Creating new streams within a stream

Question

I am new to pthread and try to understand how they work. The goal is to write a program that creates a main stream. Each new thread created creates 2 new threads. In this case, the threads should receive their id and time to wait (from 1 to 10 seconds) as parameters. New flows are created until they reach a maximum (determined by a constant). In my program, threads are created correctly. The problem is that id is not unique - each id is printed twice. For example:

 t0: Hello my id is 0 t2: Hello my id is 2 t2: Hello my id is 2 t4: Hello my id is 4 t4: Hello my id is 4

I assume that the problem lies in the general pointer to an array of parameters, but I do not know to prevent this. I block access to the array with the help of mutex and it seems like there should be no race condition. But still, the result is the id is incorrect.

 #include <stdio.h> #include <pthread.h> #include <stdlib.h> #define MAX_THREADS 10 pthread_mutex_t threads_count_mutex; pthread_mutex_t args_mutex; int wait_time_overall = 0; int threads_count = 0; void* SpawnTwoThreads(void *args) { pthread_t t1; pthread_t t2; int* thread_args = (int*)args; printf("Sleeping for %d befor thread creation\n", thread_args[1]); sleep(thread_args[1]); printf("t%d: Hello, my id is %d\n", thread_args[0], thread_args[0]); if (threads_count < MAX_THREADS) { pthread_mutex_lock(&threads_count_mutex); threads_count++; pthread_mutex_unlock(&threads_count_mutex); pthread_mutex_lock(&args_mutex); thread_args[0]++; thread_args[1] = rand() % 10; pthread_mutex_unlock(&args_mutex); pthread_create(&t1, NULL, SpawnTwoThreads, thread_args); } if (threads_count < MAX_THREADS) { pthread_mutex_lock(&threads_count_mutex); threads_count++; pthread_mutex_unlock(&threads_count_mutex); pthread_mutex_lock(&args_mutex); thread_args[0] = rand() % 10; thread_args[1]++; pthread_mutex_unlock(&args_mutex); pthread_create(&t2, NULL, SpawnTwoThreads, thread_args); } pthread_join(t1, NULL); pthread_join(t2, NULL); pthread_exit(NULL); } int main(void) { pthread_t t1; srand(time(NULL)); pthread_mutex_init(&threads_count_mutex, NULL); pthread_mutex_init(&args_mutex, NULL); int start_args[2]; start_args[0] = 0; start_args[1] = rand() % 10; pthread_create(&t1, NULL, SpawnTwoThreads, start_args); printf("In main: waiting for all threads to complete\n"); pthread_join(t1, NULL); printf("Overall waittime is %d\n", wait_time_overall); pthread_exit(NULL); }

KoVadim KoVadim 85.7k five 67 128 · Accepted Answer · 2017-11-22T13:32:07

First, the race is still there.

Here the variable is read.

 if (threads_count < MAX_THREADS) {

here it is written

 pthread_mutex_lock(&threads_count_mutex); threads_count++; pthread_mutex_unlock(&threads_count_mutex);

Yes, reading must also be wrapped in mutex. This variable a couple of times in the code is not lucky :).

But the problem is that thread_args is passed by pointer. And this seems not at all what you wanted. It is enough to make a copy of this structure before creating the stream, as everything falls into place. Plus, it seems there is a copy-paste in SpawnTwoThreads - there are two very similar thread creation codes, but the second one is the opposite. But it may be so conceived.

I thought in mutex only need to wrap the area where it is written.
Is it necessary then to also wrap each output through printf in mutex?
At the expense of reading in the if: is it then necessary to block the entire if-block using mutex?
At the expense of arguments in the second block - corrected.
Ie I can create a new array using memcpy and transfer the link to it to the stream, I understand correctly?
in this case, theoretically, reading problems should not arise - an int can be written in one instruction.
But if the structure is a bit more complicated, then it can be read while it is being written elsewhere and in an inconsistent state.
Ie I can with the help of memcpy - yes there are two int copy.
Was it possible to solve this problem more elegantly, provided that both parameters still need to be passed?

Creating new streams within a stream

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