CS110 Lecture 09: Threads

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Principles of Computer Systems

Winter 2020

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

2/5

2/10

2/12

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

  • More Practice: Race Conditions
  • Break: Announcements
  • Threads

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.

static void bat(int unused) { 
  printf("pirate\n"); 
  exit(0);
}

int main(int argc, char *argv[]) {
  signal(SIGUSR1, bat);
  pid_t pid = fork();
  if (pid == 0) { 
    printf("ghost\n"); 
    return 0;
  }
  kill(pid, SIGUSR1); 
  printf("ninja\n"); return 0;
}
  • For each of the five output orders, Place a yes if the text represents a possible output, and place a no otherwise.
    • ghost ninja pirate
    • pirate ninja
    • ninja ghost
    • ninja pirate ninja
    • ninja pirate ghost

Practice Problem 1

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.

static void bat(int unused) { 
  printf("pirate\n"); 
  exit(0);
}

int main(int argc, char *argv[]) {
  signal(SIGUSR1, bat);
  pid_t pid = fork();
  if (pid == 0) { 
    printf("ghost\n"); 
    return 0;
  }
  kill(pid, SIGUSR1); 
  printf("ninja\n"); return 0;
}
  • For each of the five output orders, Place a yes if the text represents a possible output, and place a no otherwise.
    • ghost ninja pirate.  yes
    • pirate ninja.    yes
    • ninja ghost.   no
    • ninja pirate ninja.    no
    • ninja pirate ghost.   no

Practice Problem 1

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
    • How many processes are created in total?
    • What number must be printed last?
    • What number must be printed first?
    • What number must be printed second-last?

Practice Problem 2

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
    • How many processes are created in total? 3
    • What number must be printed last?
    • What number must be printed first?
    • What number must be printed second-last?

Practice Problem 2

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
    • What number must be printed last? 
      • The last output is 5 (the grandparent that starts with counter 0 exits last, because it waits on its child)
    • What number must be printed first?
    • What number must be printed second-last?

Practice Problem 2

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
    • What number must be printed last? 
      • The last output is 5 (the grandparent that starts with counter 0 exits last, because it waits on its child)
    • What number must be printed first?
      • The parent that starts with counter 1 outputs first
    • What number must be printed second-last?

Practice Problem 2

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
    • What number must be printed last? 
      • The last output is 5 (the grandparent that starts with counter 0 exits last, because it waits on its child)
    • What number must be printed first?
      • The parent that starts with counter 1 outputs first
    • What number must be printed second-last?
      • The parent that starts with counter 1 exits second to last, after waiting for its child; the second-to-last output is 6

Practice Problem 2

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

Practice Problem 2

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
  • The second parent (1) output and the two child (2) outputs are up to the scheduler

Practice Problem 2

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.)

Practice Problem 2

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 - 6x the original

Practice 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.

static pid_t pid; // necessarily global so handler1 has access
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; 
}
  • What is the output of the program?

Practice Problem 3

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.

static pid_t pid; // necessarily global so handler1 has access
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; 
}
  • What is the output of the program?

 

​    counter = 1

  counter = 10

  counter = 1001

 

Practice Problem 3

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.

static pid_t pid; // necessarily global so handler1 has access
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; 
}
  • What are the two potential outputs of the above program if the while (true) loop is completely eliminated?

Practice Problem 3

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.

static pid_t pid; // necessarily global so handler1 has access
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; 
}
  • What are the two potential outputs of the above program if the while (true) loop is completely eliminated?
    • ​Previous output still possible via scheduling
    • 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.

    • So, another possible output would be: 

      counter = 1
counter = 1001

Practice Problem 3

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.

static pid_t pid; // necessarily global so handler1 has access
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; 
}
  • 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
      • ...

Practice Problem 3

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.

static pid_t pid; // necessarily global so handler1 has access
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; 
}
  • 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 = 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. 
      • ...

Practice Problem 3

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.

static pid_t pid; // necessarily global so handler1 has access
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; 
}
  • 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 = 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.

Practice Problem 3

Plan For Today

  • More Practice: Race Conditions
  • Break: Announcements
  • Threads

Announcements

  • Assignment 4 released later today - Stanford Shell
  • Midterm Next Friday
    • Please notify us of any OAE accommodations ASAP
    • More midterm logistics on Wednesday
  • 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: assign4/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 < stsh.cc | 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 stsh.cc
  • 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.

Assignment 4: Stanford Shell

Plan For Today

  • More Practice: Race Conditions
  • Break: Announcements
  • 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

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 (size_t i = 0; i < kNumFriends; i++) {
     friends[i].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

  • 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 {
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++)
    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
      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

How Do Mutexes Work?

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

CS110 Lecture 09: Threads (w20)

By Nick Troccoli

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CS110 Lecture 09: Threads (w20)

Winter 2020

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