Principles of Computer Systems

Autumn 2019

Stanford University

Computer Science Department

Lecturers: Chris Gregg and

                        Philip Levis

Lecture 10: Threads and Mutexes

  • Midterm on Monday, October 28, in class
  • We will also contact students with accommodations in the next few days
  • The exam will be administered using BlueBook, a computerized testing software that you will run on your laptop. If you don't have a laptop to run the program on, let us know ASAP and we will provide one.
    • You can download the BlueBook software from the main CS 110 website.
      • Make sure you test the program out before you come to the exam. We will post a basic test exam in a few days.
    • ​We will have limited power outlets for laptops, so please ensure you have a charged battery​
  • You are allowed one back/front page of 8.5 x 11in paper for any notes you would like to bring in. We will also provide a limited reference sheet with functions you may need to use for the exam.
    • Knowing the exact order of the arguments to system calls we've covered isn't expected, but knowing their semantics is

Midterm Details

  • In C, threads are a library, called pthreads, which comes with all standard UNIX installations of gcc
    • 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.
    • In the previous lecture, you saw two functions, pthread_create and pthread_join.

  • With pthreads, you pass a function with signature void* f(void* input), the library allocates a stack and runs the thread
  • Threads all share the address space of a single process: you need to be very careful about how they share data, similarly to how we did for signal handlers

pthreads in C (review of last lecture)

int pthread_create(pthread_t *thread, const pthread_attr_t *attr,
                   void *(*start_routine) (void *), void *arg);
int pthread_join(pthread_t thread, void **retval);
    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);

bug on line 2!

pthread Bug in Last Lecture

    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);













created thread stacks

main stack

  • Solve problem by passing each thread a pointer to its associated string, which doesn't change
  • Because they are a library, with pthreads you have to do everything manually (much like signals)
  • For example, a common way to implement a critical section is through a mutual exclusion variable (mutex)
    • A mutex is a lock: take the lock before entering the critical section and release it after
    • If a thread tries to take a locked lock, it waits until it is unlocked (like how we blocked signals)
    • If you forget to unlock the lock, everyone else waits forever (deadlock!)
  • C++'s greater guarantees on when things occur allow us to avoid some common errors
    • We'll start by showing you the basic APIs, so you can see how things can go wrong, then show other supported approaches that help
uint64_t increment_counter(void) {
  uint64_t val = counter;
  return val;

pthreads are great, but...

uint64_t increment_counter(void) {
  lock_guard<mutex> lg(&counter_lock);
  uint64_t val = counter;
  return val;

pthread approach

If you forget to unlock, deadlock

C++ approach

Can't forget to unlock!

static void recharge() {
    cout << oslock << "I recharge by spending time alone." << endl << osunlock; 
static const size_t kNumIntroverts = 6;
int main(int argc, char *argv[]) {
  cout << "Let's hear from " << kNumIntroverts << " introverts." << endl      
  thread introverts[kNumIntroverts]; // declare array of empty thread handles
  for (thread& introvert: introverts)
     introvert = thread(recharge);    // move anonymous threads into empty handles
  for (thread& introvert: introverts)
  cout << "Everyone's recharged!" << endl;
  return 0;

C++ Threads

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;



  • We create a thread that executes the recharge function and return a thread handle to it
  • We then move the thread handle  (via the thread's operator=(thread&& other)) into the array
    • This is a different meaning for operator=
    • After it executes, the right hand side is an empty thread
      • thread t1 = thread(func);
      • thread t2 = t1;  // t1 is no longer a handle for the thread created
    • This is an important distinction, because a traditional operator= would produce a second working copy of the same thread, which would be bad in so many ways (share a stack???)
  • The join method is equivalent to the pthread_join function we've already discussed.
  • The prototype of the thread routine—in this case, recharge—can be anything (although the return type is always ignored, so it should generally be void).

Details on the Code: It's Subtle

thread introverts[kNumIntroverts]; // declare array of empty thread handles
for (thread& introvert: introverts)
  introvert = thread(recharge);    // move anonymous threads into empty handles
for (thread& introvert: introverts)
  • operator<<, unlike printf, isn't thread-safe.
    • Jerry Cain has constructed custom stream manipulators called oslock and osunlock that can be used to acquire and release exclusive access to an ostream.
    • These manipulators—which we can use by #include-ing "ostreamlock.h"—can be used to ensure at most one thread has permission to write into a stream at any one time.

WARNING: Thread Safety and Standard I/O

  • Thread routines can accept any number of arguments using variable argument lists. (Variable argument lists—the C++ equivalent of the ellipsis in C—are supported via a recently added feature called variadic templates.)
  • Here's a slightly more involved example, where greet threads are configured to say hello a variable number of times.
static void greet(size_t id) {
  for (size_t i = 0; i < id; i++) {
    cout << oslock << "Greeter #" << id << " says 'Hello!'" << endl << osunlock;
    struct timespec ts = {
      0, random() % 1000000000
    nanosleep(&ts, NULL);
  cout << oslock << "Greeter #" << id << " has issued all of his hellos, " 
       << "so he goes home!" << endl << osunlock;

static const size_t kNumGreeters = 6;
int main(int argc, char *argv[]) {
  cout << "Welcome to Greetland!" << endl;
  thread greeters[kNumGreeters];
  for (size_t i = 0; i < kNumGreeters; i++) greeters[i] = thread(greet, i + 1);
  for (thread& greeter: greeters) greeter.join();
  cout << "Everyone's all greeted out!" << endl;
  return 0;

No More Void* Tomfoolery

  • Threads allow a process to parallelize a problem across multiple cores
  • Consider a scenario where we want to process 250 images and have 10 cores
  • Completion time is determined by the slowest thread, so we want them to have equal work
    • Static partitioning: just give each thread 25 of the images to process. Problem: what if some images take much longer than others?
    • Work queue: have each thread fetch the next unprocessed image
  • Here's our first stab at a main function.
int main(int argc, const char *argv[]) {
  thread processors[10];
  size_t remainingImages = 250;
  for (size_t i = 0; i < 10; i++)
    processors[i] = thread(process, 101 + i, ref(remainingImages));
  for (thread& proc: processors) proc.join();
  cout << "Images done!" << endl;
  return 0;

Thread-Level Parallelism

  • The processor thread routine accepts an id number (used for logging purposes) and a reference to the remainingImages.
  • It continually checks remainingImages to see if any images remain, and if so, processes the image and sends a message to cout
  • processImage execution time depends on the image.
  • Note how we can declare a function that takes a size_t and a size_t& as arguments
static void process(size_t id, size_t& remainingImages) {
  while (remainingImages > 0) {
    cout << oslock << "Thread#" << id << " processed an image (" << remainingImages 
     << " remain)." << endl << osunlock;
  cout << oslock << "Thread#" << id << " sees no remaining images and exits." 
       << endl << osunlock;

Thread Function

  • Discuss with your neighbor -- what's wrong with this code?
  • Presented below right is the abbreviated output of a imagethreads run.
  • In its current state, the program suffers from a serious race condition.
  • Why? Because remainingImages > 0 test and remainingImages-- aren't atomic
  • If a thread evaluates remainingImages > 0 to be true and commits to processing an image, the image may have been claimed by another thread.
  • This is a concurrency problem!
  • Solution? Make the test and decrement atomic with a critical section
  • Atomicity: externally, the code has either executed or not; external observers do not see any intermediate states mid-execution
myth60 ~../cs110/cthreads -> ./imagethreads
Thread# 109 processed an image, 249 remain
Thread# 102 processed an image, 248 remain
Thread# 101 processed an image, 247 remain
Thread# 104 processed an image, 246 remain
Thread# 108 processed an image, 245 remain
Thread# 106 processed an image, 244 remain
// 241 lines removed for brevity
Thread# 110 processed an image, 3 remain
Thread# 103 processed an image, 2 remain
Thread# 105 processed an image, 1 remain
Thread# 108 processed an image, 0 remain
Thread# 105 processed an image, 18446744073709551615 remain
Thread# 109 processed an image, 18446744073709551614 remain

Race Condition

  • C++ statements aren't inherently atomic. Virtually all C++ statements—even ones as simple as remainingImages--—compile to multiple assembly code instructions.
  • Assembly code instructions are atomic, but C++ statements are not.
  • g++ on the myths compiles remainingImages-- to five assembly code instructions, as with:

  • The first two lines drill through the remainingImages reference to load a copy of the remainingImages held on main's stack. The third line decrements that copy, and the last two write the decremented copy back to the remainingImages variable held on main's stack.
  • The ALU operates on registers, but registers are private to a core, so the variable needs to be loaded from and stored to memory.
    • Each thread makes a local copy of the variable before operating on it
    • What if multiple threads all load the variable at the same time: they all think there's only 128 images remaining and process 128 at the same time
0x0000000000401a9b <+36>:    mov    -0x20(%rbp),%rax
0x0000000000401a9f <+40>:    mov    (%rax),%eax
0x0000000000401aa1 <+42>:    lea    -0x1(%rax),%edx
0x0000000000401aa4 <+45>:    mov    -0x20(%rbp),%rax
0x0000000000401aa8 <+49>:    mov    %edx,(%rax)

Why Test and Decrement Is REALLY NOT Thread-Safe

  • A mutex is a type used to enforce mutual exclusion, i.e., a critical section
  • Mutexes are often called locks
    • To be very precise, mutexes are one kind of lock, there are others (read/write locks, reentrant locks, etc.), but we can just call them locks in this course, usually "lock" means "mutex"
  • When a thread locks a mutex
    • If the lock is unlocked the thread takes the lock and continues execution
    • If the lock is locked, the thread blocks and waits until the lock is unlocked
    • If multiple threads are waiting for a lock they all wait until lock is unlocked, one receives lock
  • When a thread unlocks a mutex
    • It continues normally; one waiting thread (if any) takes the lock and is scheduled to run
  • This is a subset of the C++ mutex abstraction: nicely simple!
class mutex {
  mutex();        // constructs the mutex to be in an unlocked state
  void lock();    // acquires the lock on the mutex, blocking until it's unlocked
  void unlock();  // releases the lock and wakes up another threads trying to lock it

Mutual Exclusion

  • main instantiates a mutex, which it passes (by reference!) to invocations of process.
  • The process code uses this lock to protect remainingImages.
  • Note we need to unlock on line 5 -- in complex code forgetting this is an easy bug
static void process(size_t id, size_t& remainingImages, mutex& counterLock) {
  while (true) {
    if (remainingImages == 0) {
    cout << oslock << "Thread#" << id << " processed an image (" << remainingImages 
     << " remain)." << endl << osunlock;
  cout << oslock << "Thread#" << id << " sees no remaining images and exits." 
  << endl << osunlock;

int main(int argc, const char *argv[]) {
  size_t remainingImages = 250;
  mutex  counterLock;
  thread processors[10];
  for (size_t i = 0; i < 10; i++)
    agents[i] = thread(process, 101 + i, ref(remainingImages), ref(counterLock));
  for (thread& agent: agents) agent.join();
  cout << "Done processing images!" << endl;
  return 0;

Building a Critical Section with a Mutex

  • The way we've set it up, only one thread agent can process an image at a time!
    • Image processing is actually serialized
  • We can do better: serialize deciding which image to process and parallelize the actual processing
  • Keep your critical sections as small as possible!

Critical Sections Can Be a Bottleneck

static void process(size_t id, size_t& remainingImages, mutex& counterLock) {
  while (true) {
    size_t myImage;
    counterLock.lock();    // Start of critical section
    if (remainingImages == 0) {
      counterLock.unlock(); // Rather keep it here, easier to check
    } else {
      myImage = remainingImages;
      counterLock.unlock(); // end of critical section

      cout << oslock << "Thread#" << id << " processed an image (" << remainingImages 
      << " remain)." << endl << osunlock;
  cout << oslock << "Thread#" << id << " sees no remaining images and exits." 
  << endl << osunlock;
  • What if processImage can return an error?
    • E.g., what if we need to distinguish allocating an image and processing it
    • A thread can grab the image by decrementing remainingImages but if it fails there's no way for another thread to retry
    • Because these are threads, if one thread has a SEGV the whole process will fail
    • A more complex approach might be to maintain an actual queue of images and allow threads (in a critical section) to push things back into the queue
  • What if image processing times are *highly* variable (e.g, one image takes 100x as long as the others)?
    • Might scan images to estimate execution time and try more intelligent scheduling
  • What if there's a bug in your code, such that sometimes processImage randomly enters an infinite loop?
    • Need a way to reissue an image to an idle thread
    • An infinite loop of course shouldn't occur, but when we get to networks sometimes execution time can vary by 100x for reasons outside our control

Problems That Might Arise

  • Standard mutex: what we've seen
    • If a thread holding the lock tries to re-lock it, deadlock
  • recursive_mutex
    • A thread can lock the mutex multiple times, and needs to unlock it the same number of times to release it to other threads
  • timed_mutex
    • A thread can try_lock_for / try_lock_until: if time elapses, don't take lock
    • Deadlocks if same thread tries to lock multiple times, like standard mutex
  • In this class, we'll focus on just regular mutex

Some Types of Mutexes

  • Something we've seen a few times is that you can't read and write a variable atomically
    • But a mutex does so! If the lock is unlocked, lock it
  • How does this work with caches?
    • Each core has its own cache
    • Writes are typically write-back (write to higher cache level when line is evicted), not write-through (always write to main memory) for performance
    • Caches are coherent -- if one core writes to a cache line that is also in another core's cache, the other core's cache line is invalidated: this can become a performance problem
  • Hardware provides atomic memory operations, such as compare and swap
    • cas old, new, addr
      • If addr == old, set addr to new
    • Use this as a single bit to see if the lock is held and if not, take it
    • If the lock is held already, then enqueue yourself (in a thread safe way) and tell kernel to sleep you
    • When a node unlocks, it clears the bit and wakes up a thread

How Do Mutexes Work?

Questions about threads, mutexes, race conditions, or critical sections?

  • Assignment 4 is a comprehensive test of your abilities to fork / execvp child processes and manage them through the use of signal handlers. It also tests your ability to use pipes.
  • You will be writing a shell (demo: assign3/samples/stsh_soln)
    • The shell will keep a list of all background processes, and it will have some standard shell abilities:
      • you can quit the shell (using quit or exit)
      • you can bring them to the front (using  fg)
      • you can continue a background job (using bg)
      • you can kill a set of processes in a pipeline (using slay) (this will entail learning about process groups)
      • you can stop a process (using halt)
      • you can continue a process (using cont)
      • you can get a list of jobs (using jobs)
    • You are responsible for creating pipelines that enable you to send output between programs, e.g.,
      • ls | grep stsh | cut -d- -f2
      • sort < | wc > stsh-wc.txt
    • You will also be handing off terminal control to foreground processes, which is new

Assignment 4: Stanford Shell

  • Assignment 4 contains a lot of moving parts!
  • Read through all the header files!
  • You will only need to modify
  • You can test your shell programmatically with samples/stsh-driver
  • One of the more difficult parts of the assignment is making sure you are keeping track of all the processes you've launched correctly. This involves careful use of a SIGCHLD handler.
    • ​You will also need to use a handler to capture SIGTSTP and  SIGINT to capture ctrl-Z and ctrl-C, respectively (notice that these don't affect your regular shell -- they shouldn't affect your shell, either).
  • ​​Another tricky part of the assignment is with the piping between processes. It takes time to understand what we are requiring you to accomplish
  • There is a very good list of milestones in the assignment -- try to accomplish regular milestones, and you should stay on track.
  • I understand that this is a detailed assignment, with a midterm in the middle. I suggest at least starting the assignment before the midterm and getting through a couple of milestones. But, also take the time to study for the midterm.

Assignment 4: Stanford Shell