問題是如果在 cond_wait 之前發生廣播，則此代碼不再有效。如果您取消注釋下面的睡眠，您每次運行都會遇到問題

我該怎么寫才能沒有競爭條件？我碰巧知道我可以使用 futex 解決這個問題，但我正在尋找 pthread 解決方案

//clang   -g -fsanitize=undefined,thread main.cpp && ./a.out
#include <cstdio>
#include <pthread.h>
#include <unistd.h>
pthread_mutex_t write_lock = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t checkpoint = PTHREAD_COND_INITIALIZER;

int data;

void*fn(void*) {
    pthread_mutex_lock(&write_lock);
    auto temp =   data;
    pthread_mutex_unlock(&write_lock);
    
    //This if makes it so no thread will pass until both/all threads have finished writing to data
    if (temp < 2) {
        //sleep(1); // <--------- uncomment to see problem
        while (temp < 2)
        {
            pthread_mutex_t local_mutex = PTHREAD_MUTEX_INITIALIZER;
            pthread_mutex_lock(&local_mutex);
            pthread_cond_wait(&checkpoint, &local_mutex);

            //Recheck the condition
            pthread_mutex_lock(&write_lock);
            temp = data;
            pthread_mutex_unlock(&write_lock);
        }
    } else {
        pthread_cond_broadcast(&checkpoint);
    }
    
    int sum=data; //no way to get here until both/all threads execute the sync code
    return 0;
}

int main() {
    pthread_t thread_id[2];
    for(int i=0; i<2; i  ) {
        pthread_attr_t attr;
        pthread_attr_init(&attr);
        pthread_create(&thread_id[i], &attr, &fn, 0);
    }
    for(int i=0; i<2; i  ) {
        pthread_join(thread_id[i], 0);
    }
    printf("Finish\n");
}

uj5u.com熱心網友回復：

也許我不清楚你的真正問題。這看起來像您正在嘗試堆疊您的客戶端執行緒，直到它們都超過特定的檢查點。在這種情況下，一旦data被所有執行緒碰撞，就會實作該檢查點

因此，data保持你的謂詞狀態。測驗或撰寫它是互斥保護的目的。變化檢測是 cond-var 的目的。

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

pthread_mutex_t write_lock = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t checkpoint = PTHREAD_COND_INITIALIZER;

int data;

void *fn(void *pv)
{
    // construction-passed top limit
    intptr_t n = (intptr_t)pv;

    // changing predicate, so lock mutex
    pthread_mutex_lock(&write_lock);
      data;

    // changed predicate, so tell people.
    pthread_cond_signal(&checkpoint);

    // mutex is still locked. if our predicate state isn't
    //  sufficient to move on, we wait (and release the mutex
    //  in the process)
    while (data < n)
    {
        pthread_cond_wait(&checkpoint, &write_lock);

        // TODO: woke up here. may be spurious, may be legit, but
        //  we're going to make sure on the next iteration of the
        //  loop by rechecking the predicate data, which we can do
        //  because the mutex is locked coming out of the wait.

        // ALSO: if you want off-the-mutex processing here of some
        //  kind you would do it by releasing the mutex, do your
        //  thread work, then reacquire the mutex again before the
        //  next iteration of the loop
    }

    // NOTE: still own the mutex.
    
    // no way to get here until all threads execute the sync code
    //  note: not a legit use of pthread_self, use at your peril
    printf("%p : %d\n", (void*)pthread_self(), data);

    // leaving the party, release the mtx and tell someone.
    pthread_mutex_unlock(&write_lock);
    pthread_cond_signal(&checkpoint);

    return 0;
}

#define N_THREADS   16

int main()
{
    pthread_t thread_id[N_THREADS];
    for (intptr_t i = 0; i < N_THREADS; i  )
        pthread_create(&thread_id[i], NULL, &fn, (void*)(intptr_t)N_THREADS);

    for (intptr_t i = 0; i < N_THREADS; i  )
        pthread_join(thread_id[i], 0);

    printf("Finish\n");
}

輸出（僅示例）

0x7000019e9000 : 16
0x700001448000 : 16
0x700001342000 : 16
0x700001654000 : 16
0x7000016d7000 : 16
0x7000012bf000 : 16
0x70000175a000 : 16
0x7000013c5000 : 16
0x7000017dd000 : 16
0x70000123c000 : 16
0x700001860000 : 16
0x70000154e000 : 16
0x7000018e3000 : 16
0x7000014cb000 : 16
0x700001966000 : 16
0x7000015d1000 : 16
Finish

這就是你至少看起來想要完成的事情。

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