问题
We know that two instructions can be reordered by an OoOE processor. For example, there are two global variables shared among different threads.
int data;
bool ready;
A writer thread produce data and turn on a flag ready to allow readers to consume that data.
data = 6;
ready = true;
Now, on an OoOE processor, these two instructions can be reordered (instruction fetch, execution). But what about the final commit/write-back of the results? i.e., will the store be in-order?
From what I learned, this totally depends on a processor's memory model. E.g., x86/64 has a strong memory model, and reorder of stores is disallowed. On the contrary, ARM typically has a weak model where store reordering can happen (along with several other reorderings).
Also, the gut feeling tells me that I am right because otherwise we won't need a store barrier between those two instructions as used in typical multi-threaded programs.
But, here is what our wikipedia says:
.. In the outline above, the OoOE processor avoids the stall that occurs in step (2) of the in-order processor when the instruction is not completely ready to be processed due to missing data.
OoOE processors fill these "slots" in time with other instructions that are ready, then re-order the results at the end to make it appear that the instructions were processed as normal.
I'm confused. Is it saying that the results have to be written back in-order? Really, in an OoOE processor, can store to data and ready be reordered?
回答1:
The consistency model (or memory model) for the architecture determines what memory operations can be reordered. The idea is always to achieve the best performance from the code, while preserving the semantics expected by the programmer. That is the point from wikipedia, the memory operations appear in order to the programmer, even though they may have been reordered. Reordering is generally safe when the code is single-threaded, as the processor can easily detect potential violations.
On x86, the common model is that writes are not reordered with other writes. Yet, the processor is using out of order execution (OoOE), so instructions are being reordered constantly. Generally, the processor has several additional hardware structures to support OoOE, like a reorder buffer and load-store queue. The reorder buffer ensures that all instructions appear to execute in order, such that interrupts and exceptions break a specific point in the program. The load-store queue functions similarly, in that it can restore the order of memory operations according to the memory model. The load-store queue also disambiguates addresses, so that the processor can identify when the operations are made to the same or different addresses.
Back to OoOE, the processor is executing 10s to 100s of instructions in every cycle. Loads and stores are computing their addresses, etc. The processor may prefetch the cache lines for the accesses (which may include cache coherence), but it cannot actually access the line either to read or write until it is safe (according to the memory model) to do so.
Inserting store barriers, memory fences, etc tell both the compiler and processor about further restrictions to reordering the memory operations. The compiler is part of implementing the memory model, as some languages like java have specific memory model, while others like C obey the "memory accesses should appear as if they were executed in order".
In conclusion, yes, data and ready can be reordered in an OoOE. But it depends on the memory model as to whether they actually are. So if you need a specific order, provide the appropriate indication using barriers, etc such that the compiler, processor, etc will not choose a different order for higher performance.
回答2:
The simple answer is YES on some processor types.
Before the CPU, your code faces an earlier problem, compiler reordering.
data = 6;
ready = true;
The compiler is free to rearrange these statements since, as far as it knows, they do not affect each other (it is not thread-aware).
Now down to the processor level:
1) An out-of-order processor can process these instructions in different order, including reversing the order of the stores.
2) Even if the CPU performs them in order, they memory controller may not perform them in order because it may need to flush or bring in new cache lines or do an address translation before it can write them.
3) Even if this doesn't happen, another CPU in the system may not see them in the same order. In order to observe them, it may need to bring in the modified cache lines from the core that wrote them. It may not be able to bring one cache line in earlier than another if it is held be another core or if there is contention for that line by multiple cores, and its own out of order execution will read one before the other.
4) Finally, speculative execution on other cores may read the value of data before ready was set by the writing core, and by the time it gets around to reading ready, it was already set but data was also modified.
These problems are all solved by memory barriers. Platforms with weakly-ordered memory must make use of memory barriers to ensure memory coherence for thread synchronization.
回答3:
On modern processor, the storing action itself is async (think of it like submit a change to the L1 cache and continue execution, the cache system further propagate in async manner). So the changes on two object lies on different cache block may be realised OoO from other CPU's perspective.
Furthermore, even the instruction to store those data, can be executed OoO. For example when two object is stored "at the same time", but the bus line of one object is retained/locked by other CPU or bus mastering, thus other other object may be committed earlier.
Therefore, to properly share data across threads, you need some kind of memory barrier or make use of transactional memory feature found in latest CPU like TSX.
回答4:
I think you're misinterpreting "appear that the instructions were processed as normal." What that means is that if I have:
add r1 + 7 -> r2
move r3 -> r1
and the order of those is effectively reversed by out-of-order execution, the value that participates in the add operation will still be the value of r1 that was present prior to the move. Etc. The CPU will cache register values and/or delay register stores to assure that the "meaning" of a sequential instruction stream is not changed.
This says nothing about the order of stores as visible from another processor.
来源：https://stackoverflow.com/questions/25329348/can-memory-store-be-reordered-really-in-an-oooe-processor