问题
It's my understanding that if two threads are reading from the same piece of memory, and no thread is writing to that memory, then the operation is safe. However, I'm not sure what happens if one thread is reading and the other is writing. What would happen? Is the result undefined? Or would the read just be stale? If a stale read is not a concern is it ok to have unsynchronized read-write to a variable? Or is it possible the data would be corrupted, and neither the read nor the write would be correct and one should always synchronize in this case?
I want to say that I've learned it is the later case, that a race on memory access leaves the state undefined... but I don't remember where I may have learned that and I'm having a hard time finding the answer on google. My intuition is that a variable is operated on in registers, and that true (as in hardware) concurrency is impossible (or is it), so that the worst that could happen is stale data, i.e. the following:
WriteThread: copy value from memory to register
WriteThread: update value in register
ReadThread: copy value of memory to register
WriteThread: write new value to memory
At which point the read thread has stale data.
回答1:
The result is undefined. Corrupted data is entirely possible. For an obvious example, consider a 64-bit value being manipulated by a 32-bit processor. Let's assume the value is a simple counter, and we increment it when the lower 32-bits contain 0xffffffff. The increment produces 0x00000000. When we detect that, we increment the upper word. If, however, some other thread read the value between the time the lower word was incremented and the upper word was incremented, they get a value with an un-incremented upper word, but the lower word set to 0 -- a value completely different from what it would have been either before or after the increment is complete.
回答2:
Usually memory is read or written in atomic units determined by the CPU architecture (32 bit and 64 bits item aligned on 32 bit and 64 bit boundaries is common these days).
In this case, what happens depends on the amount of data being written.
Let's consider the case of 32 bit atomic read/write cells.
If two threads write 32 bits into such an aligned cell, then it is absolutely well defined what happens: one of the two written values is retained. Unfortunately for you (well, the program), you don't know which value. By extremely clever programming, you can actually use this atomicity of reads and writes to build synchronization algorithms (e.g., Dekker's algorithm), but it is faster typically to use architecturally defined locks instead.
If two threads write more than an atomic unit (e.g., they both write a 128 bit value), then in fact the atomic unit sized pieces of the values written will be stored in a absolutely well defined way, but you won't know which pieces of which value get written in what order. So what may end up in storage is the value from the first thread, the second thread, or mixes of the bits in atomic unit sizes from both threads.
Similar ideas hold for one thread reading, and one thread writing in atomic units, and larger.
Basically, you don't want to do unsynchronized reads and writes to memory locations, because you won't know the outcome, even though it may be very well defined by the architecture.
回答3:
As I hinted in Ira Baxter's answer, CPU cache also plays a part on multicore systems. Consider the following test code:
DANGER WILL ROBISON!
The following code boosts priority to realtime to achieve somewhat more consistent results - while doing so requires admin privileges, be careful if running the code on dual- or single-core systems, since your machine will lock up for the duration of the test run.
#include <windows.h>
#include <stdio.h>
const int RUNFOR = 5000;
volatile bool terminating = false;
volatile int value;
static DWORD WINAPI CountErrors(LPVOID parm)
{
int errors = 0;
while(!terminating)
{
value = (int) parm;
if(value != (int) parm)
errors++;
}
printf("\tThread %08X: %d errors\n", parm, errors);
return 0;
}
static void RunTest(int affinity1, int affinity2)
{
terminating = false;
DWORD dummy;
HANDLE t1 = CreateThread(0, 0, CountErrors, (void*)0x1000, CREATE_SUSPENDED, &dummy);
HANDLE t2 = CreateThread(0, 0, CountErrors, (void*)0x2000, CREATE_SUSPENDED, &dummy);
SetThreadAffinityMask(t1, affinity1);
SetThreadAffinityMask(t2, affinity2);
ResumeThread(t1);
ResumeThread(t2);
printf("Running test for %d milliseconds with affinity %d and %d\n", RUNFOR, affinity1, affinity2);
Sleep(RUNFOR);
terminating = true;
Sleep(100); // let threads have a chance of picking up the "terminating" flag.
}
int main()
{
SetPriorityClass(GetCurrentProcess(), REALTIME_PRIORITY_CLASS);
RunTest(1, 2); // core 1 & 2
RunTest(1, 4); // core 1 & 3
RunTest(4, 8); // core 3 & 4
RunTest(1, 8); // core 1 & 4
}
On my Quad-core intel Q6600 system (which iirc has two sets of cores where each set share L2 cache - would explain the results anyway ;)), I get the following results:
Running test for 5000 milliseconds with affinity 1 and 2
Thread 00002000: 351883 errors
Thread 00001000: 343523 errors
Running test for 5000 milliseconds with affinity 1 and 4
Thread 00001000: 48073 errors
Thread 00002000: 59813 errors
Running test for 5000 milliseconds with affinity 4 and 8
Thread 00002000: 337199 errors
Thread 00001000: 335467 errors
Running test for 5000 milliseconds with affinity 1 and 8
Thread 00001000: 55736 errors
Thread 00002000: 72441 errors
来源：https://stackoverflow.com/questions/3580291/what-happens-if-two-threads-read-write-the-same-piece-of-memory