In-class-2

Class: CSCE-313

Notes:

Problem 1

What are the possible outputs of the following program? How can you change the program to make it print "HELPER" before "MAIN"? Note that even when a thread relinquishes CPU with a call to sched_yield(), is there a guaranteed order of execution among the main and helper threads?

#include <pthread.h>
#include <stdio.h>

void *helper(void *arg) {
    printf("HELPER");
    return NULL;
}

int main() {
    pthread_t thread;
    pthread_create(&thread, NULL, helper, NULL);
    sched_yield();
    printf("MAIN");
    return 0;
}

Answer:

• It can be:
  • MAINHELPER
  • HELPERMAIN
  • MAIN
  • ...

Notes:

• We have no control over how much each will execute before they are context switched
• Lets say HELPER ran and returned, then main thread yielded but will be scheduled again so it prints MAIN
• Lets say MAIN is printed, then it yields to the helper thread so it prints HELPER
• When your main call returns you only printed MAAIN and exit.
  • If you return from your main program, your caller will just call exit()

Problem 2

What does the following program print?

#include <pthread.h>
#include <stdio.h>

void *helper(void *arg) {
    int *num = (int *) arg;
    *num = 2;
    return NULL;
}

int main() {
    int i = 0;
    pthread_t thread;
    pthread_create(&thread, NULL, helper, &i);
    pthread_join(thread, NULL);
    printf("i is %d.\n", i);
    return 0;
}

Notes:

• You are passing an address of your stack (a pointer to an integer variable in your stack)
• Here you will block until that thread has returned
• Note that the child knows the actual type of the void *arg argument, it is able to cast it to (int *)
• Note that pthreat_join is not a busy wait, it eventually has to go to the kernel to be blocked

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