I know that both threads can use the global variable k and p and also that after the CPU-time of one thread expired the other thread gets CPU-time and that's why I get different outputs like 9, 10 but I do not understand how the outputs 10 comes from. I guess it's because of the variable y although I don't use it.

int k=2;
int* p;
   void t1_f1(void){
   int x=3;
   p=&x;
   sleep(1);
}

void t1_f2(void){
   int y=5;
   k  ;
   sleep(1);
}

void* t1_main(void* args){
   t1_f1();
   t1_f2();
   return NULL;
}

void* t2_main(void* args){
   sleep(1);
   k=k* *p;
   printf("%d \n", k);
   return NULL;
}

int main(int argc, char ** argv){
   pthread_t threads[2];
   pthread_create(threads 1, NULL, t2_main, NULL);
   pthread_create(threads, NULL, t1_main, NULL);
   pthread_join(threads[0],NULL);
   pthread_join(threads[1],NULL);
   exit(0);
}

CodePudding user response：

Your program has UB (Undefined Behavior) and therefore you cannot expect any consistent output at all.

2 Examples for UB in your code:

1. in t1_f1 you assign the global p to the address of a local variable (x). When x goes out of the scope when the function returns, you'll have a dandling pointer, and dereferencing it (*p) in t2_main is UB.
2. t2_main might execute before p is initialized at all (it is not initialized at the global scope where it is defined). In this case it's also UB to dereference it.
   Note that as commented above by @AndrewHenle, sleep() is not a threads synchronization call.