Related
As c/c++ standard said, the size of char must be 1. As my understanding, that means CPU guarantees that any read or write on a char must be done in one instruction.
Let's say we have many threads, which share a char variable:
char target = 1;
// thread a
target = 0;
// thread b
target = 1;
// thread 1
while (target == 1) {
// do something
}
// thread 2
while (target == 1) {
// do something
}
In a word, there are two kinds of threads: some of them are to set target into 0 or 1, and the others are to do some tasks if target == 1. The goal is that we can control the task-threds through modifying the value of target.
As my understanding, it doesn't seem that we need to use mutex/lock at all. But my coding experience gave me a strong feeling that we must use mutex/lock in this case.
I'm confused now. Should I use mutex/lock or not in this case?
You see, I can understand why we need mutex/lock in other cases, such as i++. Because i++ can't be done in only one instruction. So can target = 0 be done in one instruction, right? If so, does it mean that we don't need mutex/lock in this case?
Well, I know that we could use std::atomic, so my question is: is it OK to not use neither mutex/lcok nor std::atomic.
std::atomic guarantees that accessing a variable is atomic. From cppreference:
Each instantiation and full specialization of the std::atomic template
defines an atomic type. If one thread writes to an atomic object while
another thread reads from it, the behavior is well-defined (see memory
model for details on data races).
When a char actually is atomic (being size 1, is not sufficient), then std::atomic<char> needs no extra synchronization. However, on a platform where char is not atomic, std::atomic<char> guarantees that it can be read and written atomically by using a mutex or similar.
In practice, I'd expect char to be atomic, but the standard does not guarantee that.
Also consider that operations like eg += read and write the value, hence atomic reads and writes alone are not sufficient to safely call +=, while std::atomic<T> has a proper operator+=.
TL;DR
I'm confused now. Should I use mutex/lock or not in this case?
Let someone else take that decision for you. When you want something atomic, use a std::atomic<something> unless you want fine grained control over the synchronisation.
TL;DR
is it OK to not use neither mutex/lcok nor std::atomic
No
In general, it's not ok to assume things. If you need guarantees for something, then make sure you have them.
This is closely related to a common logical fallacy. Just because you cannot imagine why something could be true, that does not mean that it's true.
Longer version
As c/c++ standard said
There's no such thing as "C/C++" and definitely not "the C/C++ standard". They are two completely different languages with different standards. However, they do agree on this point. sizeof (char) is 1 in both languages.
(Sidenote: sizeof 'a' will yield different results.)
As my understanding, that means CPU guarantees that any read or write on a char must be done in one instruction.
That's not correct. The CPU has it's own specification, completely separate from the language standards. And there's nothing that says that this has to be true, even if it probably are in most or all cases.
Because i++ can't be done in only one instruction.
That is CPU dependent. The x86 architecture has an instruction for this. https://c9x.me/x86/html/file_module_x86_id_140.html
As my understanding, it doesn't seem that we need to use mutex/lock at all. But my coding experience gave me a strong feeling that we must use mutex/lock in this case.
Even if the target CPU does read and write in one instruction, which it probably does, there's nothing that says that the C or C++ code needs to be compiled to just that instruction.
The standards for both C and C++ describes the behavior of the code. Not how it is converted to assembly.
So no, you cannot make the assumptions you're doing.
In general, it cannot be assumed that reading or writing a char is an atomic operation. However, the target architecture may provide that guarantee. For embedded C programs it is common practice to rely on such underlying guarantees to avoid the overhead of synchronization mechanisms in certain situations.
In the example in the question it must be noted that even if reading/writing target is an atomic operation, the value could be changed at any time, so there is no guarantee that it will be 1 inside the while loops.
I've read this, my question is quite similar yet somewhat different.
Note, I know C++0x does not guarantee that but I'm asking particularly for a multi-core machine like x86-64.
Let's say we have 2 threads (pinned to 2 physical cores) running the following code:
// I know people may delcare volatile useless, but here I do NOT care memory reordering nor synchronization/
// I just want to suppress complier optimization of using register.
volatile int n;
void thread1() {
for (;;)
n = 0xABCD1234;
// NOTE, I know ++n is not atomic,
// but I do NOT care here.
// what I cares is whether n can be 0x00001234, i.e. in the middle of the update from core-1's cache lines to main memory,
// will core-2 see an incomplete value(like the first 2 bytes lost)?
++n;
}
}
void thread2() {
while (true) {
printf('%d', n);
}
}
Is it possible for thread 2 to see n to be something like 0x00001234, i.e. in the middle of the update from core-1's cache lines to main memory, will core-2 see an incomplete value?
I know a single 4-byte int definitely fits into a typically 128-byte-long cache line, and if that int does store inside one cache line then I believe no issues here... yet what if it acrosses the cache line boundary? i.e. will it be possbile that some char already sit inside that cache line which makes first part of the n in one cache line and the other part in the next line? If that is the case, then core-2 may have a chance seeing an incomplete value, right?
Also, I think unless making every char or short or other less-than-4-bytes types padded to be 4-byte-long, one can never guarantee a single int does not pass the cache line boundary, isn't it?
If so, would that suggest generally even setting a single int is not guaranteed to be atomic on a x86-64 multi-core machine?
I got this question because as I researched on this topic, various people in various posts seem agreed on that as long as the machine architecture is proper(e.g. x86-64) setting an int should be atomic. But as I argued above that does not hold, right?
UPDATE
I'd like to give some background of my question. I'm dealing with a real-time system, which is sampling some signal and putting the result into one global int, this is of course done in one thread. And in yet another thread I read this value and process it.
I do not care the ordering of set and get, all I need is just a complete (vs. a corrrupted integer value) value.
x86 guarantees this. C++ doesn't. If you write x86 assembly you will be fine. If you write C++ it is undefined behavior. Since you can't reason about undefined behavior (it is undefined after all) you have to go lower and look at the assembler instructions that were generated. If they do what you want then this is fine. Note, however, that compilers tend to change generated assembly when you change compilers, compiler versions, compiler flags or any code which might change the optimizer's behavior, so you will constantly have to check the assembler code to make sure it is still correct.
The easier way is to use std::atomic<int> which will guarantee that the correct assembler instructions are generated so you don't have to constantly check.
The other question talks about variables "properly aligned". If it crosses a cache-line, the variable is not properly aligned. An int will not do that unless you specifically ask the compiler to pack a struct, for example.
You also assume that using volatile int is better than atomic<int>. If volatile int is the perfect way to sync variables on your platform, surely the library implementer would also know that and store a volatile x inside atomic<x>.
There is no requirement that atomic<int> has to be extra slow just because it is standard. :-)
Why worry so much?
Rely on your implementation. std::atomic<int> will reduce to an int if int is atomic on your platform (and in x86-64 they are, if properly aligned).
I'd also be concerned about the possibility of int overflow with your code (which is undefined behaviour), if I were you.
In other words std::atomic<unsigned> is the appropriate type here.
If you're looking for atomicity guarantee, std::atomic<> is your friend. Don't rely on volatile qualifier.
The question is almost a duplicate of Why is integer assignment on a naturally aligned variable atomic on x86?. The answer there does answer everything you ask, but this question is more focused on the ABI / compiler question of whether an int (or other type?) will be sufficiently-aligned, rather than what happens when it is. There's other stuff in this question that's worth answering specifically, too.
Yes, they almost invariably will be on machines where an int fits in a single register (e.g. not AVR: an 8-bit RISC), because compilers typically choose not to use multiple store instructions when they could use 1.
Normal x86 ABIs will align an int to a 4B boundary, even inside structs (unless you use GNU C __attribute__((packed)) or the equivalent for other dialects). But beware that the i386 System V ABI only aligns double to 4 bytes; it's only outside structs that modern compilers can go beyond that and give it natural alignment, making load/store atomic.
But nothing you can legally do in C++ can ever depend on this fact (because by definition it will involve a data race on a non-atomic type so it's Undefined Behaviour). Fortunately, there are efficient ways to get the same result (i.e. about the same compiler-generated asm, without mfence instructions or other slow stuff) that don't cause undefined behaviour.
You should use atomic instead of volatile or hoping that the compiler doesn't optimize away stores or loads on a non-volatile int, because the assumption of async modification is one of the ways that volatile and atomic overlap.
I'm dealing with a real-time system, which is sampling some signal and putting the result into one global int, this is of course done in one thread. And in yet another thread I read this value and process it.
std::atomic with .store(val, std::memory_order_relaxed) and .load(std::memory_order_relaxed) will give you exactly what you want here. The HW-access thread runs free and does plain ordinary x86 store instructions into the shared variable, while the reader thread does plain ordinary x86 load instructions.
This is the C++11 way to express that this is what you want, and you should expect it to compile to the same asm as with volatile. (With maybe a couple instructions difference if you use clang, but nothing important.) If there was any case where volatile int wouldn't have sufficient alignment, or any other corner cases, atomic<int> will work (barring compiler bugs). Except maybe in a packed struct; IDK if compilers stop you from breaking atomicity by packing atomic types in structs.
In theory, you might want to use volatile std::atomic<int> to make sure the compiler doesn't optimize out multiple stores to the same variable. See Why don't compilers merge redundant std::atomic writes?. But for now, compilers don't do that kind of optimization. (volatile std::atomic<int> should still compile to the same light-weight asm.)
I know a single 4-byte int definitely fits into a typically 128-byte-long cache line, and if that int does store inside one cache line then I believe no issues here...
Cache lines are 64B on all mainstream x86 CPUs since PentiumIII; before that 32B lines were typical. (Well AMD Geode still uses 32B lines...) Pentium4 uses 64B lines, although it prefers to transfer them in pairs or something? Still, I think it's accurate to say that it really does use 64B lines, not 128B. This page lists it as 64B per line.
AFAIK, there are no x86 microarchitectures that used 128B lines in any level of cache.
Also, only Intel CPUs guarantee that cached unaligned stores / loads are atomic if they don't cross a cache-line boundary. The baseline atomicity guarantee for x86 in general (AMD/Intel/other) is don't cross an 8-byte boundary. See Why is integer assignment on a naturally aligned variable atomic on x86? for quotes from Intel/AMD manuals.
Natural alignment works on pretty much any ISA (not just x86) up to the maximum guaranteed-atomic width.
The code in your question wants a non-atomic read-modify write where the load and store are separately atomic, and impose no ordering on surrounding loads/stores.
As everyone has said, the right way to do this is with atomic<int>, but nobody has pointed out exactly how. If you just n++ on atomic_int n, you will get (for x86-64) lock add [n], 1, which will be much slower than what you get with volatile, because it makes the entire RMW operation atomic. (Perhaps this is why you were avoiding std::atomic<>?)
#include <atomic>
volatile int vcount;
std::atomic <int> acount;
static_assert(alignof(vcount) == sizeof(vcount), "under-aligned volatile counter");
void inc_volatile() {
while(1) vcount++;
}
void inc_separately_atomic() {
while(1) {
int t = acount.load(std::memory_order_relaxed);
t++;
acount.store(t, std::memory_order_relaxed);
}
}
asm output from the Godbolt compiler explorer with gcc7.2 and clang5.0
Unsurprisingly, they both compile to equivalent asm with gcc/clang for x86-32 and x86-64. gcc makes identical asm for both, except for the address to increment:
# x86-64 gcc -O3
inc_volatile:
.L2:
mov eax, DWORD PTR vcount[rip]
add eax, 1
mov DWORD PTR vcount[rip], eax
jmp .L2
inc_separately_atomic():
.L5:
mov eax, DWORD PTR acount[rip]
add eax, 1
mov DWORD PTR acount[rip], eax
jmp .L5
clang optimizes better, and uses
inc_separately_atomic():
.LBB1_1:
add dword ptr [rip + acount], 1
jmp .LBB1_1
Note the lack of a lock prefix, so inside the CPU this decodes to separate load, ALU add, and store uops. (See Can num++ be atomic for 'int num'?).
Besides smaller code-size, some of these uops can be micro-fused when they come from the same instruction, reducing front-end bottlenecks. (Totally irrelevant here; the loop bottlenecks on the 5 or 6 cycle latency of a store/reload. But if used as part of a larger loop, it would be relevant.) Unlike with a register operand, add [mem], 1 is better than inc [mem] on Intel CPUs because it micro-fuses even more: INC instruction vs ADD 1: Does it matter?.
It's interesting that clang uses the less efficient inc dword ptr [rip + vcount] for inc_volatile().
And how does an actual atomic RMW compile?
void inc_atomic_rmw() {
while(1) acount++;
}
# both gcc and clang do this:
.L7:
lock add DWORD PTR acount[rip], 1
jmp .L7
Alignment inside structs:
#include <stdint.h>
struct foo {
int a;
volatile double vdouble;
};
// will fail with -m32, in the SysV ABI.
static_assert(alignof(foo) == sizeof(double), "under-aligned volatile counter");
But atomic<double> or atomic<unsigned long long> will guarantee atomicity.
For 64-bit integer load/store on 32-bit machines, gcc uses SSE2 instructions. Some other compilers unfortunately use lock cmpxchg8b, which is far less efficient for separate stores or loads. volatile long long wouldn't give you that.
volatile double usually would normally be atomic to load/store when aligned correctly, because the normal way is already to use single 8B load/store instructions.
In general, for int num, num++ (or ++num), as a read-modify-write operation, is not atomic. But I often see compilers, for example GCC, generate the following code for it (try here):
void f()
{
int num = 0;
num++;
}
f():
push rbp
mov rbp, rsp
mov DWORD PTR [rbp-4], 0
add DWORD PTR [rbp-4], 1
nop
pop rbp
ret
Since line 5, which corresponds to num++ is one instruction, can we conclude that num++ is atomic in this case?
And if so, does it mean that so-generated num++ can be used in concurrent (multi-threaded) scenarios without any danger of data races (i.e. we don't need to make it, for example, std::atomic<int> and impose the associated costs, since it's atomic anyway)?
UPDATE
Notice that this question is not whether increment is atomic (it's not and that was and is the opening line of the question). It's whether it can be in particular scenarios, i.e. whether one-instruction nature can in certain cases be exploited to avoid the overhead of the lock prefix. And, as the accepted answer mentions in the section about uniprocessor machines, as well as this answer, the conversation in its comments and others explain, it can (although not with C or C++).
This is absolutely what C++ defines as a Data Race that causes Undefined Behaviour, even if one compiler happened to produce code that did what you hoped on some target machine. You need to use std::atomic for reliable results, but you can use it with memory_order_relaxed if you don't care about reordering. See below for some example code and asm output using fetch_add.
But first, the assembly language part of the question:
Since num++ is one instruction (add dword [num], 1), can we conclude that num++ is atomic in this case?
Memory-destination instructions (other than pure stores) are read-modify-write operations that happen in multiple internal steps. No architectural register is modified, but the CPU has to hold the data internally while it sends it through its ALU. The actual register file is only a small part of the data storage inside even the simplest CPU, with latches holding outputs of one stage as inputs for another stage, etc., etc.
Memory operations from other CPUs can become globally visible between the load and store. I.e. two threads running add dword [num], 1 in a loop would step on each other's stores. (See #Margaret's answer for a nice diagram). After 40k increments from each of two threads, the counter might have only gone up by ~60k (not 80k) on real multi-core x86 hardware.
"Atomic", from the Greek word meaning indivisible, means that no observer can see the operation as separate steps. Happening physically / electrically instantaneously for all bits simultaneously is just one way to achieve this for a load or store, but that's not even possible for an ALU operation. I went into a lot more detail about pure loads and pure stores in my answer to Atomicity on x86, while this answer focuses on read-modify-write.
The lock prefix can be applied to many read-modify-write (memory destination) instructions to make the entire operation atomic with respect to all possible observers in the system (other cores and DMA devices, not an oscilloscope hooked up to the CPU pins). That is why it exists. (See also this Q&A).
So lock add dword [num], 1 is atomic. A CPU core running that instruction would keep the cache line pinned in Modified state in its private L1 cache from when the load reads data from cache until the store commits its result back into cache. This prevents any other cache in the system from having a copy of the cache line at any point from load to store, according to the rules of the MESI cache coherency protocol (or the MOESI/MESIF versions of it used by multi-core AMD/Intel CPUs, respectively). Thus, operations by other cores appear to happen either before or after, not during.
Without the lock prefix, another core could take ownership of the cache line and modify it after our load but before our store, so that other store would become globally visible in between our load and store. Several other answers get this wrong, and claim that without lock you'd get conflicting copies of the same cache line. This can never happen in a system with coherent caches.
(If a locked instruction operates on memory that spans two cache lines, it takes a lot more work to make sure the changes to both parts of the object stay atomic as they propagate to all observers, so no observer can see tearing. The CPU might have to lock the whole memory bus until the data hits memory. Don't misalign your atomic variables!)
Note that the lock prefix also turns an instruction into a full memory barrier (like MFENCE), stopping all run-time reordering and thus giving sequential consistency. (See Jeff Preshing's excellent blog post. His other posts are all excellent, too, and clearly explain a lot of good stuff about lock-free programming, from x86 and other hardware details to C++ rules.)
On a uniprocessor machine, or in a single-threaded process, a single RMW instruction actually is atomic without a lock prefix. The only way for other code to access the shared variable is for the CPU to do a context switch, which can't happen in the middle of an instruction. So a plain dec dword [num] can synchronize between a single-threaded program and its signal handlers, or in a multi-threaded program running on a single-core machine. See the second half of my answer on another question, and the comments under it, where I explain this in more detail.
Back to C++:
It's totally bogus to use num++ without telling the compiler that you need it to compile to a single read-modify-write implementation:
;; Valid compiler output for num++
mov eax, [num]
inc eax
mov [num], eax
This is very likely if you use the value of num later: the compiler will keep it live in a register after the increment. So even if you check how num++ compiles on its own, changing the surrounding code can affect it.
(If the value isn't needed later, inc dword [num] is preferred; modern x86 CPUs will run a memory-destination RMW instruction at least as efficiently as using three separate instructions. Fun fact: gcc -O3 -m32 -mtune=i586 will actually emit this, because (Pentium) P5's superscalar pipeline didn't decode complex instructions to multiple simple micro-operations the way P6 and later microarchitectures do. See the Agner Fog's instruction tables / microarchitecture guide for more info, and the x86 tag wiki for many useful links (including Intel's x86 ISA manuals, which are freely available as PDF)).
Don't confuse the target memory model (x86) with the C++ memory model
Compile-time reordering is allowed. The other part of what you get with std::atomic is control over compile-time reordering, to make sure your num++ becomes globally visible only after some other operation.
Classic example: Storing some data into a buffer for another thread to look at, then setting a flag. Even though x86 does acquire loads/release stores for free, you still have to tell the compiler not to reorder by using flag.store(1, std::memory_order_release);.
You might be expecting that this code will synchronize with other threads:
// int flag; is just a plain global, not std::atomic<int>.
flag--; // Pretend this is supposed to be some kind of locking attempt
modify_a_data_structure(&foo); // doesn't look at flag, and the compiler knows this. (Assume it can see the function def). Otherwise the usual don't-break-single-threaded-code rules come into play!
flag++;
But it won't. The compiler is free to move the flag++ across the function call (if it inlines the function or knows that it doesn't look at flag). Then it can optimize away the modification entirely, because flag isn't even volatile.
(And no, C++ volatile is not a useful substitute for std::atomic. std::atomic does make the compiler assume that values in memory can be modified asynchronously similar to volatile, but there's much more to it than that. (In practice there are similarities between volatile int to std::atomic with mo_relaxed for pure-load and pure-store operations, but not for RMWs). Also, volatile std::atomic<int> foo is not necessarily the same as std::atomic<int> foo, although current compilers don't optimize atomics (e.g. 2 back-to-back stores of the same value) so volatile atomic wouldn't change the code-gen.)
Defining data races on non-atomic variables as Undefined Behaviour is what lets the compiler still hoist loads and sink stores out of loops, and many other optimizations for memory that multiple threads might have a reference to. (See this LLVM blog for more about how UB enables compiler optimizations.)
As I mentioned, the x86 lock prefix is a full memory barrier, so using num.fetch_add(1, std::memory_order_relaxed); generates the same code on x86 as num++ (the default is sequential consistency), but it can be much more efficient on other architectures (like ARM). Even on x86, relaxed allows more compile-time reordering.
This is what GCC actually does on x86, for a few functions that operate on a std::atomic global variable.
See the source + assembly language code formatted nicely on the Godbolt compiler explorer. You can select other target architectures, including ARM, MIPS, and PowerPC, to see what kind of assembly language code you get from atomics for those targets.
#include <atomic>
std::atomic<int> num;
void inc_relaxed() {
num.fetch_add(1, std::memory_order_relaxed);
}
int load_num() { return num; } // Even seq_cst loads are free on x86
void store_num(int val){ num = val; }
void store_num_release(int val){
num.store(val, std::memory_order_release);
}
// Can the compiler collapse multiple atomic operations into one? No, it can't.
# g++ 6.2 -O3, targeting x86-64 System V calling convention. (First argument in edi/rdi)
inc_relaxed():
lock add DWORD PTR num[rip], 1 #### Even relaxed RMWs need a lock. There's no way to request just a single-instruction RMW with no lock, for synchronizing between a program and signal handler for example. :/ There is atomic_signal_fence for ordering, but nothing for RMW.
ret
inc_seq_cst():
lock add DWORD PTR num[rip], 1
ret
load_num():
mov eax, DWORD PTR num[rip]
ret
store_num(int):
mov DWORD PTR num[rip], edi
mfence ##### seq_cst stores need an mfence
ret
store_num_release(int):
mov DWORD PTR num[rip], edi
ret ##### Release and weaker doesn't.
store_num_relaxed(int):
mov DWORD PTR num[rip], edi
ret
Notice how MFENCE (a full barrier) is needed after a sequential-consistency stores. x86 is strongly ordered in general, but StoreLoad reordering is allowed. Having a store buffer is essential for good performance on a pipelined out-of-order CPU. Jeff Preshing's Memory Reordering Caught in the Act shows the consequences of not using MFENCE, with real code to show reordering happening on real hardware.
Re: discussion in comments on #Richard Hodges' answer about compilers merging std::atomic num++; num-=2; operations into one num--; instruction:
A separate Q&A on this same subject: Why don't compilers merge redundant std::atomic writes?, where my answer restates a lot of what I wrote below.
Current compilers don't actually do this (yet), but not because they aren't allowed to. C++ WG21/P0062R1: When should compilers optimize atomics? discusses the expectation that many programmers have that compilers won't make "surprising" optimizations, and what the standard can do to give programmers control. N4455 discusses many examples of things that can be optimized, including this one. It points out that inlining and constant-propagation can introduce things like fetch_or(0) which may be able to turn into just a load() (but still has acquire and release semantics), even when the original source didn't have any obviously redundant atomic ops.
The real reasons compilers don't do it (yet) are: (1) nobody's written the complicated code that would allow the compiler to do that safely (without ever getting it wrong), and (2) it potentially violates the principle of least surprise. Lock-free code is hard enough to write correctly in the first place. So don't be casual in your use of atomic weapons: they aren't cheap and don't optimize much. It's not always easy easy to avoid redundant atomic operations with std::shared_ptr<T>, though, since there's no non-atomic version of it (although one of the answers here gives an easy way to define a shared_ptr_unsynchronized<T> for gcc).
Getting back to num++; num-=2; compiling as if it were num--:
Compilers are allowed to do this, unless num is volatile std::atomic<int>. If a reordering is possible, the as-if rule allows the compiler to decide at compile time that it always happens that way. Nothing guarantees that an observer could see the intermediate values (the num++ result).
I.e. if the ordering where nothing becomes globally visible between these operations is compatible with the ordering requirements of the source
(according to the C++ rules for the abstract machine, not the target architecture), the compiler can emit a single lock dec dword [num] instead of lock inc dword [num] / lock sub dword [num], 2.
num++; num-- can't disappear, because it still has a Synchronizes With relationship with other threads that look at num, and it's both an acquire-load and a release-store which disallows reordering of other operations in this thread. For x86, this might be able to compile to an MFENCE, instead of a lock add dword [num], 0 (i.e. num += 0).
As discussed in PR0062, more aggressive merging of non-adjacent atomic ops at compile time can be bad (e.g. a progress counter only gets updated once at the end instead of every iteration), but it can also help performance without downsides (e.g. skipping the atomic inc / dec of ref counts when a copy of a shared_ptr is created and destroyed, if the compiler can prove that another shared_ptr object exists for entire lifespan of the temporary.)
Even num++; num-- merging could hurt fairness of a lock implementation when one thread unlocks and re-locks right away. If it's never actually released in the asm, even hardware arbitration mechanisms won't give another thread a chance to grab the lock at that point.
With current gcc6.2 and clang3.9, you still get separate locked operations even with memory_order_relaxed in the most obviously optimizable case. (Godbolt compiler explorer so you can see if the latest versions are different.)
void multiple_ops_relaxed(std::atomic<unsigned int>& num) {
num.fetch_add( 1, std::memory_order_relaxed);
num.fetch_add(-1, std::memory_order_relaxed);
num.fetch_add( 6, std::memory_order_relaxed);
num.fetch_add(-5, std::memory_order_relaxed);
//num.fetch_add(-1, std::memory_order_relaxed);
}
multiple_ops_relaxed(std::atomic<unsigned int>&):
lock add DWORD PTR [rdi], 1
lock sub DWORD PTR [rdi], 1
lock add DWORD PTR [rdi], 6
lock sub DWORD PTR [rdi], 5
ret
Without many complications an instruction like add DWORD PTR [rbp-4], 1 is very CISC-style.
It perform three operations: load the operand from memory, increment it, store the operand back to memory.
During these operations the CPU acquire and release the bus twice, in between any other agent can acquire it too and this violates the atomicity.
AGENT 1 AGENT 2
load X
inc C
load X
inc C
store X
store X
X is incremented only once.
...and now let's enable optimisations:
f():
rep ret
OK, let's give it a chance:
void f(int& num)
{
num = 0;
num++;
--num;
num += 6;
num -=5;
--num;
}
result:
f(int&):
mov DWORD PTR [rdi], 0
ret
another observing thread (even ignoring cache synchronisation delays) has no opportunity to observe the individual changes.
compare to:
#include <atomic>
void f(std::atomic<int>& num)
{
num = 0;
num++;
--num;
num += 6;
num -=5;
--num;
}
where the result is:
f(std::atomic<int>&):
mov DWORD PTR [rdi], 0
mfence
lock add DWORD PTR [rdi], 1
lock sub DWORD PTR [rdi], 1
lock add DWORD PTR [rdi], 6
lock sub DWORD PTR [rdi], 5
lock sub DWORD PTR [rdi], 1
ret
Now, each modification is:-
observable in another thread, and
respectful of similar modifications happening in other threads.
atomicity is not just at the instruction level, it involves the whole pipeline from processor, through the caches, to memory and back.
Further info
Regarding the effect of optimisations of updates of std::atomics.
The c++ standard has the 'as if' rule, by which it is permissible for the compiler to reorder code, and even rewrite code provided that the outcome has the exact same observable effects (including side-effects) as if it had simply executed your code.
The as-if rule is conservative, particularly involving atomics.
consider:
void incdec(int& num) {
++num;
--num;
}
Because there are no mutex locks, atomics or any other constructs that influence inter-thread sequencing, I would argue that the compiler is free to rewrite this function as a NOP, eg:
void incdec(int&) {
// nada
}
This is because in the c++ memory model, there is no possibility of another thread observing the result of the increment. It would of course be different if num was volatile (might influence hardware behaviour). But in this case, this function will be the only function modifying this memory (otherwise the program is ill-formed).
However, this is a different ball game:
void incdec(std::atomic<int>& num) {
++num;
--num;
}
num is an atomic. Changes to it must be observable to other threads that are watching. Changes those threads themselves make (such as setting the value to 100 in between the increment and decrement) will have very far-reaching effects on the eventual value of num.
Here is a demo:
#include <thread>
#include <atomic>
int main()
{
for (int iter = 0 ; iter < 20 ; ++iter)
{
std::atomic<int> num = { 0 };
std::thread t1([&] {
for (int i = 0 ; i < 10000000 ; ++i)
{
++num;
--num;
}
});
std::thread t2([&] {
for (int i = 0 ; i < 10000000 ; ++i)
{
num = 100;
}
});
t2.join();
t1.join();
std::cout << num << std::endl;
}
}
sample output:
99
99
99
99
99
100
99
99
100
100
100
100
99
99
100
99
99
100
100
99
The add instruction is not atomic. It references memory, and two processor cores may have different local cache of that memory.
IIRC the atomic variant of the add instruction is called lock xadd
Since line 5, which corresponds to num++ is one instruction, can we conclude that num++ is atomic in this case?
It is dangerous to draw conclusions based on "reverse engineering" generated assembly. For example, you seem to have compiled your code with optimization disabled, otherwise the compiler would have thrown away that variable or loaded 1 directly to it without invoking operator++. Because the generated assembly may change significantly, based on optimization flags, target CPU, etc., your conclusion is based on sand.
Also, your idea that one assembly instruction means an operation is atomic is wrong as well. This add will not be atomic on multi-CPU systems, even on the x86 architecture.
Even if your compiler always emitted this as an atomic operation, accessing num from any other thread concurrently would constitute a data race according to the C++11 and C++14 standards and the program would have undefined behavior.
But it is worse than that. First, as has been mentioned, the instruction generated by the compiler when incrementing a variable may depend on the optimization level. Secondly, the compiler may reorder other memory accesses around ++num if num is not atomic, e.g.
int main()
{
std::unique_ptr<std::vector<int>> vec;
int ready = 0;
std::thread t{[&]
{
while (!ready);
// use "vec" here
});
vec.reset(new std::vector<int>());
++ready;
t.join();
}
Even if we assume optimistically that ++ready is "atomic", and that the compiler generates the checking loop as needed (as I said, it's UB and therefore the compiler is free to remove it, replace it with an infinite loop, etc.), the compiler might still move the pointer assignment, or even worse the initialization of the vector to a point after the increment operation, causing chaos in the new thread. In practice, I would not be surprised at all if an optimizing compiler removed the ready variable and the checking loop completely, as this does not affect observable behavior under language rules (as opposed to your private hopes).
In fact, at last year's Meeting C++ conference, I've heard from two compiler developers that they very gladly implement optimizations that make naively written multi-threaded programs misbehave, as long as language rules allow it, if even a minor performance improvement is seen in correctly written programs.
Lastly, even if you didn't care about portability, and your compiler was magically nice, the CPU you are using is very likely of a superscalar CISC type and will break down instructions into micro-ops, reorder and/or speculatively execute them, to an extent only limited by synchronizing primitives such as (on Intel) the LOCK prefix or memory fences, in order to maximize operations per second.
To make a long story short, the natural responsibilities of thread-safe programming are:
Your duty is to write code that has well-defined behavior under language rules (and in particular the language standard memory model).
Your compiler's duty is to generate machine code which has the same well-defined (observable) behavior under the target architecture's memory model.
Your CPU's duty is to execute this code so that the observed behavior is compatible with its own architecture's memory model.
If you want to do it your own way, it might just work in some cases, but understand that the warranty is void, and you will be solely responsible for any unwanted outcomes. :-)
PS: Correctly written example:
int main()
{
std::unique_ptr<std::vector<int>> vec;
std::atomic<int> ready{0}; // NOTE the use of the std::atomic template
std::thread t{[&]
{
while (!ready);
// use "vec" here
});
vec.reset(new std::vector<int>());
++ready;
t.join();
}
This is safe because:
The checks of ready cannot be optimized away according to language rules.
The ++ready happens-before the check that sees ready as not zero, and other operations cannot be reordered around these operations. This is because ++ready and the check are sequentially consistent, which is another term described in the C++ memory model and that forbids this specific reordering. Therefore the compiler must not reorder the instructions, and must also tell the CPU that it must not e.g. postpone the write to vec to after the increment of ready. Sequentially consistent is the strongest guarantee regarding atomics in the language standard. Lesser (and theoretically cheaper) guarantees are available e.g. via other methods of std::atomic<T>, but these are definitely for experts only, and may not be optimized much by the compiler developers, because they are rarely used.
On a single-core x86 machine, an add instruction will generally be atomic with respect to other code on the CPU1. An interrupt can't split a single instruction down the middle.
Out-of-order execution is required to preserve the illusion of instructions executing one at a time in order within a single core, so any instruction running on the same CPU will either happen completely before or completely after the add.
Modern x86 systems are multi-core, so the uniprocessor special case doesn't apply.
If one is targeting a small embedded PC and has no plans to move the code to anything else, the atomic nature of the "add" instruction could be exploited. On the other hand, platforms where operations are inherently atomic are becoming more and more scarce.
(This doesn't help you if you're writing in C++, though. Compilers don't have an option to require num++ to compile to a memory-destination add or xadd without a lock prefix. They could choose to load num into a register and store the increment result with a separate instruction, and will likely do that if you use the result.)
Footnote 1: The lock prefix existed even on original 8086 because I/O devices operate concurrently with the CPU; drivers on a single-core system need lock add to atomically increment a value in device memory if the device can also modify it, or with respect to DMA access.
Back in the day when x86 computers had one CPU, the use of a single instruction ensured that interrupts would not split the read/modify/write and if the memory would not be used as a DMA buffer too, it was atomic in fact (and C++ did not mention threads in the standard, so this wasn’t addressed).
When it was rare to have a dual processor (e.g. dual-socket Pentium Pro) on a customer desktop, I effectively used this to avoid the LOCK prefix on a single-core machine and improve performance.
Today, it would only help against multiple threads that were all set to the same CPU affinity, so the threads you are worried about would only come into play via time slice expiring and running the other thread on the same CPU (core). That is not realistic.
With modern x86/x64 processors, the single instruction is broken up into several micro ops and furthermore the memory reading and writing is buffered. So different threads running on different CPUs will not only see this as non-atomic but may see inconsistent results concerning what it reads from memory and what it assumes other threads have read to that point in time: you need to add memory fences to restore sane behavior.
No.
https://www.youtube.com/watch?v=31g0YE61PLQ
(That's just a link to the "No" scene from "The Office")
Do you agree that this would be a possible output for the program:
sample output:
100
100
100
100
100
100
100
100
100
100
100
100
100
100
100
100
100
100
100
100
If so, then the compiler is free to make that the only possible output for the program, in whichever way the compiler wants. ie a main() that just puts out 100s.
This is the "as-if" rule.
And regardless of output, you can think of thread synchronization the same way - if thread A does num++; num--; and thread B reads num repeatedly, then a possible valid interleaving is that thread B never reads between num++ and num--. Since that interleaving is valid, the compiler is free to make that the only possible interleaving. And just remove the incr/decr entirely.
There are some interesting implications here:
while (working())
progress++; // atomic, global
(ie imagine some other thread updates a progress bar UI based on progress)
Can the compiler turn this into:
int local = 0;
while (working())
local++;
progress += local;
probably that is valid. But probably not what the programmer was hoping for :-(
The committee is still working on this stuff. Currently it "works" because compilers don't optimize atomics much. But that is changing.
And even if progress was also volatile, this would still be valid:
int local = 0;
while (working())
local++;
while (local--)
progress++;
:-/
That a single compiler's output, on a specific CPU architecture, with optimizations disabled (since gcc doesn't even compile ++ to add when optimizing in a quick&dirty example), seems to imply incrementing this way is atomic doesn't mean this is standard-compliant (you would cause undefined behavior when trying to access num in a thread), and is wrong anyways, because add is not atomic in x86.
Note that atomics (using the lock instruction prefix) are relatively heavy on x86 (see this relevant answer), but still remarkably less than a mutex, which isn't very appropriate in this use-case.
Following results are taken from clang++ 3.8 when compiling with -Os.
Incrementing an int by reference, the "regular" way :
void inc(int& x)
{
++x;
}
This compiles into :
inc(int&):
incl (%rdi)
retq
Incrementing an int passed by reference, the atomic way :
#include <atomic>
void inc(std::atomic<int>& x)
{
++x;
}
This example, which is not much more complex than the regular way, just gets the lock prefix added to the incl instruction - but caution, as previously stated this is not cheap. Just because assembly looks short doesn't mean it's fast.
inc(std::atomic<int>&):
lock incl (%rdi)
retq
Yes, but...
Atomic is not what you meant to say. You're probably asking the wrong thing.
The increment is certainly atomic. Unless the storage is misaligned (and since you left alignment to the compiler, it is not), it is necessarily aligned within a single cache line. Short of special non-caching streaming instructions, each and every write goes through the cache. Complete cache lines are being atomically read and written, never anything different.
Smaller-than-cacheline data is, of course, also written atomically (since the surrounding cache line is).
Is it thread-safe?
This is a different question, and there are at least two good reasons to answer with a definite "No!".
First, there is the possibility that another core might have a copy of that cache line in L1 (L2 and upwards is usually shared, but L1 is normally per-core!), and concurrently modifies that value. Of course that happens atomically, too, but now you have two "correct" (correctly, atomically, modified) values -- which one is the truly correct one now?
The CPU will sort it out somehow, of course. But the result may not be what you expect.
Second, there is memory ordering, or worded differently happens-before guarantees. The most important thing about atomic instructions is not so much that they are atomic. It's ordering.
You have the possibility of enforcing a guarantee that everything that happens memory-wise is realized in some guaranteed, well-defined order where you have a "happened before" guarantee. This ordering may be as "relaxed" (read as: none at all) or as strict as you need.
For example, you can set a pointer to some block of data (say, the results of some calculation) and then atomically release the "data is ready" flag. Now, whoever acquires this flag will be led into thinking that the pointer is valid. And indeed, it will always be a valid pointer, never anything different. That's because the write to the pointer happened-before the atomic operation.
When your compiler uses only a single instruction for the increment and your machine is single-threaded, your code is safe. ^^
Try compiling the same code on a non-x86 machine, and you'll quickly see very different assembly results.
The reason num++ appears to be atomic is because on x86 machines, incrementing a 32-bit integer is, in fact, atomic (assuming no memory retrieval takes place). But this is neither guaranteed by the c++ standard, nor is it likely to be the case on a machine that doesn't use the x86 instruction set. So this code is not cross-platform safe from race conditions.
You also don't have a strong guarantee that this code is safe from Race Conditions even on an x86 architecture, because x86 doesn't set up loads and stores to memory unless specifically instructed to do so. So if multiple threads tried to update this variable simultaneously, they may end up incrementing cached (outdated) values
The reason, then, that we have std::atomic<int> and so on is so that when you're working with an architecture where the atomicity of basic computations is not guaranteed, you have a mechanism that will force the compiler to generate atomic code.
So I'm aware that nothing is atomic in C++. But I'm trying to figure out if there are any "pseudo-atomic" assumptions I can make. The reason is that I want to avoid using mutexes in some simple situations where I only need very weak guarantees.
1) Suppose I have globally defined volatile bool b, which
initially I set true. Then I launch a thread which executes a loop
while(b) doSomething();
Meanwhile, in another thread, I execute b=true.
Can I assume that the first thread will continue to execute? In other words, if b starts out as true, and the first thread checks the value of b at the same time as the second thread assigns b=true, can I assume that the first thread will read the value of b as true? Or is it possible that at some intermediate point of the assignment b=true, the value of b might be read as false?
2) Now suppose that b is initially false. Then the first thread executes
bool b1=b;
bool b2=b;
if(b1 && !b2) bad();
while the second thread executes b=true. Can I assume that bad() never gets called?
3) What about an int or other builtin types: suppose I have volatile int i, which is initially (say) 7, and then I assign i=7. Can I assume that, at any time during this operation, from any thread, the value of i will be equal to 7?
4) I have volatile int i=7, and then I execute i++ from some thread, and all other threads only read the value of i. Can I assume that i never has any value, in any thread, except for either 7 or 8?
5) I have volatile int i, from one thread I execute i=7, and from another I execute i=8. Afterwards, is i guaranteed to be either 7 or 8 (or whatever two values I have chosen to assign)?
There are no threads in standard C++, and Threads cannot be implemented as a library.
Therefore, the standard has nothing to say about the behaviour of programs which use threads. You must look to whatever additional guarantees are provided by your threading implementation.
That said, in threading implementations I've used:
(1) yes, you can assume that irrelevant values aren't written to variables. Otherwise the whole memory model goes out the window. But be careful that when you say "another thread" never sets b to false, that means anywhere, ever. If it does, that write could perhaps be re-ordered to occur during your loop.
(2) no, the compiler can re-order the assignments to b1 and b2, so it is possible for b1 to end up true and b2 false. In such a simple case I don't know why it would re-order, but in more complex cases there might be very good reasons.
[Edit: oops, by the time I got to answering (2) I'd forgotten that b was volatile. Reads from a volatile variable won't be re-ordered, sorry, so yes on a typical threading implementation (if there is any such thing), you can assume that you won't end up with b1 true and b2 false.]
(3) same as 1. volatile in general has nothing to do with threading at all. However, it is quite exciting in some implementations (Windows), and might in effect imply memory barriers.
(4) on an architecture where int writes are atomic yes, although volatile has nothing to do with it. See also...
(5) check the docs carefully. Likely yes, and again volatile is irrelevant, because on almost all architectures int writes are atomic. But if int write is not atomic, then no (and no for the previous question), even if it's volatile you could in principle get a different value. Given those values 7 and 8, though, we're talking a pretty weird architecture for the byte containing the relevant bits to be written in two stages, but with different values you could more plausibly get a partial write.
For a more plausible example, suppose that for some bizarre reason you have a 16 bit int on a platform where only 8bit writes are atomic. Odd, but legal, and since int must be at least 16 bits you can see how it could come about. Suppose further that your initial value is 255. Then increment could legally be implemented as:
read the old value
increment in a register
write the most significant byte of the result
write the least significant byte of the result.
A read-only thread which interrupted the incrementing thread between the third and fourth steps of that, could see the value 511. If the writes are in the other order, it could see 0.
An inconsistent value could be left behind permanently if one thread is writing 255, another thread is concurrently writing 256, and the writes get interleaved. Impossible on many architectures, but to know that this won't happen you need to know at least something about the architecture. Nothing in the C++ standard forbids it, because the C++ standard talks about execution being interrupted by a signal, but otherwise has no concept of execution being interrupted by another part of the program, and no concept of concurrent execution. That's why threads aren't just another library - adding threads fundamentally changes the C++ execution model. It requires the implementation to do things differently, as you'll eventually discover if for example you use threads under gcc and forget to specify -pthreads.
The same could happen on a platform where aligned int writes are atomic, but unaligned int writes are permitted and not atomic. For example IIRC on x86, unaligned int writes are not guaranteed atomic if they cross a cache line boundary. x86 compilers will not mis-align a declared int variable, for this reason and others. But if you play games with structure packing you could probably provoke an example.
So: pretty much any implementation will give you the guarantees you need, but might do so in quite a complicated way.
In general, I've found that it is not worth trying to rely on platform-specific guarantees about memory access, that I don't fully understand, in order to avoid mutexes. Use a mutex, and if that's too slow use a high-quality lock-free structure (or implement a design for one) written by someone who really knows the architecture and compiler. It will probably be correct, and subject to correctness will probably outperform anything I invent myself.
Most of the answers correctly address the CPU memory ordering issues you're going to experience, but none have detailed how the compiler can thwart your intentions by re-ordering your code in ways that break your assumptions.
Consider an example taken from this post:
volatile int ready;
int message[100];
void foo(int i)
{
message[i/10] = 42;
ready = 1;
}
At -O2 and above, recent versions of GCC and Intel C/C++ (don't know about VC++) will do the store to ready first, so it can be overlapped with computation of i/10 (volatile does not save you!):
leaq _message(%rip), %rax
movl $1, _ready(%rip) ; <-- whoa Nelly!
movq %rsp, %rbp
sarl $2, %edx
subl %edi, %edx
movslq %edx,%rdx
movl $42, (%rax,%rdx,4)
This isn't a bug, it's the optimizer exploiting CPU pipelining. If another thread is waiting on ready before accessing the contents of message then you have a nasty and obscure race.
Employ compiler barriers to ensure your intent is honored. An example that also exploits the relatively strong ordering of x86 are the release/consume wrappers found in Dmitriy Vyukov's Single-Producer Single-Consumer queue posted here:
// load with 'consume' (data-dependent) memory ordering
// NOTE: x86 specific, other platforms may need additional memory barriers
template<typename T>
T load_consume(T const* addr)
{
T v = *const_cast<T const volatile*>(addr);
__asm__ __volatile__ ("" ::: "memory"); // compiler barrier
return v;
}
// store with 'release' memory ordering
// NOTE: x86 specific, other platforms may need additional memory barriers
template<typename T>
void store_release(T* addr, T v)
{
__asm__ __volatile__ ("" ::: "memory"); // compiler barrier
*const_cast<T volatile*>(addr) = v;
}
I suggest that if you are going to venture into the realm of concurrent memory access, use a library that will take care of these details for you. While we all wait for n2145 and std::atomic check out Thread Building Blocks' tbb::atomic or the upcoming boost::atomic.
Besides correctness, these libraries can simplify your code and clarify your intent:
// thread 1
std::atomic<int> foo; // or tbb::atomic, boost::atomic, etc
foo.store(1, std::memory_order_release);
// thread 2
int tmp = foo.load(std::memory_order_acquire);
Using explicit memory ordering, foo's inter-thread relationship is clear.
May be this thread is ancient, but the C++ 11 standard DOES have a thread library and also a vast atomic library for atomic operations. The purpose is specifically for concurrency support and avoid data races.
The relevant header is atomic
It's generally a really, really bad idea to depend on this, as you could end up with bad things happening and only one some architectures. The best solution would be to use a guaranteed atomic API, for example the Windows Interlocked api.
If your C++ implementation supplies the library of atomic operations specified by n2145 or some variant thereof, you can presumably rely on it. Otherwise, you cannot in general rely on "anything" about atomicity at the language level, since multitasking of any kind (and therefore atomicity, which deals with multitasking) is not specified by the existing C++ standard.
Volatile in C++ do not plays the same role than in Java. All the cases are undefined behavior as Steve saids. Some cases can be Ok for a compiler, oa given processor architecture and with a multi-threading system, but switching the optimization flags can make your program behave differently, as the C++03 compilers don't know about threads.
C++0x defines the rules that avoid race conditions and the operations that help you to master that, but to may knowledge there is no yet a compiler that implement yet all the part of the standard related to this subject.
My answer is going to be frustrating: No, No, No, No, and No.
1-4) The compiler is allowed to do ANYTHING it pleases with a variable it writes to. It may store temporary values in it, so long as ends up doing something that would do the same thing as that thread executing in a vacuum. ANYTHING is valid
5) Nope, no guarantee. If a variable is not atomic, and you write to it on one thread, and read or write to it on another, it is a race case. The spec declares such race cases to be undefined behavior, and absolutely anything goes. That being said, you will be hard pressed to find a compiler that does not give you 7 or 8, but it IS legal for a compiler to give you something else.
I always refer to this highly comical explanation of race cases.
http://software.intel.com/en-us/blogs/2013/01/06/benign-data-races-what-could-possibly-go-wrong
I have two threads, one updating an int and one reading it. This is a statistic value where the order of the reads and writes is irrelevant.
My question is, do I need to synchronize access to this multi-byte value anyway? Or, put another way, can part of the write be complete and get interrupted, and then the read happen.
For example, think of a value = 0x0000FFFF that gets incremented value of 0x00010000.
Is there a time where the value looks like 0x0001FFFF that I should be worried about? Certainly the larger the type, the more possible something like this to happen.
I've always synchronized these types of accesses, but was curious what the community thinks.
Boy, what a question. The answer to which is:
Yes, no, hmmm, well, it depends
It all comes down to the architecture of the system. On an IA32 a correctly aligned address will be an atomic operation. Unaligned writes might be atomic, it depends on the caching system in use. If the memory lies within a single L1 cache line then it is atomic, otherwise it's not. The width of the bus between the CPU and RAM can affect the atomic nature: a correctly aligned 16bit write on an 8086 was atomic whereas the same write on an 8088 wasn't because the 8088 only had an 8 bit bus whereas the 8086 had a 16 bit bus.
Also, if you're using C/C++ don't forget to mark the shared value as volatile, otherwise the optimiser will think the variable is never updated in one of your threads.
At first one might think that reads and writes of the native machine size are atomic but there are a number of issues to deal with including cache coherency between processors/cores. Use atomic operations like Interlocked* on Windows and the equivalent on Linux. C++0x will have an "atomic" template to wrap these in a nice and cross-platform interface. For now if you are using a platform abstraction layer it may provide these functions. ACE does, see the class template ACE_Atomic_Op.
IF you're reading/writing 4-byte value AND it is DWORD-aligned in memory AND you're running on the I32 architecture, THEN reads and writes are atomic.
Yes, you need to synchronize accesses. In C++0x it will be a data race, and undefined behaviour. With POSIX threads it's already undefined behaviour.
In practice, you might get bad values if the data type is larger than the native word size. Also, another thread might never see the value written due to optimizations moving the read and/or write.
You must synchronize, but on certain architectures there are efficient ways to do it.
Best is to use subroutines (perhaps masked behind macros) so that you can conditionally replace implementations with platform-specific ones.
The Linux kernel already has some of this code.
On Windows, Interlocked***Exchange***Add is guaranteed to be atomic.
To echo what everyone said upstairs, the language pre-C++0x cannot guarantee anything about shared memory access from multiple threads. Any guarantees would be up to the compiler.
No, they aren't (or at least you can't assume they are). Having said that, there are some tricks to do this atomically, but they typically aren't portable (see Compare-and-swap).
I agree with many and especially Jason. On windows, one would likely use InterlockedAdd and its friends.
Asside from the cache issue mentioned above...
If you port the code to a processor with a smaller register size it will not be atomic anymore.
IMO, threading issues are too thorny to risk it.
Lets take this example
int x;
x++;
x=x+5;
The first statement is assumed to be atomic because it translates to a single INC assembly directive that takes a single CPU cycle. However, the second assignment requires several operations so it's clearly not an atomic operation.
Another e.g,
x=5;
Again, you have to disassemble the code to see what exactly happens here.
tc,
I think the moment you use a constant ( like 6) , the instruction wouldn't be completed in one machine cycle.
Try to see the instruction set of x+=6 as compared to x++
Some people think that ++c is atomic, but have a eye on the assembly generated. For example with 'gcc -S' :
movl cpt.1586(%rip), %eax
addl $1, %eax
movl %eax, cpt.1586(%rip)
To increment an int, the compiler first load it into a register, and stores it back into the memory. This is not atomic.
Definitively NO !
That answer from our highest C++ authority, M. Boost:
Operations on "ordinary" variables are not guaranteed to be atomic.
The only portable way is to use the sig_atomic_t type defined in signal.h header for your compiler. In most C and C++ implementations, that is an int. Then declare your variable as "volatile sig_atomic_t."
Reads and writes are atomic, but you also need to worry about the compiler re-ordering your code. Compiler optimizations may violate happens-before relationship of statements in your code. By using atomic you don't have to worry about that.
...
atomic i;
soap_status = GOT_RESPONSE ;
i = 1
In the above example, the variable 'i' will only be set to 1 after we get a soap response.