Is this situation considered as race condition?

Question

Considering the following code:

#define _XOPEN_SOURCE 600
#define _DEFAULT_SOURCE

#include <pthread.h>
#include <stdatomic.h>
#include <stdint.h>

#include <unistd.h>

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

#define ENTRY_NUM 100

struct value {
    pthread_mutex_t mutex;
    int i;
};

struct entry {
    atomic_uintptr_t val;
};

struct entry entries[ENTRY_NUM];

void* thread1(void *arg)
{
    for (int i = 0; i != ENTRY_NUM; ++i) {
        struct value *val = (struct value*) atomic_load(&entries[i].val);
        if (val == NULL)
            continue;

        pthread_mutex_lock(&val->mutex);
        printf("%d\n", val->i);
        pthread_mutex_unlock(&val->mutex);
    }

    return NULL;
}

void* thread2(void *arg)
{
    /*
     * Do some costy operations before continuing.
     */
    usleep(1);

    for (int i = 0; i != ENTRY_NUM; ++i) {
        struct value *val = (struct value*) atomic_load(&entries[i].val);

        pthread_mutex_lock(&val->mutex);
        atomic_store(&entries[i].val, (uintptr_t) NULL);
        pthread_mutex_unlock(&val->mutex);

        pthread_mutex_destroy(&val->mutex);
        free(val);
    }

    return NULL;
}

int main() {
    for (int i = 0; i != ENTRY_NUM; ++i) {
        struct value *val = malloc(sizeof(struct value));
        pthread_mutex_init(&val->mutex, NULL);
        val->i = i;
        atomic_store(&entries[i].val, (uintptr_t) val);
    }

    pthread_t ids[2];

    pthread_create(&ids[0], NULL, thread1, NULL);
    pthread_create(&ids[1], NULL, thread2, NULL);

    pthread_join(ids[0], NULL);
    pthread_join(ids[1], NULL);

    return 0;
}

Suppose in function thread1, entries[i].val is loaded, then the scheduler schedule the process to sleep.

Then thread2 awakes from usleep, since ((struct val*) entries[0].val)->mutex aren't locked, thread2 locks it, stores NULL to entries[0].val and free the original of entries[0].val.

Now, is that a race condition? If so, how to avoid this without locking entries or entries[0]?

Answer 1

You are correct my friend, there is indeed a race condition in such code.

Let me open by overall and saying that thread is prawn to race conditions by definition and this is also correct for any other library implemented thread which is unacknowledged by your compiler ahead of compilation time.

Regarding your specific example, yes as you've explained yourself since we can not assume when does your scheduler go into action, thread1 could atomic load your entries, context switch to thread2, which will then free these entries before thread1 gets processor time again. How do you prevent or avoid such race conditions? avoid accessing them without locking them, even though atomic load is an "atomic read" you are logically allowing other threads to access these entries. the entire code scope of both thread1 and thread2 should be protected with a mutex. despite using atomic_load, you are just guaranteeing that at that atomic time, no other accesses to that entry will be made, but during the time between the atomic_load and your first calling to pthread_mutex_lock context switches can indeed occur! as you've mentioned yourself, this is both bad practice and logically wrong. - hence as I've already stated, you should protect the entire scope with pthread_mutex_lock

In general, as I've stated in the beginning of this, considering that your compiler is unaware to the concept of threads during compilation times, it is very sensitive to race conditions that you may not even be aware of, - e.g.: when accessing different areas of some shared memory, the compiler does not take into consideration that other threads may exist and accesses some memory as it desires, and may affect different areas of memories during, even though logically the code itself doesn't, and the "correctness" of the code is valid. there was some paper published on this called Threads Cannot be Implemented as a Library by Hans-J Boehm I highly suggest you read it, I promise it will increase your understanding of race conditions and threads using pthread at general!