这是实现没有互斥对象的异步消息队列的安全方法吗？

Question

Below is how I implemented an asynchronous message queue with simple checks using while loops. I think this is a better way than using mutexes if it works okay. It seems to work okay by running some tests in my machine, but I am really not sure about whether this is safe because I have not much experience in programming asynchronous systems for multiple threads/processes. Is my work below safely preventing race conditions? or will it crash with some heavier loads or in any other condition?

typedef struct MessageQueueElement {
    Message message;
    struct MessageQueueElement *next;
} MessageQueueElement;

typedef struct MessageQueue { //singly-linked list as a queue
    MessageQueueElement *first;
    MessageQueueElement *last;
    bool sending;
} MessageQueue;

void createMessageQueue(MessageQueue *this) {
    this->first = malloc(sizeof(MessageQueueElement));
    this->last = this->first;
    this->sending = false;
}

void sendMessage(MessageQueue *this, Message *message) {
    while (this->sending);
    //do nothing while this function is called from another thread

    this->sending = true;
    this->last->message = *message;
    this->last = this->last->next = malloc(sizeof(MessageQueueElement));
    //add a message to the queue

    this->sending = false;
}

int waitMessage(MessageQueue *this, int (*readMessage)(unsigned, unsigned, void *)) {
    while (this->first == this->last);
    //do nothing while the queue is empty

    int n = readMessage(this->first->message.type, this->first->message.code, this->first->message.data);
    MessageQueueElement *temp = this->first;
    this->first = this->first->next;
    free(temp);
    return n;
}

See below for the whole context and some test code.

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

#define EXIT_MESSAGE 0
#define THREAD_MESSAGE 1
#define EXIT 0
#define CONTINUE 1

int readMessage(size_t type, size_t code, void *data) {
    if (type == THREAD_MESSAGE) {
        printf("message from thread %d: %s\n", code, (char *)data);
        free(data);
    } else {
        return EXIT;
    }
    return CONTINUE;
}

MessageQueue mq;
int nThreads;
int counter = 0;

void *worker(void *p) {
    double pi = 0.0;
    for (int i = 0; i < 10; i += 1) {
        for (int j = 0; j < 100000; j += 1) {
            double n = i * 100000.0 + j;
            pi += (4.0 / (8.0 * n + 1.0) - 2.0 / (8.0 * n + 4.0) - 1.0 / (8.0 * n + 5.0) - 1.0 / (8.0 * n + 6.0)) / pow(16.0, n);
        }
        char *s = malloc(100);
        sprintf(s, "calculating pi... %d percent complete", (i + 1) * 10);
        sendMessage(&mq, &(Message){.type = THREAD_MESSAGE, .code = (int)p, .data = s});
    }
    char *s = malloc(100);
    sprintf(s, "pi equals %.8f", pi);
    sendMessage(&mq, &(Message){.type = THREAD_MESSAGE, .code = (int)p, .data = s});
    counter += 1;
    if (counter == nThreads) {
        sendMessage(&mq, &(Message){.type = EXIT_MESSAGE});
    }
}


int main(int argc, char **argv) {
    nThreads = atoi(argv[1]);
    createMessageQueue(&mq);

    pthread_t threads[nThreads];
    for (int i = 0; i < nThreads; i += 1) {
        pthread_create(&threads[i], NULL, worker, (void *)i);
    }
    while (waitMessage(&mq, readMessage));
    for (int i = 0; i < nThreads; i += 1) {
        pthread_join(threads[i], NULL);
    }
    return 0;
}

Answer 1

Obviously not Ok. For starters, if two threads are waiting for the queue to become empty, they are likely to both find out that sending == false at exactly the same time, and will both jump in, and then things go WRONG in a bad way. That's exactly what mutexes are there for. So this doesn't work.

It's also awfully bad form to busy wait on a variable. If you have a quad core CPU, there is a good chance that four cores spend 100% of available CPU just waiting for a variable to change. Not good.

And since sending isn't volatile, your compiler will not see the slightest reason to generate code to set it to true, so again this isn't going to work.

To get this working, you would need to do all the things that a mutex does. And you'd have to do everything exactly right. Which is highly dependent on the exact processor that you are using. You need to be aware of cache consistency, ordering of read and write operations, memory barriers and so on, and if you haven't even heard of these, then you have no chance to get it right.