Lecture 09: Introduction to Threads

Principles of Computer Systems

Spring 2019

Stanford University

Computer Science Department

Instructors: Chris Gregg and

Phil Levis

PDF of this presentation

Let's go through another example that is the kind of signals problem you may see on the midterm exam.
- Consider this program and its execution. Assume that all processes run to completion, all system and printf calls succeed, and that all calls to printf are atomic. Assume nothing about scheduling or time slice durations.

Practice Midterm Problem 2 (we saw Problem 1 last lecture)

int main(int argc, char *argv[]) {
    pid_t pid;
    int counter = 0;
    while (counter < 2) {
        pid = fork();
        if (pid > 0) break;
        counter++;
        printf("%d", counter);
    }
    if (counter > 0) printf("%d", counter);
    if (pid > 0) {
        waitpid(pid, NULL, 0);
        counter += 5;
        printf("%d", counter);
    }
    return 0;
}

List all possible outputs

Let's go through another example that is the kind of signals problem you may see on the midterm exam.
- Consider this program and its execution. Assume that all processes run to completion, all system and printf calls succeed, and that all calls to printf are atomic. Assume nothing about scheduling or time slice durations.

Practice Midterm Problem 2

int main(int argc, char *argv[]) {
    pid_t pid;
    int counter = 0;
    while (counter < 2) {
        pid = fork();
        if (pid > 0) break;
        counter++;
        printf("%d", counter);
    }
    if (counter > 0) printf("%d", counter);
    if (pid > 0) {
        waitpid(pid, NULL, 0);
        counter += 5;
        printf("%d", counter);
    }
    return 0;
}

List all possible outputs

Possible Output 1: 112265 Possible Output 2: 121265 Possible Output 3: 122165

If the > of the counter > 0 test is changed to a >=, then counter values of zeroes would be included in each possible output. How many different outputs are now possible? (No need to list the outputs—just present the number.)

Let's go through another example that is the kind of signals problem you may see on the midterm exam.
- Consider this program and its execution. Assume that all processes run to completion, all system and printf calls succeed, and that all calls to printf are atomic. Assume nothing about scheduling or time slice durations.

int main(int argc, char *argv[]) {
    pid_t pid;
    int counter = 0;
    while (counter < 2) {
        pid = fork();
        if (pid > 0) break;
        counter++;
        printf("%d", counter);
    }
    if (counter > 0) printf("%d", counter);
    if (pid > 0) {
        waitpid(pid, NULL, 0);
        counter += 5;
        printf("%d", counter);
    }
    return 0;
}

List all possible outputs

Possible Output 1: 112265 Possible Output 2: 121265 Possible Output 3: 122165

If the > of the counter > 0 test is changed to a >=, then counter values of zeroes would be included in each possible output. How many different outputs are now possible? (No need to list the outputs—just present the number.)

18 outputs now (6 x the first number)

Practice Midterm Problem 2

Consider the following program. Assume that each call to printf flushes its output to the console in full, and further assume that none of the system calls fail in any unpredictable way (e.g. fork never fails, and waitpid only returns -1 because there aren’t any child processes at the moment it decides on its return value).

static pid_t pid; // necessarily global so handler1 has
                  // access to it 
static int counter = 0;
static void handler1(int unused) { 
	counter++;
	printf("counter = %d\n", counter); 
	kill(pid, SIGUSR1);
}
static void handler2(int unused) { 
	counter += 10;
	printf("counter = %d\n", counter); 
	exit(0);
}
int main(int argc, char *argv[]) { 
	signal(SIGUSR1, handler1);
	if ((pid = fork()) == 0) {
		signal(SIGUSR1, handler2); 
		kill(getppid(), SIGUSR1); 
		while (true) {}
	}
	if (waitpid(-1, NULL, 0) > 0) { 
		counter += 1000;
		printf("counter = %d\n", counter);
	}
	return 0; 
}

Practice Midterm Problem 3

What is the output of the program?
What are the two potential outputs of the above program if the while (true) loop is completely eliminated?
Describe how the two processes would need to be scheduled in order for each of the two outputs to be presented.
Now further assume the call to exit(0) has also been removed from the handler2 function . Are there any other potential program outputs? If not, explain why. If so, what are they?

Consider the following program. Assume that each call to printf flushes its output to the console in full, and further assume that none of the system calls fail in any unpredictable way (e.g. fork never fails, and waitpid only returns -1 because there aren’t any child processes at the moment it decides on its return value).

Practice Midterm Problem 3

What is the output of the program?

    counter = 1
    counter = 10
    counter = 1001

This is the only possible output based on the program's logic

static pid_t pid; // necessarily global so handler1 has
                  // access to it 
static int counter = 0;
static void handler1(int unused) { 
	counter++;
	printf("counter = %d\n", counter); 
	kill(pid, SIGUSR1);
}
static void handler2(int unused) { 
	counter += 10;
	printf("counter = %d\n", counter); 
	exit(0);
}
int main(int argc, char *argv[]) { 
	signal(SIGUSR1, handler1);
	if ((pid = fork()) == 0) {
		signal(SIGUSR1, handler2); 
		kill(getppid(), SIGUSR1); 
		while (true) {}
	}
	if (waitpid(-1, NULL, 0) > 0) { 
		counter += 1000;
		printf("counter = %d\n", counter);
	}
	return 0; 
}

Consider the following program. Assume that each call to printf flushes its output to the console in full, and further assume that none of the system calls fail in any unpredictable way (e.g. fork never fails, and waitpid only returns -1 because there aren’t any child processes at the moment it decides on its return value).

static pid_t pid; // necessarily global so handler1 has
                  // access to it 
static int counter = 0;
static void handler1(int unused) { 
	counter++;
	printf("counter = %d\n", counter); 
	kill(pid, SIGUSR1);
}
static void handler2(int unused) { 
	counter += 10;
	printf("counter = %d\n", counter); 
	exit(0);
}
int main(int argc, char *argv[]) { 
	signal(SIGUSR1, handler1);
	if ((pid = fork()) == 0) {
		signal(SIGUSR1, handler2); 
		kill(getppid(), SIGUSR1); 
		while (true) {}
	}
	if (waitpid(-1, NULL, 0) > 0) { 
		counter += 1000;
		printf("counter = %d\n", counter);
	}
	return 0; 
}

Practice Midterm Problem 3

What are the two potential outputs of the above program if the while (true) loop is completely eliminated?
- The output from before (the 1 / 10 / 1001) output is still possible, because the child process can be swapped out just after the kill(getppid(), SIGUSR1) call, and effectively emulate the stall that came with the while (true) loop when it was present.
- Now, though, the child process could complete and exit normally before the parent process—via its handler1 function— has the opportunity to signal the child. That would mean handler2 wouldn’t even execute, and we wouldn’t expect to see counter = 10. (Note that the child process’s call to waitpid returns -1, since it itself has no grandchild processes of its own).

So, another possible output would be:
counter = 1 counter = 1001

Consider the following program. Assume that each call to printf flushes its output to the console in full, and further assume that none of the system calls fail in any unpredictable way (e.g. fork never fails, and waitpid only returns -1 because there aren’t any child processes at the moment it decides on its return value).

Practice Midterm Problem 3

Now further assume the call to exit(0) has also been removed from the handler2 function . Are there any other potential program outputs? If not, explain why. If so, what are they?
No other potential outputs, because:
- counter = 1 is still printed exactly once, just in the parent, before the parent fires a SIGUSR1 signal at the child (which may or may not have run to completion).
- counter = 10 is potentially printed if the child is still running at the time the parent fires that SIGUSR1 signal at it. The 10 can only appear after the 1, and if it appears, it must appear before the 1001.
- counter = 1001 is always printed last, after the child process exits. It’s possible that the child existed at the time the parent signaled it to inspire handler2 to print a 10, but that would happen before the 1001 is printed.

Note that the child process either prints nothing at all, or it prints a 10. The child process can never print 1001, because its waitpid call would return -1 and circumvent the code capable of printing the 1001.

static pid_t pid; // necessarily global so handler1 has
                  // access to it 
static int counter = 0;
static void handler1(int unused) { 
	counter++;
	printf("counter = %d\n", counter); 
	kill(pid, SIGUSR1);
}
static void handler2(int unused) { 
	counter += 10;
	printf("counter = %d\n", counter); 
	exit(0);
}
int main(int argc, char *argv[]) { 
	signal(SIGUSR1, handler1);
	if ((pid = fork()) == 0) {
		signal(SIGUSR1, handler2); 
		kill(getppid(), SIGUSR1); 
		while (true) {}
	}
	if (waitpid(-1, NULL, 0) > 0) { 
		counter += 1000;
		printf("counter = %d\n", counter);
	}
	return 0; 
}

A thread is an independent execution sequence within a single process.
- Operating systems and programming languages generally allow processes to run two or more functions simultaneously via threading.
- The stack segment is subdivided into multiple miniature stacks, one for each thread.
- The thread manager time slices and switches between threads in much the same way that the OS scheduler switches between processes.
  - In fact, threads are often called lightweight processes.
  - Each thread maintains its own stack, but all threads share the same text, data, and heap segments.
    - Pro: it's easier to support communication between threads, because they run in the same virtual address space.
    - Con: there's no memory protection, since virtual address space is shared. Race conditions and deadlock threats need to be mitigated, and debugging can be diﬃcult. Many bugs are hard to reproduce, since thread scheduling isn't predictable.
    - Pro and con: Multiple threads can access the same globals.
    - Pro and con: One thread can share its stack space (via pointers) with others.

Lecture 09: Introduction to Threads

ANSI C doesn't provide native support for threads.
- But pthreads, which comes with all standard UNIX and Linux installations of gcc, provides thread support, along with other related concurrency directives..
- The primary pthreads data type is the pthread_t, which is an opaque type used to manage the execution of a function within its own thread of execution.
- The only pthreads functions we'll need (before formally transitioning to C++ threads) are pthread_create and pthread_join.
- Here's a very small program illustrating how pthreads work (see next slide for live demo).

static void *recharge(void *args) {
    printf("I recharge by spending time alone.\n");
    return NULL;
}

static const size_t kNumIntroverts = 6;
int main(int argc, char *argv[]) {
    printf("Let's hear from %zu introverts.\n", kNumIntroverts);
    pthread_t introverts[kNumIntroverts];
    for (size_t i = 0; i < kNumIntroverts; i++)
        pthread_create(&introverts[i], NULL, recharge, NULL);
    for (size_t i = 0; i < kNumIntroverts; i++)
        pthread_join(introverts[i], NULL);
    printf("Everyone's recharged!\n");
    return 0;
}

Lecture 09: Introduction to Threads

The program on the prior slide declares an array of six pthread_t handles.
The program initializes each pthread_t (via pthread_create) by installing recharge as the thread routine each pthread_t should execute.
All thread routines take a void * and return a void *. That's the best C can do to support generic programming.
The second argument to pthread_create is used to set a thread priority and other attributes. We can just pass in NULL if all threads should have the same priority. That's what we do here.
The fourth argument is passed verbatim to the thread routine as each thread is launched. In this case, there are no meaningful arguments, so we just pass in NULL.
Each of the six recharge threads is eligible for processor time the instant the surrounding pthread_t has been initialized.
The six threads compete for thread manager's attention, and we have very little control over what choices it makes when deciding what thread to run next.
pthread_join is to threads what waitpid is to processes.
The main thread of execution blocks until the child threads all exit.
The second argument to pthread_join can be used to catch a thread routine's return value. If we don't care to receive it, we can pass in NULL to ignore it.

Lecture 09: Introduction to Threads

When you introduce any form of concurrency, you need to be careful to avoid concurrency issues like race conditions and deadlock.
Here's a slightly more involved program where friends meet up (see next slide for live demo):

static const char *kFriends[] = {
    "Langston", "Manan", "Edward", "Jordan", "Isabel", "Anne",
    "Imaginary"
};

static const size_t kNumFriends = sizeof(kFriends)/sizeof(kFriends[0]) - 1; // count excludes imaginary friend!

static void *meetup(void *args) {
    const char *name = kFriends[*(size_t *)args];
    printf("Hey, I'm %s.  Empowered to meet you.\n", name);
    return NULL;
}

int main() {
    printf("Let's hear from %zu friends.\n", kNumFriends);
    pthread_t friends[kNumFriends];
    for (size_t i = 0; i < kNumFriends; i++)
        pthread_create(&friends[i], NULL, meetup, &i);
    for (size_t j = 0; j < kNumFriends; j++)
        pthread_join(friends[j], NULL);
    printf("Is everyone accounted for?\n");
    return 0;
}

Lecture 09: Introduction to Threads

Here are a few sample runs that clearly illustrate that the program on the previous slide is severely broken.

cgregg@myth63$ ./friends
Let's hear from 6 friends.
Hey, I'm Jordan.  Empowered to meet you.
Hey, I'm Anne.  Empowered to meet you.
Hey, I'm Anne.  Empowered to meet you.
Hey, I'm Anne.  Empowered to meet you.
Hey, I'm Anne.  Empowered to meet you.
Hey, I'm Imaginary.  Empowered to meet you.
Is everyone accounted for?
cgregg@myth63$ ./friends
Let's hear from 6 friends.
Hey, I'm Jordan.  Empowered to meet you.
Hey, I'm Anne.  Empowered to meet you.
Hey, I'm Imaginary.  Empowered to meet you.
Hey, I'm Imaginary.  Empowered to meet you.
Hey, I'm Imaginary.  Empowered to meet you.
Hey, I'm Imaginary.  Empowered to meet you.
Is everyone accounted for?
cgregg@myth63$ ./friendsLet's hear from 6 friends.
Hey, I'm Jordan.  Empowered to meet you.
Hey, I'm Isabel.  Empowered to meet you.
Hey, I'm Isabel.  Empowered to meet you.
Hey, I'm Anne.  Empowered to meet you.
Hey, I'm Anne.  Empowered to meet you.
Hey, I'm Imaginary.  Empowered to meet you.
Is everyone accounted for?

Lecture 09: Introduction to Threads

Clearly something is wrong, but why?
- Note that meetup references its incoming parameter now, and that pthread_create accepts the address of the surrounding loop's index variable i via its fourth parameter. pthread_create's fourth argument is always passed verbatim as the single argument to the thread routine.
- The problem? The main thread advances i without regard for the fact that i's address was shared with six child threads.
  - At ﬁrst glance, it's easy to absentmindedly assume that pthread_create captures not just the address of i, but the value of i itself. That assumption of course, is incorrect, as it captures the address and nothing else.
  - The address of i (even after it goes out of scope) is constant, but its contents evolve in parallel with the execution of the six meetup threads. *(size_t *)args takes a snapshot of whatever i happens to contain at the time it's evaluated.
  - Often, the majority of the meetup threads only execute after the main thread has worked through all of its ﬁrst for loop. The space at &i is left with a 6, and that's why Imaginary is printed so often.
- This is another example of a race condition, and is typical of the types of problems that come up when multiple threads share access to the same data.

Lecture 09: Introduction to Threads

Fortunately, the ﬁx is simple.
We just pass the relevant const char * instead. Snapshots of the const char * pointers are passed verbatim to meetup. The strings themselves are constants.
Full program illustrating the ﬁx can be found right here.

static const char *kFriends[] = {
  "Langston", "Manan", "Edward", "Jordan", "Isabel", "Anne",
  "Imaginary"
};

static const size_t kNumFriends = sizeof(kFriends)/sizeof(kFriends[0]) - 1; // count excludes imaginary friend!

static void *meetup(void *args) {
  const char *name = args;
  printf("Hey, I'm %s.  Empowered to meet you.\n", name);
  return NULL;
}

int main() {
  printf("%zu friends meet.\n", kNumFriends);
  pthread_t friends[kNumFriends];
  for (size_t i = 0; i < kNumFriends; i++)
    pthread_create(&friends[i], NULL, meetup, (void *) kFriends[i]); // this line is different than before, too
  for (size_t i = 0; i < kNumFriends; i++)
    pthread_join(friends[i], NULL);
  printf("All friends are real!\n");
  return 0;
}

Lecture 09: Introduction to Threads

Here are a few test runs just so you see that it's ﬁxed. Race conditions are often quite
complicated, and avoiding them won't always be this trivial.

cgregg@myth63$ ./friends
Let's hear from 6 friends.
Hey, I'm Langston.  Empowered to meet you.
Hey, I'm Manan.  Empowered to meet you.
Hey, I'm Edward.  Empowered to meet you.
Hey, I'm Jordan.  Empowered to meet you.
Hey, I'm Isabel.  Empowered to meet you.
Hey, I'm Anne.  Empowered to meet you.
Is everyone accounted for?
cgregg@myth63$ ./friends
Let's hear from 6 friends.
Hey, I'm Langston.  Empowered to meet you.
Hey, I'm Manan.  Empowered to meet you.
Hey, I'm Edward.  Empowered to meet you.
Hey, I'm Isabel.  Empowered to meet you.
Hey, I'm Jordan.  Empowered to meet you.
Hey, I'm Anne.  Empowered to meet you.
Is everyone accounted for?
cgregg@myth63$ ./friends
Let's hear from 6 friends.
Hey, I'm Langston.  Empowered to meet you.
Hey, I'm Edward.  Empowered to meet you.
Hey, I'm Manan.  Empowered to meet you.
Hey, I'm Anne.  Empowered to meet you.
Hey, I'm Jordan.  Empowered to meet you.
Hey, I'm Isabel.  Empowered to meet you.
Is everyone accounted for?

Lecture 09: Introduction to Threads

Lecture 09: Introduction to Threading

By Chris Gregg

Lecture 09: Introduction to Threading

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Lecture 09: Introduction to Threads

Practice Midterm Problem 2 (we saw Problem 1 last lecture)

Practice Midterm Problem 2

Practice Midterm Problem 2

Practice Midterm Problem 3

Practice Midterm Problem 3

Practice Midterm Problem 3

Practice Midterm Problem 3

Lecture 09: Introduction to Threads

Lecture 09: Introduction to Threads

Lecture 09: Introduction to Threads

Lecture 09: Introduction to Threads

Lecture 09: Introduction to Threads

Lecture 09: Introduction to Threads

Lecture 09: Introduction to Threads

Lecture 09: Introduction to Threads

Lecture 09: Introduction to Threads

Lecture 09: Introduction to Threads

Lecture 09: Introduction to Threads

Lecture 09: Introduction to Threading

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