If I have some code that looks something like:
typedef struct {
bool some_flag;
pthread_cond_t c;
pthread_mutex_t m;
} foo_t;
// I assume the mutex has already been locked, and will be unlocked
// some time after this function returns. For clarity. Definitely not
// out of laziness ;)
void check_flag(foo_t* f) {
while(f->flag)
pthread_cond_wait(&f->c, &f->m);
}
Is there anything in the C standard preventing an optimizer from rewriting check_flag as:
void check_flag(foo_t* f) {
bool cache = f->flag;
while(cache)
pthread_cond_wait(&f->c, &f->m);
}
In other words, does the generated code have to follow the f pointer every time through the loop, or is the compiler free to pull the dereference out?
If it is free to pull it out, is there any way to prevent this? Do I need to sprinkle a volatile keyword somewhere? It can't be check_flag's parameter because I plan on having other variables in this struct that I don't mind the compiler optimizing like this.
Might I have to resort to:
void check_flag(foo_t* f) {
volatile bool* cache = &f->some_flag;
while(*cache)
pthread_cond_wait(&f->c, &f->m);
}
In the general case, even if multi-threading wasn't involved and your loop looked like:
void check_flag(foo_t* f) {
while(f->flag)
foo(&f->c, &f->m);
}
the compiler would be unable to to cache the f->flag test. That's because the compiler can't know whether or not a function (like foo() above) might change whatever object f is pointing to.
Under special circumstances (foo() is visible to the compiler, and all pointers passed to the check_flag() are known not to be aliased or otherwise modifiable by foo()) the compiler might be able to optimize the check.
However, pthread_cond_wait() must be implemented in a way that would prevent that optimization.
See Does guarding a variable with a pthread mutex guarantee it's also not cached?:
You might also be interested in Steve Jessop's answer to: Can a C/C++ compiler legally cache a variable in a register across a pthread library call?
But how far you want to take the issues raised by Boehm's paper in your own work is up to you. As far as I can tell, if you want to take the stand that pthreads doesn't/can't make the guarantee, then you're in essence taking the stand that pthreads is useless (or at least provides no safety guarantees, which I think by reduction has the same outcome). While this might be true in the strictest sense (as addressed in the paper), it's also probably not a useful answer. I'm not sure what option you'd have other than pthreads on Unix-based platforms.
Normally, you should try to lock the pthread mutex before waiting on the condition object as the pthread_cond_wait call release the mutex (and reacquire it before returning). So, your check_flag function should be rewritten like that to conform to the semantic on the pthread condition.
void check_flag(foo_t* f) {
pthread_mutex_lock(&f->m);
while(f->flag)
pthread_cond_wait(&f->c, &f->m);
pthread_mutex_unlock(&f->m);
}
Concerning the question of whether or not the compiler is allowed to optimize the reading of the flagfield, this answer explains it in more detail than I can.
Basically, the compiler know about the semantic of pthread_cond_wait, pthread_mutex_lock and pthread_mutex_unlock. He know that he can't optimize memory reading in those situation (the call to pthread_cond_wait in this exemple). There is no notion of memory barrier here, just a special knowledge of certain function, and some rule to follow in their presence.
There is another thing protecting you from optimization performed by the processor. Your average processor is capable of reordering memory access (read / write) provided that the semantic is conserved, and it is always doing it (as it allow to increase performance). However, this break when more than one processor can access the same memory address. A memory barrier is just an instruction to the processor telling it that it can move the read / write that were issued before the barrier and execute them after the barrier. It has finish them now.
As written, the compiler is free to cache the result as you describe or even in a more subtle way - by putting it into a register. You can prevent this optimization from taking place by making the variable volatile. But that is not necessarily enough - you should not code it this way! You should use condition variables as prescribed (lock, wait, unlock).
Trying to do work around the library is bad, but it gets worse. Perhaps reading Hans Boehm's paper on the general topic from PLDI 2005 ("Threads Cannot be Implemented as a Library"), or many of his follow-on articles (which lead up to work on a revised C++ memory model) will put the fear of God in you and steer you back to the straight and narrow :).
Volatile is for this purpose. Relying on the compiler to know about pthread coding practices seems a little nuts to me, although; compilers are pretty smart these days. In fact, the compiler probably sees that you are looping to test a variable and won't cache it in a register for that reason, not because it sees you using pthreads. Just use volatile if you really care.
Kind of funny little note. We have a VOLATILE #define that is either "volatile" (when we think the bug can't possibly be our code...) or blank. When we think we have a crash due to the optimizer killing us, we #define it "volatile" which puts volatile in front of almost everything. We then test to see if the problem goes away. So far... the bugs have been the developer and not the compiler! who'd have thought!? We have developed a high performance "non locking" and "non blocking" threading library. We have a test platform that hammers it to the point of thousands of races per second. So fare, we have never detected a problem needing volatile! So far gcc has never cached a shared variable in a register. yah...we are surprised too. We are still waiting for our chance to use volatile!
Related
I have a thread which basically does:
int changed; //global variable
..
for (;;) {
pthread_mutex_lock(&mtx);
if (changed) {
do_changes();
changed = 0;
}
pthread_mutex_unlock(&mtx);
do_stuff();
}
The loop is run a few hundred thousand times a second, while the global changed variable will be set rarely (a few times a day) by another thread.
With a change to
volatile int changed; //global variable
..
for (;;) {
if (changed) {
pthread_mutex_lock(&mtx);
do_changes();
changed = 0;
pthread_mutex_unlock(&mtx);
}
do_stuff();
}
I can measure 3-4% performance increase of the loop with this approach, which is worth while pursuing.
However volatile variables seems to be heavily discouraged.
Are there any drawbacks with the approach here ? Any corner cases that might cause the 2. version to not work as intended ?
volatile does not make your variable thread safe or atomic. You may like to use C11 atomics for that.
You basically have two threads changing changed variable thus overwriting previous values causing data race.
I cannot recommend enough watching atomic Weapons: The C++ Memory Model and Modern Hardware. (It applies to C as well).
If you are using a C11 environment that support them you can use atomic variables. If your system supports it, they use special instructions to achieve atomicity, instead of locks. If your system don't support that, they use locks (the flag type is always lockless).
If you don't have C11, but you have a GCC-compatible compiler see the sync family of functions. It's similar (but older) to the C11 atomic variables but if your system don't support them they generate a function call.
Author: However volatile variables seems to be heavily discouraged. Are there any drawbacks with the approach here ? Any corner cases that might cause the 2. version to not work as intended ?
First of all: Whenever you are tempted to use volatile think again if you don't need atomics. Now see what may happen:
1) It is not safe if you have the loop in multiple threads. You can think about duplicating the check:
if (changed) { // quick check
pthread_mutex_lock(&mtx);
if (changed) { // another thread could do the work
...
2) If your code is critical to see it was changed, you need to use atomics, because that if(changed) before pthread_mutex_lock may not see it because of cache.
3) It may work on x86(_64) with strong memory ordering and atomic int accesses, but fail on other architecture. That is the reason why volatile is discouraged, use atomics (and make a habbit of it). volatile does not force atomic usage or any other synchronization. Atomics do (read-modify-write instructions).
std::atomic_flag validated;
std::mutex mx; struct MyData { ... } data;
void change() {
lock_guard<mutex> lock(mx);
data.something();
validated.clear();
}
void validate() {
if(!validated.test_and_set()) {
lock_guard<mutex> lock(mx);
data.update();
}
}
NOTE: You will never know for sure if data is valid or not unless you hold the lock and use another variable for it.
4) Just try your original code with pthread_spinlock_t
5) Little advice: don't play God with synchronization unless you really know, what you are doing. You can switch from mutex to spinlock (written by somebody else) and do some benchmarking.
About edits and comments: The original answer just started with 1) nothing before. As it turned out, some people neither read the full question nor the full answer. Pitty those quick downvoters. This site is not facebook and those votes should be eihter is helpful or is not helpful, these up/downvotes are no likes and dislikes like on facebook! I may disagree with some parts of the other answers, but still think they are helpful, although incomplete (not aswering the full question, but only a part of it) or partially incorrent (there is nothing bad in writing same value to variable from many threads if we know that we can do that without problems).
It is a good idea to make it volatile to ensure that the compiler is not doing unwanted optimizations, but it doesn't make it atomic.
If there are more threads reading "changed" it could happen that one thread updates "changed" to 0 when there's another thread waiting to execute do_changes(), which will happen once the mutex is released because the condition was already evaluated.
If you want to avoid this then move the if statement inside the mutex protected space.
Hope this helps.
Carles.
In a codebase I reviewed, I found the following idiom.
void notify(struct actor_t act) {
write(act.pipe, "M", 1);
}
// thread A sending data to thread B
void send(byte *data) {
global.data = data;
notify(threadB);
}
// in thread B event loop
read(this.sock, &cmd, 1);
switch (cmd) {
case 'M': use_data(global.data);break;
...
}
"Hold it", I said to the author, a senior member of my team, "there's no memory barrier here! You don't guarantee that global.data will be flushed from the cache to main memory. If thread A and thread B will run in two different processors - this scheme might fail".
The senior programmer grinned, and explained slowly, as if explaining his five years old boy how to tie his shoelaces: "Listen young boy, we've seen here many thread related bugs, in high load testing, and in real clients", he paused to scratch his longish beard, "but we've never had a bug with this idiom".
"But, it says in the book..."
"Quiet!", he hushed me promptly, "Maybe theoretically, it's not guaranteed, but in practice, the fact you used a function call is effectively a memory barrier. The compiler will not reorder the instruction global.data = data, since it can't know if anyone using it in the function call, and the x86 architecture will ensure that the other CPUs will see this piece of global data by the time thread B reads the command from the pipe. Rest assured, we have ample real world problems to worry about. We don't need to invest extra effort in bogus theoretical problems.
"Rest assured my boy, in time you'll understand to separate the real problem from the I-need-to-get-a-PhD non-problems."
Is he correct? Is that really a non-issue in practice (say x86, x64 and ARM)?
It's against everything I learned, but he does have a long beard and a really smart looks!
Extra points if you can show me a piece of code proving him wrong!
Memory barriers aren't just to prevent instruction reordering. Even if instructions aren't reordered it can still cause problems with cache coherence. As for the reordering - it depends on your compiler and settings. ICC is particularly agressive with reordering. MSVC w/ whole program optimization can be, too.
If your shared data variable is declared as volatile, even though it's not in the spec most compilers will generate a memory variable around reads and writes from the variable and prevent reordering. This is not the correct way of using volatile, nor what it was meant for.
(If I had any votes left, I'd +1 your question for the narration.)
In practice, a function call is a compiler barrier, meaning that the compiler will not move global memory accesses past the call. A caveat to this is functions which the compiler knows something about, e.g. builtins, inlined functions (keep in mind IPO!) etc.
So a processor memory barrier (in addition to a compiler barrier) is in theory needed to make this work. However, since you're calling read and write which are syscalls that change the global state, I'm quite sure that the kernel issues memory barriers somewhere in the implementation of those. There is no such guarantee though, so in theory you need the barriers.
The basic rule is: the compiler must make the global state appear to be exactly as you coded it, but if it can prove that a given function doesn't use global variables then it can implement the algorithm any way it chooses.
The upshot is that traditional compilers always treated functions in another compilation unit as a memory barrier because they couldn't see inside those functions. Increasingly, modern compilers are growing "whole program" or "link time" optimization strategies which break down these barriers and will cause poorly written code to fail, even though it's been working fine for years.
If the function in question is in a shared library then it won't be able to see inside it, but if the function is one defined by the C standard then it doesn't need to -- it already knows what the function does -- so you have to be careful of those also. Note that a compiler will not recognise a kernel call for what it is, but the very act of inserting something that the compiler can't recognise (inline assembler, or a function call to an assembler file) will create a memory barrier in itself.
In your case, notify will either be a black box the compiler can't see inside (a library function) or else it will contain a recognisable memory barrier, so you are most likely safe.
In practice, you have to write very bad code to fall over this.
In practice, he's correct and a memory barrier is implied in this specific case.
But the point is that if its presence is "debatable", the code is already too complex and unclear.
Really guys, use a mutex or other proper constructs. It's the only safe way to deal with threads and to write maintainable code.
And maybe you'll see other errors, like that the code is unpredictable if send() is called more than one time.
I'm referring to the main static languages today (C, C++, java, C#,). I've heard some contradicting answers about this, so I wanted to know:
If I have some code such as:
loop(...) {
type x = val;
...
}
('loop' is some type of loop, e.g. for, while)
Will it cause memory allocation in each iteration of the loop, or just once? Is it different from writing this:
type x;
loop(...) {
x = val;
...
}
where memory is only allocated once for x?
The strictly correct answer is that it depends on the implementation, as both are semantically correct. No language specification would require or prohibit such implementation details.
That said, any implementation worth its salt will be able to reuse the same stack slot or even CPU register (with native compilation, especially likely in presence of a JIT). Even the bytecode will likely be completely identical.
And finally, there's that thing with premature optimization... Unless proven otherwise, you shouldn't even bother thinking about low-level details like this (if you think knowledge and control over such issues matters, perhaps you should just program in assembler), because:
Unless you're doing a microbenchmark (or a really huge number-crunching task - but how many people freaking out about performance actually do those?), you won't even notice any difference even if it isn't optimized. If you're doing anything of interest in the loop body, it will dwarf the difference (again, if any). Especially if you're doing any I/O.
Even if there is memory allocation, it boils down to pushing and popping a few bytes on the native stack, which in turn boils down to adding an integer constant to a hardware register. All C and C++ programs use that stack for their local variables, and non of those ever complained about its performance... if you have to reserve space, you can't get faster than using the stack.
If you have to ask this kind of question, you're not someone who could do anything about it. Those people know to just (1) measure it, (2) look at the generated code and (3) look for large-scale optimizations before even thinking on this level ;)
I need to have a string as a global variable. There is a possibility for multiple threads to set the global variable. Should I have to go for mutex for this? Or will OS handle such actions.
Going for mutex affects the application performance.
I am not concerned about the order of actions happening. I am afraid of the data corruption.
Could somebody let me know about this.
It sounds like you understand all of the concerns. If the global variable can be corrupt you definitely need to lock it in a mutex. This will affect performance, since this part is by definition now going to be synchronous. That being said, you will want to lock the smallest part of the code as necessary, to minimize the time that synchronous code is being called.
What's your global variable? A pointer to the string buffer, or the buffer itself?
On many architectures (including AFAIR 32-bit x86) overwriting a single pointer is atomic.
This example might work:
volatile char **global_var;
void set_var(char *str) {
char *tmp = strdup(str);
global_var = &tmp;
}
You can use Thread-Local Storage for this.
Unfortunately, its not specified in current C99 standart, but possible will be there in C1X. For now, you can use compiler-specific implementations (GCC, ICC and Visual C have it).
As far as the standards are concerned, yes, you must use a mutex. Failure to do so results in undefined behavior. In practice, most machine architectures will have no problem with this. Future versions of the C standard (C1x) will have atomic types which, if used here, would definitely make the assignment without lock safe (albeit possibly using an internal lock, on broken archs that lack real atomics).
In the C programming language and Pthreads as the threading library; do variables/structures that are shared between threads need to be declared as volatile? Assuming that they might be protected by a lock or not (barriers perhaps).
Does the pthread POSIX standard have any say about this, is this compiler-dependent or neither?
Edit to add: Thanks for the great answers. But what if you're not using locks; what if you're using barriers for example? Or code that uses primitives such as compare-and-swap to directly and atomically modify a shared variable...
As long as you are using locks to control access to the variable, you do not need volatile on it. In fact, if you're putting volatile on any variable you're probably already wrong.
https://software.intel.com/en-us/blogs/2007/11/30/volatile-almost-useless-for-multi-threaded-programming/
The answer is absolutely, unequivocally, NO. You do not need to use 'volatile' in addition to proper synchronization primitives. Everything that needs to be done are done by these primitives.
The use of 'volatile' is neither necessary nor sufficient. It's not necessary because the proper synchronization primitives are sufficient. It's not sufficient because it only disables some optimizations, not all of the ones that might bite you. For example, it does not guarantee either atomicity or visibility on another CPU.
But unless you use volatile, the compiler is free to cache the shared data in a register for any length of time... if you want your data to be written to be predictably written to actual memory and not just cached in a register by the compiler at its discretion, you will need to mark it as volatile. Alternatively, if you only access the shared data after you have left a function modifying it, you might be fine. But I would suggest not relying on blind luck to make sure that values are written back from registers to memory.
Right, but even if you do use volatile, the CPU is free to cache the shared data in a write posting buffer for any length of time. The set of optimizations that can bite you is not precisely the same as the set of optimizations that 'volatile' disables. So if you use 'volatile', you are relying on blind luck.
On the other hand, if you use sychronization primitives with defined multi-threaded semantics, you are guaranteed that things will work. As a plus, you don't take the huge performance hit of 'volatile'. So why not do things that way?
I think one very important property of volatile is that it makes the variable be written to memory when modified, and reread from memory each time it accessed. The other answers here mix volatile and synchronization, and it is clear from some other answers than this that volatile is NOT a sync primitive (credit where credit is due).
But unless you use volatile, the compiler is free to cache the shared data in a register for any length of time... if you want your data to be written to be predictably written to actual memory and not just cached in a register by the compiler at its discretion, you will need to mark it as volatile. Alternatively, if you only access the shared data after you have left a function modifying it, you might be fine. But I would suggest not relying on blind luck to make sure that values are written back from registers to memory.
Especially on register-rich machines (i.e., not x86), variables can live for quite long periods in registers, and a good compiler can cache even parts of structures or entire structures in registers. So you should use volatile, but for performance, also copy values to local variables for computation and then do an explicit write-back. Essentially, using volatile efficiently means doing a bit of load-store thinking in your C code.
In any case, you positively have to use some kind of OS-level provided sync mechanism to create a correct program.
For an example of the weakness of volatile, see my Decker's algorithm example at http://jakob.engbloms.se/archives/65, which proves pretty well that volatile does not work to synchronize.
There is a widespread notion that the keyword volatile is good for multi-threaded programming.
Hans Boehm points out that there are only three portable uses for volatile:
volatile may be used to mark local variables in the same scope as a setjmp whose value should be preserved across a longjmp. It is unclear what fraction of such uses would be slowed down, since the atomicity and ordering constraints have no effect if there is no way to share the local variable in question. (It is even unclear what fraction of such uses would be slowed down by requiring all variables to be preserved across a longjmp, but that is a separate matter and is not considered here.)
volatile may be used when variables may be "externally modified", but the modification in fact is triggered synchronously by the thread itself, e.g. because the underlying memory is mapped at multiple locations.
A volatile sigatomic_t may be used to communicate with a signal handler in the same thread, in a restricted manner. One could consider weakening the requirements for the sigatomic_t case, but that seems rather counterintuitive.
If you are multi-threading for the sake of speed, slowing down code is definitely not what you want. For multi-threaded programming, there two key issues that volatile is often mistakenly thought to address:
atomicity
memory consistency, i.e. the order of a thread's operations as seen by another thread.
Let's deal with (1) first. Volatile does not guarantee atomic reads or writes. For example, a volatile read or write of a 129-bit structure is not going to be atomic on most modern hardware. A volatile read or write of a 32-bit int is atomic on most modern hardware, but volatile has nothing to do with it. It would likely be atomic without the volatile. The atomicity is at the whim of the compiler. There's nothing in the C or C++ standards that says it has to be atomic.
Now consider issue (2). Sometimes programmers think of volatile as turning off optimization of volatile accesses. That's largely true in practice. But that's only the volatile accesses, not the non-volatile ones. Consider this fragment:
volatile int Ready;
int Message[100];
void foo( int i ) {
Message[i/10] = 42;
Ready = 1;
}
It's trying to do something very reasonable in multi-threaded programming: write a message and then send it to another thread. The other thread will wait until Ready becomes non-zero and then read Message. Try compiling this with "gcc -O2 -S" using gcc 4.0, or icc. Both will do the store to Ready first, so it can be overlapped with the computation of i/10. The reordering is not a compiler bug. It's an aggressive optimizer doing its job.
You might think the solution is to mark all your memory references volatile. That's just plain silly. As the earlier quotes say, it will just slow down your code. Worst yet, it might not fix the problem. Even if the compiler does not reorder the references, the hardware might. In this example, x86 hardware will not reorder it. Neither will an Itanium(TM) processor, because Itanium compilers insert memory fences for volatile stores. That's a clever Itanium extension. But chips like Power(TM) will reorder. What you really need for ordering are memory fences, also called memory barriers. A memory fence prevents reordering of memory operations across the fence, or in some cases, prevents reordering in one direction.Volatile has nothing to do with memory fences.
So what's the solution for multi-threaded programming? Use a library or language extension that implements the atomic and fence semantics. When used as intended, the operations in the library will insert the right fences. Some examples:
POSIX threads
Windows(TM) threads
OpenMP
TBB
Based on article by Arch Robison (Intel)
In my experience, no; you just have to properly mutex yourself when you write to those values, or structure your program such that the threads will stop before they need to access data that depends on another thread's actions. My project, x264, uses this method; threads share an enormous amount of data but the vast majority of it doesn't need mutexes because its either read-only or a thread will wait for the data to become available and finalized before it needs to access it.
Now, if you have many threads that are all heavily interleaved in their operations (they depend on each others' output on a very fine-grained level), this may be a lot harder--in fact, in such a case I'd consider revisiting the threading model to see if it can possibly be done more cleanly with more separation between threads.
NO.
Volatile is only required when reading a memory location that can change independently of the CPU read/write commands. In the situation of threading, the CPU is in full control of read/writes to memory for each thread, therefore the compiler can assume the memory is coherent and optimizes the CPU instructions to reduce unnecessary memory access.
The primary usage for volatile is for accessing memory-mapped I/O. In this case, the underlying device can change the value of a memory location independently from CPU. If you do not use volatile under this condition, the CPU may use a previously cached memory value, instead of reading the newly updated value.
POSIX 7 guarantees that functions such as pthread_lock also synchronize memory
https://pubs.opengroup.org/onlinepubs/9699919799/basedefs/V1_chap04.html#tag_04_11 "4.12 Memory Synchronization" says:
The following functions synchronize memory with respect to other threads:
pthread_barrier_wait()
pthread_cond_broadcast()
pthread_cond_signal()
pthread_cond_timedwait()
pthread_cond_wait()
pthread_create()
pthread_join()
pthread_mutex_lock()
pthread_mutex_timedlock()
pthread_mutex_trylock()
pthread_mutex_unlock()
pthread_spin_lock()
pthread_spin_trylock()
pthread_spin_unlock()
pthread_rwlock_rdlock()
pthread_rwlock_timedrdlock()
pthread_rwlock_timedwrlock()
pthread_rwlock_tryrdlock()
pthread_rwlock_trywrlock()
pthread_rwlock_unlock()
pthread_rwlock_wrlock()
sem_post()
sem_timedwait()
sem_trywait()
sem_wait()
semctl()
semop()
wait()
waitpid()
Therefore if your variable is guarded between pthread_mutex_lock and pthread_mutex_unlock then it does not need further synchronization as you might attempt to provide with volatile.
Related questions:
Does guarding a variable with a pthread mutex guarantee it's also not cached?
Does pthread_mutex_lock contains memory fence instruction?
Volatile would only be useful if you need absolutely no delay between when one thread writes something and another thread reads it. Without some sort of lock, though, you have no idea of when the other thread wrote the data, only that it's the most recent possible value.
For simple values (int and float in their various sizes) a mutex might be overkill if you don't need an explicit synch point. If you don't use a mutex or lock of some sort, you should declare the variable volatile. If you use a mutex you're all set.
For complicated types, you must use a mutex. Operations on them are non-atomic, so you could read a half-changed version without a mutex.
Volatile means that we have to go to memory to get or set this value. If you don't set volatile, the compiled code might store the data in a register for a long time.
What this means is that you should mark variables that you share between threads as volatile so that you don't have situations where one thread starts modifying the value but doesn't write its result before a second thread comes along and tries to read the value.
Volatile is a compiler hint that disables certain optimizations. The output assembly of the compiler might have been safe without it but you should always use it for shared values.
This is especially important if you are NOT using the expensive thread sync objects provided by your system - you might for example have a data structure where you can keep it valid with a series of atomic changes. Many stacks that do not allocate memory are examples of such data structures, because you can add a value to the stack then move the end pointer or remove a value from the stack after moving the end pointer. When implementing such a structure, volatile becomes crucial to ensure that your atomic instructions are actually atomic.
The underlying reason is that the C language semantic is based upon a single-threaded abstract machine. And the compiler is within its own right to transform the program as long as the program's 'observable behaviors' on the abstract machine stay unchanged. It can merge adjacent or overlapping memory accesses, redo a memory access multiple times (upon register spilling for example), or simply discard a memory access, if it thinks the program's behaviors, when executed in a single thread, doesn't change. Therefore as you may suspect, the behaviors do change if the program is actually supposed to be executing in a multi-threaded way.
As Paul Mckenney pointed out in a famous Linux kernel document:
It _must_not_ be assumed that the compiler will do what you want
with memory references that are not protected by READ_ONCE() and
WRITE_ONCE(). Without them, the compiler is within its rights to
do all sorts of "creative" transformations, which are covered in
the COMPILER BARRIER section.
READ_ONCE() and WRITE_ONCE() are defined as volatile casts on referenced variables. Thus:
int y;
int x = READ_ONCE(y);
is equivalent to:
int y;
int x = *(volatile int *)&y;
So, unless you make a 'volatile' access, you are not assured that the access happens exactly once, no matter what synchronization mechanism you are using. Calling an external function (pthread_mutex_lock for example) may force the compiler do memory accesses to global variables. But this happens only when the compiler fails to figure out whether the external function changes these global variables or not. Modern compilers employing sophisticated inter-procedure analysis and link-time optimization make this trick simply useless.
In summary, you should mark variables shared by multiple threads volatile or access them using volatile casts.
As Paul McKenney has also pointed out:
I have seen the glint in their eyes when they discuss optimization techniques that you would not want your children to know about!
But see what happens to C11/C++11.
Some people obviously are assuming that the compiler treats the synchronization calls as memory barriers. "Casey" is assuming there is exactly one CPU.
If the sync primitives are external functions and the symbols in question are visible outside the compilation unit (global names, exported pointer, exported function that may modify them) then the compiler will treat them -- or any other external function call -- as a memory fence with respect to all externally visible objects.
Otherwise, you are on your own. And volatile may be the best tool available for making the compiler produce correct, fast code. It generally won't be portable though, when you need volatile and what it actually does for you depends a lot on the system and compiler.
No.
First, volatile is not necessary. There are numerous other operations that provide guaranteed multithreaded semantics that don't use volatile. These include atomic operations, mutexes, and so on.
Second, volatile is not sufficient. The C standard does not provide any guarantees about multithreaded behavior for variables declared volatile.
So being neither necessary nor sufficient, there's not much point in using it.
One exception would be particular platforms (such as Visual Studio) where it does have documented multithreaded semantics.
Variables that are shared among threads should be declared 'volatile'. This tells the
compiler that when one thread writes to such variables, the write should be to memory
(as opposed to a register).