问题
Assume:
A. C++ under WIN32.
B. A properly aligned volatile integer incremented and decremented using InterlockedIncrement() and InterlockedDecrement().
__declspec (align(8)) volatile LONG _ServerState = 0;
If I want to simply read _ServerState, do I need to read the variable via an InterlockedXXX function?
For instance, I have seen code such as:
LONG x = InterlockedExchange(&_ServerState, _ServerState);
and
LONG x = InterlockedCompareExchange(&_ServerState, _ServerState, _ServerState);
The goal is to simply read the current value of _ServerState.
Can't I simply say:
if (_ServerState == some value)
{
// blah blah blah
}
There seems to be some confusion WRT this subject. I understand register-sized reads are atomic in Windows, so I would assume the InterlockedXXX function is unnecessary.
Matt J.
Okay, thanks for the responses. BTW, this is Visual C++ 2005 and 2008.
If it's true I should use an InterlockedXXX function to read the value of _ServerState, even if just for the sake of clarity, what's the best way to go about that?
LONG x = InterlockedExchange(&_ServerState, _ServerState);
This has the side effect of modifying the value, when all I really want to do is read it. Not only that, but there is a possibility that I could reset the flag to the wrong value if there is a context switch as the value of _ServerState is pushed on the stack in preparation of calling InterlockedExchange().
LONG x = InterlockedCompareExchange(&_ServerState, _ServerState, _ServerState);
I took this from an example I saw on MSDN.
See http://msdn.microsoft.com/en-us/library/ms686355(VS.85).aspx
All I need is something along the lines:
lock mov eax, [_ServerState]
In any case, the point, which I thought was clear, is to provide thread-safe access to a flag without incurring the overhead of a critical section. I have seen LONGs used this way via the InterlockedXXX() family of functions, hence my question.
Okay, we are thinking a good solution to this problem of reading the current value is:
LONG Cur = InterlockedCompareExchange(&_ServerState, 0, 0);
回答1:
It depends on what you mean by "goal is to simply read the current value of _ServerState" and it depends on what set of tools and the platform you use (you specify Win32 and C++, but not which C++ compiler, and that may matter).
If you simply want to read the value such that the value is uncorrupted (ie., if some other processor is changing the value from 0x12345678 to 0x87654321 your read will get one of those 2 values and not 0x12344321) then simply reading will be OK as long as the variable is :
- marked
volatile, - properly aligned, and
- read using a single instruction with a word size that the processor handles atomically
None of this is promised by the C/C++ standard, but Windows and MSVC do make these guarantees, and I think that most compilers that target Win32 do as well.
However, if you want your read to be synchronized with behavior of the other thread, there's some additional complexity. Say that you have a simple 'mailbox' protocol:
struct mailbox_struct {
uint32_t flag;
uint32_t data;
};
typedef struct mailbox_struct volatile mailbox;
// the global - initialized before wither thread starts
mailbox mbox = { 0, 0 };
//***************************
// Thread A
while (mbox.flag == 0) {
/* spin... */
}
uint32_t data = mbox.data;
//***************************
//***************************
// Thread B
mbox.data = some_very_important_value;
mbox.flag = 1;
//***************************
The thinking is Thread A will spin waiting for mbox.flag to indicate mbox.data has a valid piece of information. Thread B will write some data into mailbox.data then will set mbox.flag to 1 as a signal that mbox.data is valid.
In this case a simple read in Thread A of mbox.flag might get the value 1 even though a subsequent read of mbox.data in Thread A does not get the value written by Thread B.
This is because even though the compiler will not reorder the Thread B writes to mbox.data and mbox.flag, the processor and/or cache might. C/C++ guarantees that the compiler will generate code such that Thread B will write to mbox.data before it writes to mbox.flag, but the processor and cache might have a different idea - special handling called 'memory barriers' or 'acquire and release semantics' must be used to ensure ordering below the level of the thread's stream of instructions.
I'm not sure if compilers other than MSVC make any claims about ordering below the instruction level. However MS does guarantee that for MSVC volatile is enough - MS specifies that volatile writes have release semantics and volatile reads have acquire semantics - though I'm not sure at which version of MSVC this applies - see http://msdn.microsoft.com/en-us/library/12a04hfd.aspx?ppud=4.
I have also seen code like you describe that uses Interlocked APIs to perform simple reads and writes to shared locations. My take on the matter is to use the Interlocked APIs. Lock free inter-thread communication is full of very difficult to understand and subtle pitfalls, and trying to take a shortcut on a critical bit of code that may end up with a very difficult to diagnose bug doesn't seem like a good idea to me. Also, using an Interlocked API screams to anyone maintaining the code, "this is data access that needs to be shared or synchronized with something else - tread carefully!".
Also when using the Interlocked API you're taking the specifics of the hardware and the compiler out of the picture - the platform makes sure all of that stuff is dealt with properly - no more wondering...
Read Herb Sutter's Effective Concurrency articles on DDJ (which happen to be down at the moment, for me at least) for good information on this topic.
回答2:
Your way is good:
LONG Cur = InterlockedCompareExchange(&_ServerState, 0, 0);
I'm using similar solution:
LONG Cur = InterlockedExchangeAdd(&_ServerState, 0);
回答3:
Interlocked instructions provide atomicity and inter-processor synchronization. Both writes and reads must be synchronized, so yes, you should be using interlocked instructions to read a value that is shared between threads and not protected by a lock. Lock-free programming (and that's what you're doing) is a very tricky area, so you might consider using locks instead. Unless this is really one of your program's bottlenecks that must be optimized?
回答4:
32-bit read operations are already atomic on some 32-bit systems (Intel spec says these operations are atomic, but there's no guarantee that this will be true on other x86-compatible platforms). So you shouldn't use this for threads synchronization.
If you need a flag some sort you should consider using Event object and WaitForSingleObject function for that purpose.
回答5:
To anyone who has to revisit this thread I want to add to what was well explained by Bartosz that _InterlockedCompareExchange() is a good alternative to standard atomic_load() if standard atomics are not available. Here is the code for atomically reading my_uint32_t_var in C on i86 Win64. atomic_load() is included as a benchmark:
long debug_x64_i = std::atomic_load((const std::_Atomic_long *)&my_uint32_t_var);
00000001401A6955 mov eax,dword ptr [rbp+30h]
00000001401A6958 xor edi,edi
00000001401A695A mov dword ptr [rbp-0Ch],eax
debug_x64_i = _InterlockedCompareExchange((long*)&my_uint32_t_var, 0, 0);
00000001401A695D xor eax,eax
00000001401A695F lock cmpxchg dword ptr [rbp+30h],edi
00000001401A6964 mov dword ptr [rbp-0Ch],eax
debug_x64_i = _InterlockedOr((long*)&my_uint32_t_var, 0);
00000001401A6967 prefetchw [rbp+30h]
00000001401A696B mov eax,dword ptr [rbp+30h]
00000001401A696E xchg ax,ax
00000001401A6970 mov ecx,eax
00000001401A6972 lock cmpxchg dword ptr [rbp+30h],ecx
00000001401A6977 jne foo+30h (01401A6970h)
00000001401A6979 mov dword ptr [rbp-0Ch],eax
long release_x64_i = std::atomic_load((const std::_Atomic_long *)&my_uint32_t_var);
00000001401A6955 mov eax,dword ptr [rbp+30h]
release_x64_i = _InterlockedCompareExchange((long*)&my_uint32_t_var, 0, 0);
00000001401A6958 mov dword ptr [rbp-0Ch],eax
00000001401A695B xor edi,edi
00000001401A695D mov eax,dword ptr [rbp-0Ch]
00000001401A6960 xor eax,eax
00000001401A6962 lock cmpxchg dword ptr [rbp+30h],edi
00000001401A6967 mov dword ptr [rbp-0Ch],eax
release_x64_i = _InterlockedOr((long*)&my_uint32_t_var, 0);
00000001401A696A prefetchw [rbp+30h]
00000001401A696E mov eax,dword ptr [rbp+30h]
00000001401A6971 mov ecx,eax
00000001401A6973 lock cmpxchg dword ptr [rbp+30h],ecx
00000001401A6978 jne foo+31h (01401A6971h)
00000001401A697A mov dword ptr [rbp-0Ch],eax
回答6:
you should be okay. It's volatile, so the optimizer shouldn't savage you, and it's a 32-bit value so it should be at least approximately atomic. The one possible surprise is if the instruction pipeline can get around that.
On the other hand, what's the additional cost of using the guarded routines?
回答7:
Current value reading may not need any lock.
回答8:
The Interlocked* functions prevent two different processors from accessing the same piece of memory. In a single processor system you are going to be ok. If you have a dual-core system where you have threads on different cores both accessing this value, you might have problems doing what you think is atomic without the Interlocked*.
回答9:
Your initial understanding is basically correct. According to the memory model which Windows requires on all MP platforms it supports (or ever will support), reads from a naturally-aligned variable marked volatile are atomic as long as they are smaller than the size of a machine word. Same with writes. You don't need a 'lock' prefix.
If you do the reads without using an interlock, you are subject to processor reordering. This can even occur on x86, in a limited circumstance: reads from a variable may be moved above writes of a different variable. On pretty much every non-x86 architecture that Windows supports, you are subject to even more complicated reordering if you don't use explicit interlocks.
There's also a requirement that if you're using a compare exchange loop, you must mark the variable you're compare exchanging on as volatile. Here's a code example to demonstrate why:
long g_var = 0; // not marked 'volatile' -- this is an error
bool foo () {
long oldValue;
long newValue;
long retValue;
// (1) Capture the original global value
oldValue = g_var;
// (2) Compute a new value based on the old value
newValue = SomeTransformation(oldValue);
// (3) Store the new value if the global value is equal to old?
retValue = InterlockedCompareExchange(&g_var,
newValue,
oldValue);
if (retValue == oldValue) {
return true;
}
return false;
}
What can go wrong is that the compiler is well within its rights to re-fetch oldValue from g_var at any time if it's not volatile. This 'rematerialization' optimization is great in many cases because it can avoid spilling registers to the stack when register pressure is high.
Thus, step (3) of the function would become:
// (3) Incorrectly store new value regardless of whether the global
// is equal to old.
retValue = InterlockedCompareExchange(&g_var,
newValue,
g_var);
回答10:
Read is fine. A 32-bit value is always read as a whole as long as it's not split on a cache line. Your align 8 guarantees that it's always within a cache line so you'll be fine.
Forget about instructions reordering and all that non-sense. Results are always retired in-order. It would be a processor recall otherwise!!!
Even for a dual CPU machine (i.e. shared via the slowest FSBs), you'll still be fine as the CPUs guarantee cache coherency via MESI Protocol. The only thing you're not guaranteed is the value you read may not be the absolute latest. BUT, what is the latest anyway? That's something you likely won't need to know in most situations if you're not writing back to the location based on the value of that read. Otherwise, you'd have used interlocked ops to handle it in the first place.
In short, you gain nothing by using Interlocked ops on a read (except perhaps reminding the next person maintaining your code to tread carefully - then again, that person may not be qualified to maintain your code to begin with).
EDIT: In response to a comment left by Adrian McCarthy.
You're overlooking the effect of compiler optimizations. If the compiler thinks it has the value already in a register, then it's going to re-use that value instead of re-reading it from memory. Also, the compiler may do instruction re-ordering for optimization if it believes there are no observable side effects.
I did not say reading from a non-volatile variable is fine. All the question was asking was if interlocked was required. In fact, the variable in question was clearly declared with volatile. Or were you overlooking the effect of the keyword volatile?
来源：https://stackoverflow.com/questions/779996/reading-interlocked-variables