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Here we need to repeatedly print the 1 2 3 sequence infinite times using threads in C programming language.
Let’s take a look at the sample output of our code:
1 2 3 1 2 3 1 2 3 1 2 3
For this we will need to use three Threads running in parallel in C programming language. You also need a variable that is initialized to 1 in the first thread and has its value updated based on its previous value. Then run an infinite loop inside the function.
Let's look at the program that implements our solution:
#include <stdio.h> #include <pthread.h> pthread_cond_t cond1 = PTHREAD_COND_INITIALIZER; pthread_cond_t cond2 = PTHREAD_COND_INITIALIZER; pthread_cond_t cond3 = PTHREAD_COND_INITIALIZER; pthread_mutex_t lock = PTHREAD_MUTEX_INITIALIZER; int value = 1; void *foo(void *n){ while(1) { pthread_mutex_lock(&lock); if (value != (int)*(int*)n) { if ((int)*(int*)n == 1) { pthread_cond_wait(&cond1, &lock); } else if ((int)*(int*)n == 2) { pthread_cond_wait(&cond2, &lock); } else { pthread_cond_wait(&cond3, &lock); } } printf("%d ", *(int*)n); if (value == 3) { value = 1; pthread_cond_signal(&cond1); } else if(value == 1) { value = 2; pthread_cond_signal(&cond2); } else if (value == 2) { value = 3; pthread_cond_signal(&cond3); } pthread_mutex_unlock(&lock); } return NULL; } int main(){ pthread_t tid1, tid2, tid3; int n1 = 1, n2 = 2, n3 = 3; pthread_create(&tid1, NULL, foo, (void *)&n1); pthread_create(&tid2, NULL, foo, (void *)&n2); pthread_create(&tid3, NULL, foo, (void *)&n3); while(1); return 0; }
1 2 3 1 2 3 1 2 3 1 2 3 1 2 3….
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