Rust contains loop, which is specially designed for infinite looping:
Using the keyword loop, Rust provides a way to loop indefinitely until
some terminating statement is reached (like program exit)
I can't think of a single scenario when I would like my loops to run forever. Are there any scenarios when an infinite loop can be useful?
loop is often used for loops where we want to break in the middle of the loop's body. That is, you want to do something before testing any condition, then exit the loop if some condition is met, then do something else that must only be done after testing the condition, then repeat. In other languages, this is often rendered as while (true) or for (;;). This situation is common enough that Rust decided to embrace this pattern by reserving a keyword to declare loops that have no entry condition.
Also, Rust doesn't have the equivalent of C's do..while loop, which tests a condition at the end of an iteration, but not before the first iteration. In Rust, you'd emulate it with a loop and an if condition { break } statement at the end of the loop. do..while loops are relatively rare in practice, rarer than "break in the middle" loops in my experience.
I often use loop in programs where I expect constant user input, e.g. in command-line games:
fn play() {
let mut game = Game::new();
loop { game.turn() }
}
It is also useful due to possible input errors - during each turn they can be printed to the player and the game can be continued as long as the player wants until a victory, failure or resignation breaks the loop.
The main loop in embedded programs. Embedded devices often expect to run for as long as they are switched on. The two main pieces of code are usually event handlers and the main loop. If there is a main loop, it runs forever.
The main loop may have a "wait for interrupt" or it may just constantly dedicate its CPU resources to figuring out a better way to achieve its goal.
Related
Currently I am doing something like this
use tokio::time::timeout;
while let Ok(option_element) = timeout(Duration::from_nanos(1), stream.next()).await {
...
}
to drain the items already in the rx buffer of the stream. I don't want to wait for the next element that has not been received.
I think the timeout would slow down the while loop.
I am wondering that is there a better way to do this without the use of the timeout?
Possibly like this https://github.com/async-rs/async-std/issues/579 but for the streams in futures/tokio.
The direct answer to your question is to use the FutureExt::now_or_never method from the futures crate as in stream.next().now_or_never().
However it is important to avoid writing a busy loop that waits on several things by calling now_or_never on each thing in a loop. This is bad because it is blocking the thread, and you should prefer a different solution such as tokio::select! to wait for multiple things. For the special case of this where you are constantly checking whether the task should shut down, see this other question instead.
On the other hand, an example where using now_or_never is perfectly fine is when you want to empty a queue for the items available now so you can batch process them in some manner. This is fine because the now_or_never loop will stop spinning as soon as it has emptied the queue.
Beware that if the stream is empty, then now_or_never will succeed because next() immediately returns None in this case.
I'm discovering the pthread library (in C) and I'm having some trouble understanding well a few things.
First of all, I understand what a mutex is, I understand how it works, ok, I also understand the concept of the cond, but I can't manage to use it properly (I don't really get how to combine the mutex and the cond)
This is, in pseudo-code, what I want to do :
thread :
loop :
// do something
end loop
end thread
So there is n threads, but each thread uses the same function. I want the inside of the loop to be executed in parallel by all the threads BUT each thread must be in the same iteration of the loop, meaning I don't care in what order the instructions inside the loop are executed between threads, but to start iteration 2 of a thread, all the other threads must have finished iteration 1 (etc).
So my question is : how do you do that ? Not particularly in a specific example, but theoretically.
EDIT
I manage to do it, I don't know if it's the proper way, but it's working :
global nbOfThreads
global nbOfIterations
thread :
lock(mutex0)
unlock(mutex0)
loop :
// Do something
lock(mutex1)
nbOfIterations++
if (nbOfIterations == nbOfThread) :
nbOfIterations = 0
broadcast(cond)
unlock(mutex1)
continue
end if
wait(cond, mutex1)
unlock(mutex1)
end loop
end thread
main (n) :
nbOfThreads = n
nbOfIterations = 0
lock(mutex0)
do nbOfThreads times : create(thread)
unlock(mutex0)
end main
I obviously tried to understand myself, but there are some things I don't understand :
The main one : WHY does a cond need to be pair with a mutex
In some examples I saw something like this :
// thread A :
while (!condition)
wait(&cond)
// thread B :
if (condition)
signal(&cond)
well I really don't get the point of this while loop, I thought wait put the thread in pause until the condition is true (until the other thread send the signal). I mean I would get it if it was an if instead of a while.
Thank you
WHY does a cond need.... because the (!condition) you reference almost certainly depends upon some bits of the object not changing while you reference them. Correspondingly, modifying the state of the object should be done in such a way as to appear atomic to any observer; thus a mutex. While you could rely on too-clever-by-half hackery like atomic types, there is also the problem of ‘what if it was modified just after you checked it’ -- a race condition. Thus the idiomatic lock(); while (!cond) { wait(); }.
The point of the while... The signal+wait is not a handoff of control; after the signal, any number of things could happen to the object before a particular thread returns from wait. Even though the condition might have been in the correct state, by the time thread A examines it, it may no longer be. At the point of exiting the while loop, thread A knows: The condition is in the state I desire, and I have exclusive access to the object.
Condition variables can have spurious wake-ups. The condition might not actually be true when the wait function returns.
Depending on your task, a different synchronization primitive, such as a barrier (see pthread_barrier_init) or a semaphore (sem_init) might be easier to use.
so I've got this multithreaded, recursive application. It's coded in Pharo Smalltalk but the logical solution to the issue is likely to be the same across most languages.
I have 4 of the same process running relatively simultaneously. It's the last iteration of a recursive call. I'd like to print the result calculated by my recursive function (it's a dictionary being modified in the argument of the recursive function/message). The issue I'm facing right now is that the print is called in the base case terminator of the recursion, so the result is printed 4 times.
I tried setting a global variable which allows for me to print the result of the process which finishes first, but of course that means that the result is wrong. It needs to print the result of the last process to execute of all the processes in that last iteration of the recursion.
How could I go about this without going too deep into the Process class? Thanks for any help.
Do you know the number of threads? (Supposedly, 4)
Then you can use an atomic long (in java, for example):
AtomicLong myAtomicLong = new AtomicLong(0);
...
...
// do my work
if (totalThreadCount == myAtomicLong.getAndIncrement() -1)
{
//my print
}
The increment and get is atomic, so the last thread to want to print, will get there and the condition will be true after all other threads have finished their jobs. Please notice that it is important to place the increment and check after the job, is done.
I have this POSIX thread:
void subthread(void)
{
while(!quit_thread) {
// do something
...
// don't waste cpu cycles
if(!quit_thread) usleep(500);
}
// free resources
...
// tell main thread we're done
quit_thread = FALSE;
}
Now I want to terminate subthread() from my main thread. I've tried the following:
quit_thread = TRUE;
// wait until subthread() has cleaned its resources
while(quit_thread);
But it does not work! The while() clause does never exit although my subthread clearly sets quit_thread to FALSE after having freed its resources!
If I modify my shutdown code like this:
quit_thread = TRUE;
// wait until subthread() has cleaned its resources
while(quit_thread) usleep(10);
Then everything is working fine! Could someone explain to me why the first solution does not work and why the version with usleep(10) suddenly works? I know that this is not a pretty solution. I could use semaphores/signals for this but I'd like to learn something about multithreading, so I'd like to know why my first solution doesn't work.
Thanks!
Without a memory fence, there is no guarantee that values written in one thread will appear in another. Most of the pthread primitives introduce a barrier, as do several system calls such as usleep. Using a mutex around both the read and write introduces a barrier, and more generally prevents multi-byte values being visible in partially written state.
You also need to separate the idea of asking a thread to stop executing, and reporting that it has stopped, and appear to be using the same variable for both.
What's most likely to be happening is that your compiler is not aware that quit_thread can be changed by another thread (because C doesn't know about threads, at least at the time this question was asked). Because of that, it's optimising the while loop to an infinite loop.
In other words, it looks at this code:
quit_thread = TRUE;
while(quit_thread);
and thinks to itself, "Hah, nothing in that loop can ever change quit_thread to FALSE, so the coder obviously just meant to write while (TRUE);".
When you add the call to usleep, the compiler has another think about it and assumes that the function call may change the global, so it plays it safe and doesn't optimise it.
Normally you would mark the variable as volatile to stop the compiler from optimising it but, in this case, you should use the facilities provided by pthreads and join to the thread after setting the flag to true (and don't have the sub-thread reset it, do that in the main thread after the join if it's necessary). The reason for that is that a join is likely to be more efficient than a continuous loop waiting for a variable change since the thread doing the join will most likely not be executed until the join needs to be done.
In your spinning solution, the joining thread will most likely continue to run and suck up CPU grunt.
In other words, do something like:
Main thread Child thread
------------------- -------------------
fStop = false
start Child Initialise
Do some other stuff while not fStop:
fStop = true Do what you have to do
Finish up and exit
join to Child
Do yet more stuff
And, as an aside, you should technically protect shared variables with mutexes but this is one of the few cases where it's okay, one-way communication where half-changed values of a variable don't matter (false/not-false).
The reason you normally mutex-protect a variable is to stop one thread seeing it in a half-changed state. Let's say you have a two-byte integer for a count of some objects, and it's set to 0x00ff (255).
Let's further say that thread A tries to increment that count but it's not an atomic operation. It changes the top byte to 0x01 but, before it gets a chance to change the bottom byte to 0x00, thread B swoops in and reads it as 0x01ff.
Now that's not going to be very good if thread B want to do something with the last element counted by that value. It should be looking at 0x0100 but will instead try to look at 0x01ff, the effect of which will be wrong, if not catastrophic.
If the count variable were protected by a mutex, thread B wouldn't be looking at it until thread A had finished updating it, hence no problem would occur.
The reason that doesn't matter with one-way booleans is because any half state will also be considered as true or false so, if thread A was halfway between turning 0x0000 into 0x0001 (just the top byte), thread B would still see that as 0x0000 (false) and keep going (until thread A finishes its update next time around).
And if thread A was turning the boolean into 0xffff, the half state of 0xff00 would still be considered true by thread B so it would do its thing before thread A had finished updating the boolean.
Neither of those two possibilities is bad simply because, in both, thread A is in the process of changing the boolean and it will finish eventually. Whether thread B detects it a tiny bit earlier or a tiny bit later doesn't really matter.
The while(quite_thread); is using the value quit_thread was set to on the line before it. Calling a function (usleep) induces the compiler to reload the value on each test.
In any case, this is the wrong way to wait for a thread to complete. Use pthread_join instead.
You're "learning" multhithreading the wrong way. The right way is to learn to use mutexes and condition variables; any other solution will fail under some circumstances.
I understand about race conditions and how with multiple threads accessing the same variable, updates made by one can be ignored and overwritten by others, but what if each thread is writing the same value (not different values) to the same variable; can even this cause problems? Could this code:
GlobalVar.property = 11;
(assuming that property will never be assigned anything other than 11), cause problems if multiple threads execute it at the same time?
The problem comes when you read that state back, and do something about it. Writing is a red herring - it is true that as long as this is a single word most environments guarantee the write will be atomic, but that doesn't mean that a larger piece of code that includes this fragment is thread-safe. Firstly, presumably your global variable contained a different value to begin with - otherwise if you know it's always the same, why is it a variable? Second, presumably you eventually read this value back again?
The issue is that presumably, you are writing to this bit of shared state for a reason - to signal that something has occurred? This is where it falls down: when you have no locking constructs, there is no implied order of memory accesses at all. It's hard to point to what's wrong here because your example doesn't actually contain the use of the variable, so here's a trivialish example in neutral C-like syntax:
int x = 0, y = 0;
//thread A does:
x = 1;
y = 2;
if (y == 2)
print(x);
//thread B does, at the same time:
if (y == 2)
print(x);
Thread A will always print 1, but it's completely valid for thread B to print 0. The order of operations in thread A is only required to be observable from code executing in thread A - thread B is allowed to see any combination of the state. The writes to x and y may not actually happen in order.
This can happen even on single-processor systems, where most people do not expect this kind of reordering - your compiler may reorder it for you. On SMP even if the compiler doesn't reorder things, the memory writes may be reordered between the caches of the separate processors.
If that doesn't seem to answer it for you, include more detail of your example in the question. Without the use of the variable it's impossible to definitively say whether such a usage is safe or not.
It depends on the work actually done by that statement. There can still be some cases where Something Bad happens - for example, if a C++ class has overloaded the = operator, and does anything nontrivial within that statement.
I have accidentally written code that did something like this with POD types (builtin primitive types), and it worked fine -- however, it's definitely not good practice, and I'm not confident that it's dependable.
Why not just lock the memory around this variable when you use it? In fact, if you somehow "know" this is the only write statement that can occur at some point in your code, why not just use the value 11 directly, instead of writing it to a shared variable?
(edit: I guess it's better to use a constant name instead of the magic number 11 directly in the code, btw.)
If you're using this to figure out when at least one thread has reached this statement, you could use a semaphore that starts at 1, and is decremented by the first thread that hits it.
I would expect the result to be undetermined. As in it would vary from compiler to complier, langauge to language and OS to OS etc. So no, it is not safe
WHy would you want to do this though - adding in a line to obtain a mutex lock is only one or two lines of code (in most languages), and would remove any possibility of problem. If this is going to be two expensive then you need to find an alternate way of solving the problem
In General, this is not considered a safe thing to do unless your system provides for atomic operation (operations that are guaranteed to be executed in a single cycle).
The reason is that while the "C" statement looks simple, often there are a number of underlying assembly operations taking place.
Depending on your OS, there are a few things you could do:
Take a mutual exclusion semaphore (mutex) to protect access
in some OS, you can temporarily disable preemption, which guarantees your thread will not swap out.
Some OS provide a writer or reader semaphore which is more performant than a plain old mutex.
Here's my take on the question.
You have two or more threads running that write to a variable...like a status flag or something, where you only want to know if one or more of them was true. Then in another part of the code (after the threads complete) you want to check and see if at least on thread set that status... for example
bool flag = false
threadContainer tc
threadInputs inputs
check(input)
{
...do stuff to input
if(success)
flag = true
}
start multiple threads
foreach(i in inputs)
t = startthread(check, i)
tc.add(t) // Keep track of all the threads started
foreach(t in tc)
t.join( ) // Wait until each thread is done
if(flag)
print "One of the threads were successful"
else
print "None of the threads were successful"
I believe the above code would be OK, assuming you're fine with not knowing which thread set the status to true, and you can wait for all the multi-threaded stuff to finish before reading that flag. I could be wrong though.
If the operation is atomic, you should be able to get by just fine. But I wouldn't do that in practice. It is better just to acquire a lock on the object and write the value.
Assuming that property will never be assigned anything other than 11, then I don't see a reason for assigment in the first place. Just make it a constant then.
Assigment only makes sense when you intend to change the value unless the act of assigment itself has other side effects - like volatile writes have memory visibility side-effects in Java. And if you change state shared between multiple threads, then you need to synchronize or otherwise "handle" the problem of concurrency.
When you assign a value, without proper synchronization, to some state shared between multiple threads, then there's no guarantees for when the other threads will see that change. And no visibility guarantees means that it it possible that the other threads will never see the assignt.
Compilers, JITs, CPU caches. They're all trying to make your code run as fast as possible, and if you don't make any explicit requirements for memory visibility, then they will take advantage of that. If not on your machine, then somebody elses.