CS110 Lecture 09: Threads
Principles of Computer Systems
Winter 2021
Stanford University
Computer Science Department
Instructors: Chris Gregg and
Nick Troccoli
CS110 Topic 3: How can we have concurrency within a single process?
Learning About Processes
Introduction to Threads
Mutexes and Condition Variables
Condition Variables and Semaphores
Multithreading Patterns
Today
Lecture 10
Lecture 11
Lecture 12
Today's Learning Goals
- Learn about how threads allow for concurrency within a single process
- Understand the differences between threads and processes
- Discover some of the pitfalls of threads sharing the same virtual address space
Plan For Today
- Threads
- C threads
- C++ threads
A thread is an independent execution sequence within a single process.
- Most common: assign each thread to execute a single function in parallel
- Each thread operates within the same process, so they share global data (!) (text, data, and heap segments)
- They each have their own stack (e.g. for calls within a single thread)
- Execution alternates between threads as it does for processes
- Many similarities between threads and processes; in fact, threads are often called lightweight processes.
Threads
Processes:
- isolate virtual address spaces (good: security and stability, bad: harder to share info)
- can run external programs easily (fork-exec) (good)
- harder to coordinate multiple tasks within the same program (bad)
Threads:
- share virtual address space (bad: security and stability, good: easier to share info)
- can't run external programs easily (bad)
- easier to coordinate multiple tasks within the same program (good)
Threads vs. Processes
- 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).
- But
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
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
(viapthread_create
) by installingrecharge
as the thread routine eachpthread_t
should execute. - All thread routines take a
void *
and return avoid *
. 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 inNULL
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 surroundingpthread_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 whatwaitpid
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 inNULL
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: concurrency!
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 thatpthread_create
accepts the address of the surrounding loop's index variablei
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 thati
'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 ofi
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 sixmeetup
threads.*(size_t *)args
takes a snapshot of whateveri
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.
- At first glance, it's easy to absentmindedly assume that pthread_create captures not just the address of
- 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.
- Note that
Lecture 09: Introduction to Threads
- Fortunately, the fix is simple.
- We just pass the relevant
const char *
instead. Snapshots of theconst char *
pointers are passed verbatim tomeetup
. 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; // note that we get the name directly instead of through an index
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 now provides a variable that won't change
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
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
C++ Threads
static void greeting() {
cout << oslock << "Hello, world!" << endl << osunlock;
}
static const size_t kNumFriends = 6;
int main(int argc, char *argv[]) {
cout << "Let's hear from " << kNumFriends << " threads." << endl;
thread friends[kNumFriends]; // declare array of empty thread handles
// Spawn threads
for (size_t i = 0; i < kNumFriends; i++) {
friends[i] = thread(greeting);
}
// Wait for threads
for (thread &f : friends) {
f.join();
}
cout << "Everyone's said hello!" << endl;
return 0;
}
-
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
- 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) {
processImage(remainingImages);
remainingImages--;
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
- Is there anything 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!
- Can we solve this by decrementing immediately in the while loop?
- We can try, and we will likely get a solution that seems to work just fine.
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
static void process(size_t id, size_t& remainingImages) {
while (remainingImages > 0) {
remainingImages--;
processImage(remainingImages);
cout << oslock << "Thread#" << id << " processed an image (" << remainingImages
<< " remain)." << endl << osunlock;
}
cout << oslock << "Thread#" << id << " sees no remaining images and exits."
<< endl << osunlock;
}
- But, we still have a subtle concurrency problem!
- Between the line that performs the while loop calculation, and the decrement, another thread could also get into the while loop, and we would have a similar problem.
- But, it goes even deeper! Even the decrement instruction itself could have a race condition!
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)
- 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
- We need a way to ensure that we can have a critical section where only one thread can check the value of remainingImages at a time, and perform the decrement as well.
- The way we will do it is through the idea of mutual exclusion, as described on the next slide.
Race Condition
- 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 {
public:
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) {
counterLock.lock();
if (remainingImages == 0) {
counterLock.unlock();
break;
}
processImage(remainingImages);
remainingImages--;
cout << oslock << "Thread#" << id << " processed an image (" << remainingImages
<< " remain)." << endl << osunlock;
counterLock.unlock();
}
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++)
processors[i] = thread(process, 101 + i, ref(remainingImages), ref(counterLock));
for (thread& process: processes) process.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
break;
} else {
myImage = remainingImages;
remainingImages--;
counterLock.unlock(); // end of critical section
processImage(myImage);
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
- cas old, new, addr
How Do Mutexes Work?
CS110 Lecture 09: Threads (w21)
By Chris Gregg
CS110 Lecture 09: Threads (w21)
Winter 2020
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