On the one hand, Wikipedia writes about the steps of the out-of-order execution:
Instruction fetch.
Instruction dispatch to an instruction queue (also called instruction buffer or reservation stations).
The instruction waits in the queue until its input operands are available. The instruction is then allowed to leave the queue before
earlier, older instructions.
The instruction is issued to the appropriate functional unit and executed by that unit.
The results are queued.
Only after all older instructions have their results written back to the register file, then this result is written back to the register file. This is called the graduation or retire stage.
The similar information can be found in the "Computer Organization and Design" book:
To make programs behave as if they were running on a simple in-order
pipeline, the instruction fetch and decode unit is required to issue
instructions in order, which allows dependences to be tracked, and the
commit unit is required to write results to registers and memory in
program fetch order. This conservative mode is called in-order
commit... Today, all dynamically scheduled pipelines use in-order commit.
So, as far as I understand, even if the instructions execution is done in the out-of-order manner, the results of their executions are preserved in the reorder buffer and then committed to the memory/registers in a deterministic order.
On the other hand, there is a known fact that modern CPUs can reorder memory operations for the performance acceleration purposes (for example, two adjacent independent load instructions can be reordered). Wikipedia writes about it here.
Could you please shed some light on this discrepancy?
TL:DR: memory ordering is not the same thing as out of order execution. It happens even on in-order pipelined CPUs.
In-order commit is necessary1 for precise exceptions that can roll-back to exactly the instruction that faulted, without any instructions after that having already retired. The cardinal rule of out-of-order execution is don't break single-threaded code. If you allowed out-of-order commit (retirement) without any kind of other mechanism, you could have a page-fault happen while some later instructions had already executed once, and/or some earlier instructions hadn't executed yet. This would make restarting execution after handing a page-fault impossible the normal way.
(In-order issue/rename and dependency-tracking takes care of correct execution in the normal case of no exceptions.)
Memory ordering is all about what other cores see. Also notice that what you quoted is only talking about committing results to the register file, not to memory.
(Footnote 1: Kilo-instruction Processors: Overcoming the Memory Wall is a theoretical paper about checkpointing state to allow rollback to a consistent machine state at some point before an exception, allowing much larger out-of-order windows without a gigantic ROB of that size. AFAIK, no mainstream commercial designs have used that, but it shows that there are in theory approaches other than strictly in-order retirement to building a usable CPU.
Apple's M1 reportedly has a significantly larger out-of-order window than its x86 contemporaries, but I haven't seen any definite info that it uses anything other than a very large ROB.)
Since each core's private L1 cache is coherent with all the other data caches in the system, memory ordering is a question of when instructions read or write cache. This is separate from when they retire from the out-of-order core.
Loads become globally visible when they read their data from cache. This is more or less when they "execute", and definitely way before they retire (aka commit).
Stores become globally visible when their data is committed to cache. This has to wait until they're known to be non-speculative, i.e. that no exceptions or interrupts will cause a roll-back that has to "undo" the store. So a store can commit to L1 cache as early as when it retires from the out-of-order core.
But even in-order CPUs use a store queue or store buffer to hide the latency of stores that miss in L1 cache. The out-of-order machinery doesn't need to keep tracking a store once it's known that it will definitely happen, so a store insn/uop can retire even before it commits to L1 cache. The store buffer holds onto it until L1 cache is ready to accept it. i.e. when it owns the cache line (Exclusive or Modified state of the MESI cache coherency protocol), and the memory-ordering rules allow the store to become globally visible now.
See also my answer on Write Allocate / Fetch on Write Cache Policy
As I understand it, a store's data is added to the store queue when it "executes" in the out-of-order core, and that's what a store execution unit does. (Store-address writing the address, and store-data writing the data into the store-buffer entry reserved for it at allocation/rename time, so either of those parts can execute first on CPUs where those parts are scheduled separately, e.g. Intel.)
Loads have to probe the store queue so that they see recently-stored data.
For an ISA like x86, with strong ordering, the store queue has to preserve the memory-ordering semantics of the ISA. i.e. stores can't reorder with other stores, and stores can't become globally visible before earlier loads. (LoadStore reordering isn't allowed (nor is StoreStore or LoadLoad), only StoreLoad reordering).
David Kanter's article on how TSX (transactional memory) could be implemented in different ways than what Haswell does provides some insight into the Memory Order Buffer, and how it's a separate structure from the ReOrder Buffer (ROB) that tracks instruction/uop reordering. He starts by describing how things currently work, before getting into how it could be modified to track a transaction that can commit or abort as a group.
Related
I came across this code emission for x64 were "Atomic Load" is using a simple movq whereas "Atomic Store" is using xchgq.
This link explains that Atomic Load/Stores on aligned addresses are atomic by default. I'm assuming that's why Atomic Load in the above link is using a simple movq.
I have the following questions;
Is Atomic Store using a xchgq (which enables LOCK by default) to fix any issues with cache lines? essentially it's making sure all cache lines are updated properly? If cache line wasn't an issue they could have just used movq?
Does it also mean cache coherency is only an issue when Storing? As Load above is not using a locked instruction?
No, seq_cst stores use xchg (or mov + mfence but that's slower on recent CPUs) for ordering wrt. other operations. release or relaxed atomic stores can just use mov and will still be promptly visible to other cores. (Not before later loads in this thread might have executed, though.)
Cache coherence isn't the cause of memory-reordering, that's local to each core. (For x86, the memory model is program order + a store buffer with store-forwarding. It's the store buffer that causes stores to not become visible until after the store instruction has retired from out-of-order exec.)
The answer you linked which says "if I set this to true (or false), no other thread will read a different value after I've set it" (that's not quite such a certainty - you need a "lock" prefix to guarantee that). is somewhat misleading. They mean that (implicit-lock) xchg includes a full memory barrier, so no code in the storing thread can access memory until after the store is actually committed to cache, globally visible.
A clearer way to state that is that it makes this thread wait without doing anything until the store is visible. i.e. stall this thread until the store buffer has finished committing all previous stores. That would eventually happen on its own. So it's really about ordering of this thread relative to store visibility, not other threads. Other threads (cores) can locally do their own early loading / late storing, although on x86 all loads happen in program order. That's why I commented on that answer you linked to disagree with the way it was presenting things.
Can a speculatively executed CPU branch contain opcodes that access RAM? (What a store buffer does)
C++ How is release-and-acquire achieved on x86 only using MOV? discusses cache-coherency and limits on local reordering being enough to give release/acquire synchronization.
Why does a std::atomic store with sequential consistency use XCHG?
Acquire-release on x86
Which is a better write barrier on x86: lock+addl or xchgl? - shows in more detail why we need xchg or a separate memory barrier for a seq_cst store.
https://software.rajivprab.com/2018/04/29/myths-programmers-believe-about-cpu-caches/ (It's talking about Java volatile, which is like C++ std::atomic with memory_order_seq_cst.
How does memory reordering help processors and compilers?
Does atomic read guarantees reading of the latest value? - people often get hung up on "latest value" guarantees. Don't. Acquire/release just works, and stronger orders or memory barriers don't make stores visible to other cores sooner in any significant way.
I know store buffer and invalidate queues are reasons that cause memory reordering. What I don't know is if Out-of-Order-Execution can cause memory reordering.
In my opinion, Out-of-Order-Execution can't cause reordering because the results are always retired in-order as mentioned in this question.
To make my question more clear, let's say we have such an relax memory consistency architecture:
It doesn't have store buffer and invalidate queues
It can do Out-of-Order-Execution
Can memory reordering still happen in this architecture?
Does memory barrier has two functions, one is forbidding the Out-of-Order execution, the other is flushing invalidation queue and draining store buffer?
Yes, out of order execution can definitely cause memory reordering, such as load/load re-ordering
It is not so much a question of the loads being retired in order, as of when the load value is bound to the load instruction. Eg Load1 may precede Load2 in program order, Load2 gets its value from memory before Load1 does, and eg if there is an intervening store to the location read by Load2, then Load/load reordering has occurred.
However, certain systems, such as Intel P6 family systems, have additional mechanisms to detect such conditions to obtain stronger memory order models.
In these systems all loads are buffered until retirement, and if a possible store is detected to such a buffered but not yet retired load, then the load and program order instructions are “nuked”, and execution is resumed art, e.g., Load2.
I call this Freye’s Rule snooping, after I learned that Brad Freye at IBM had invented it many years before I thought I had. I believe the standard academic reference is Gharachorloo.
I.e. it is not so much buffering loads until retirement, as it is providing such a detection and correction mechanism associated with buffering loads until retirement. Many CPUs provide buffering until retirement but do not provide this detection mechanism.
Note also that this requires something like snoop based cache coherence. Many systems, including Intel systems that have such mechanisms also support noncoherent memory, e.g. memory that may be cached but which is managed by software. If speculative loads are allowed to such cacheable but non-coherent memory regions, the Freye’s Rule mechanism will not work and memory will be weakly ordered.
Note: I said “buffer until retirement”, but if you think about it you can easily come up with ways of buffering not quite until retirement. E.g. you can stop this snooping when all earlier loads have them selves been bound, and there is no longer any possibility of an intervening store being observed even transitively.
This can be important, because there is quite a lot of performance to be gained by “early retirement“, removing instructions such as loads from buffering and repair mechanisms before all earlier instructions have retired. Early retirement can greatly reduce the cost of out of order hardware mechanisms.
The Intel optimization manual talks about the number of store buffers that exist in many parts of the processor, but do not seem to talk about the size of the store buffers. Is this public information or is the size of a store buffer kept as a microarchitectural detail?
The processors I am looking into are primarily Broadwell and Skylake, but information about others would be nice as well.
Also, what do store buffers do, exactly?
Related: what is a store buffer? and a beginner-friendly (but detailed) intro to the concept of buffers in Can a speculatively executed CPU branch contain opcodes that access RAM? which I highly recommend reading for CPU-architecture background on why we need them and what they do (decouple execution from commit to L1d / cache misses, and allow speculative exec of stores without making speculation visible in coherent cache.)
Also How do the store buffer and Line Fill Buffer interact with each other? has a good description of the steps in executing a store instruction and how it eventually commits to L1d cache.
The store buffer as a whole is composed of multiple entries.
Each core has its own store buffer1 to decouple execution and retirement from commit into L1d cache. Even an in-order CPU benefits from a store buffer to avoid stalling on cache-miss stores, because unlike loads they just have to become visible eventually. (No practical CPUs use a sequential-consistency memory model, so at least StoreLoad reordering is allowed, even in x86 and SPARC-TSO).
For speculative / out-of-order CPUs, it also makes it possible roll back a store after detecting an exception or other mis-speculation in an older instruction, without speculative stores ever being globally visible. This is obviously essential for correctness! (You can't roll back other cores, so you can't let them see your store data until it's known to be non-speculative.)
When both logical cores are active (hyperthreading), Intel partitions the store buffer in two; each logical core gets half. Loads from one logical core only snoop its own half of the store buffer2. What will be used for data exchange between threads are executing on one Core with HT?
The store buffer commits data from retired store instructions into L1d as fast as it can, in program order (to respect x86's strongly-ordered memory model3). Requiring stores to commit as they retire would unnecessarily stall retirement for cache-miss stores. Retired stores still in the store buffer are definitely going to happen and can't be rolled back, so they can actually hurt interrupt latency. (Interrupts aren't technically required to be serializing, but any stores done by an IRQ handler can't become visible until after existing pending stores are drained. And iret is serializing, so even in the best case the store buffer drains before returning.)
It's a common(?) misconception that it has to be explicitly flushed for data to become visible to other threads. Memory barriers don't cause the store buffer to be flushed, full barriers make the current core wait until the store buffer drains itself, before allowing any later loads to happen (i.e. read L1d). Atomic RMW operations have to wait for the store buffer to drain before they can lock a cache line and do both their load and store to that line without allowing it to leave MESI Modified state, thus stopping any other agent in the system from observing it during the atomic operation.
To implement x86's strongly ordered memory model while still microarchitecturally allowing early / out-of-order loads (and later checking if the data is still valid when the load is architecturally allowed to happen), load buffer + store buffer entries collectively form the Memory Order Buffer (MOB). (If a cache line isn't still present when the load was allowed to happen, that's a memory-order mis-speculation.) This structure is presumably where mfence and locked instructions can put a barrier that blocks StoreLoad reordering without blocking out-of-order execution. (Although mfence on Skylake does block OoO exec of independent ALU instructions, as an implementation detail.)
movnt cache-bypassing stores (like movntps) also go through the store buffer, so they can be treated as speculative just like everything else in an OoO exec CPU. But they commit directly to an LFB (Line Fill Buffer), aka write-combining buffer, instead of to L1d cache.
Store instructions on Intel CPUs decode to store-address and store-data uops (micro-fused into one fused-domain uop). The store-address uop just writes the address (and probably the store width) into the store buffer, so later loads can set up store->load forwarding or detect that they don't overlap. The store-data uop writes the data.
Store-address and store-data can execute in either order, whichever is ready first: the allocate/rename stage that writes uops from the front-end into the ROB and RS in the back end also allocates a load or store buffer for load or store uops at issue time. Or stalls until one is available. Since allocation and commit happen in-order, that probably means older/younger is easy to keep track of because it can just be a circular buffer that doesn't have to worry about old long-lived entries still being in use after wrapping around. (Unless cache-bypassing / weakly-ordered NT stores can do that? They can commit to an LFB (Line Fill Buffer) out of order. Unlike normal stores, they commit directly to an LFB for transfer off-core, rather than to L1d.)
but what is the size of an entry?
Store buffer sizes are measured in entries, not bits.
Narrow stores don't "use less space" in the store buffer, they still use exactly 1 entry.
Skylake's store buffer has 56 entries (wikichip), up from 42 in Haswell/Broadwell, and 36 in SnB/IvB (David Kanter's HSW writeup on RealWorldTech has diagrams). You can find numbers for most earlier x86 uarches in Kanter's writeups on RWT, or Wikichip's diagrams, or various other sources.
SKL/BDW/HSW also have 72 load buffer entries, SnB/IvB have 64. This is the number of in-flight load instructions that either haven't executed or are waiting for data to arrive from outer caches.
The size in bits of each entry is an implementation detail that has zero impact on how you optimize software. Similarly, we don't know the size in bits of of a uop (in the front-end, in the ROB, in the RS), or TLB implementation details, or many other things, but we do know how many ROB and RS entries there are, and how many TLB entries of different types there are in various uarches.
Intel doesn't publish circuit diagrams for their CPU designs and (AFAIK) these sizes aren't generally known, so we can't even satisfy our curiosity about design details / tradeoffs.
Write coalescing in the store buffer:
Back-to-back narrow stores to the same cache line can (probably?) be combined aka coalesced in the store buffer before they commit, so it might only take one cycle on a write port of L1d cache to commit multiple stores.
We know for sure that some non-x86 CPUs do this, and we have some evidence / reason to suspect that Intel CPUs might do this. But if it happens, it's limited. #BeeOnRope and I currently think Intel CPUs probably don't do any significant merging. And if they do, the most plausible case is that entries at the end of the store buffer (ready to commit to L1d) that all go to the same cache line might merge into one buffer, optimizing commit if we're waiting for an RFO for that cache line. See discussion in comments on Are two store buffer entries needed for split line/page stores on recent Intel?. I proposed some possible experiments but haven't done them.
Earlier stuff about possible store-buffer merging:
See discussion starting with this comment: Are write-combining buffers used for normal writes to WB memory regions on Intel?
And also Unexpectedly poor and weirdly bimodal performance for store loop on Intel Skylake may be relevant.
We know for sure that some weakly-ordered ISAs like Alpha 21264 did store coalescing in their store buffer, because the manual documents it, along with its limitations on what it can commit and/or read to/from L1d per cycle. Also PowerPC RS64-II and RS64-III, with less detail, in docs linked from a comment here: Are there any modern CPUs where a cached byte store is actually slower than a word store?
People have published papers on how to do (more aggressive?) store coalescing in TSO memory models (like x86), e.g. Non-Speculative Store Coalescing in Total Store Order
Coalescing could allow a store-buffer entry to be freed before its data commits to L1d (presumably only after retirement), if its data is copied to a store to the same line. This could only happen if no stores to other lines separate them, or else it would cause stores to commit (become globally visible) out of program order, violating the memory model. But we think this can happen for any 2 stores to the same line, even the first and last byte.
A problem with this idea is that SB entry allocation is probably a ring buffer, like the ROB. Releasing entries out of order would mean hardware would need to scan every entry to find a free one, and then if they're reallocated out of order then they're not in program order for later stores. That could make allocation and store-forwarding much harder so it's probably not plausible.
As discussed in
Are two store buffer entries needed for split line/page stores on recent Intel?, it would make sense for an SB entry to hold all of one store even if it spans a cache-line boundary. Cache line boundaries become relevant when committing to L1d cache on leaving the SB. We know that store-forwarding can work for stores that split across a cache line. That seems unlikely if they were split into multiple SB entries in the store ports.
Terminology: I've been using "coalescing" to talk about merging in the store buffer, vs. "write combining" to talk about NT stores that combine in an LFB before (hopefully) doing a full-line write with no RFO. Or stores to WC memory regions which do the same thing.
This distinction / convention is just something I made up. According to discussion in comments, this might not be standard computer architecture terminology.
Intel's manuals (especially the optimization manual) are written over many years by different authors, and also aren't consistent in their terminology. Take most parts of the optimization manual with a grain of salt especially if it talks about Pentium4. The new sections about Sandybridge and Haswell are reliable, but older parts might have stale advice that's only / mostly relevant for P4 (e.g. inc vs. add 1), or the microarchitectural explanations for some optimization rules might be confusing / wrong. Especially section 3.6.10 Write Combining. The first bullet point about using LFBs to combine stores while waiting for lines to arrive for cache-miss stores to WB memory just doesn't seem plausible, because of memory-ordering rules. See discussion between me and BeeOnRope linked above, and in comments here.
Footnote 1:
A write-combining cache to buffer write-back (or write-through) from inner caches would have a different name. e.g. Bulldozer-family uses 16k write-through L1d caches, with a small 4k write-back buffer. (See Why do L1 and L2 Cache waste space saving the same data? for details and links to even more details. See Cache size estimation on your system? for a rewrite-an-array microbenchmark that slows down beyond 4k on a Bulldozer-family CPU.)
Footnote 2: Some POWER CPUs let other SMT threads snoop retired stores in the store buffer: this can cause different threads to disagree about the global order of stores from other threads. Will two atomic writes to different locations in different threads always be seen in the same order by other threads?
Footnote 3: non-x86 CPUs with weak memory models can commit retired stores in any order, allowing more aggressive coalescing of multiple stores to the same line, and making a cache-miss store not stall commit of other stores.
If new CPUs had a cache buffer which was only committed to the actual CPU cache if the instructions are ever committed would attacks similar to Meltdown still be possible?
The proposal is to make speculative execution be able to load from memory, but not write to the CPU caches until they are actually committed.
TL:DR: yes I think it would solve Spectre (and Meltdown) in their current form (using a flush+read cache-timing side channel to copy the secret data from a physical register), but probably be too expensive (in power cost, and maybe also performance) to be a likely implementation.
But with hyperthreading (or more generally any SMT), there's also an ALU / port-pressure side-channel if you can get mis-speculation to run data-dependent ALU instructions with the secret data, instead of using it as an array index. The Meltdown paper discusses this possibility before focusing on the flush+reload cache-timing side-channel. (It's more viable for Meltdown than Spectre, because you have much better control of the timing of when the the secret data is used).
So modifying cache behaviour doesn't block the attacks. It would take away the reliable side-channel for getting the secret data into the attacking process, though. (i.e. ALU timing has higher noise and thus lower bandwidth to get the same reliability; Shannon's noisy channel theorem), and you have to make sure your code runs on the same physical core as the code under attack.
On CPUs without SMT (e.g. Intel's desktop i5 chips), the ALU timing side-channel is very hard to use with Spectre, because you can't directly use perf counters on code you don't have privilege for. (But Meltdown could still be exploited by timing your own ALU instructions with Linux perf, for example).
Meltdown specifically is much easier to defend against, microarchitecturally, with simpler and cheaper changes to the hard-wired parts of the CPU that microcode updates can't rewire.
You don't need to block speculative loads from affecting cache; the change could be as simple as letting speculative execution continue after a TLB-hit load that will fault if it reaches retirement, but with the value used by speculative execution of later instructions forced to 0 because of the failed permission check against the TLB entry.
So the mis-speculated (after the faulting load of secret) touch array[secret*4096] load would always make the same cache line hot, with no secret-data-dependent behaviour. The secret data itself would enter cache, but not a physical register. (And this stops ALU / port-pressure side-channels, too.)
Stopping the faulting load from even bringing the "secret" line into cache in the first place could make it harder to tell the difference between a kernel mapping and an unmapped page, which could possibly help protect against user-space trying to defeat KASLR by finding which virtual addresses the kernel has mapped. But that's not Meltdown.
Spectre
Spectre is the hard one because the mis-speculated instructions that make data-dependent modifications to microarchitectural state do have permission to read the secret data. Yes, a "load queue" that works similarly to the store queue could do the trick, but implementing it efficiently could be expensive. (Especially given the cache coherency problem that I didn't think of when I wrote this first section.)
(There are other ways of implementing the your basic idea; maybe there's even a way that's viable. But extra bits on L1D lines to track their status has downsides and isn't obviously easier.)
The store queue tracks stores from execution until they commit to L1D cache. (Stores can't commit to L1D until after they retire, because that's the point at which they're known to be non-speculative, and thus can be made globally visible to other cores).
A load queue would have to store whole incoming cache lines, not just the bytes that were loaded. (But note that Skylake-X can do 64-byte ZMM stores, so its store-buffer entries do have to be the size of a cache line. But if they can borrow space from each other or something, then there might not be 64 * entries bytes of storage available, i.e. maybe only the full number of entries is usable with scalar or narrow-vector stores. I've never read anything about a limitation like this, so I don't think there is one, but it's plausible)
A more serious problem is that Intel's current L1D design has 2 read ports + 1 write port. (And maybe another port for writing lines that arrive from L2 in parallel with committing a store? There was some discussion about that on Unexpectedly poor and weirdly bimodal performance for store loop on Intel Skylake.)
If your loaded data can't enter L1D until after the loads retire, then they're probably going to be competing for the same write port that stores use.
Loads that hit in L1D can still come directly from L1D, though, and loads that hit in the memory-order-buffer could still be executed at 2 per clock. (The MOB would now include this new load queue as well as the usual store queue + markers for loads to maintain x86 memory ordering semantics). You still need both L1D read ports to maintain performance for code that doesn't touch a lot of new memory, and mostly is reloading stuff that's been hot in L1D for a while.
This would make the MOB about twice as large (in terms of data storage), although it doesn't need any more entries. As I understand it, the MOB in current Intel CPUs is composed of the individual load-buffer and store-buffer entries. (Haswell has 72 and 42 respectively).
Hmm, a further complication is that the load data in the MOB has to maintain cache coherency with other cores. This is very different from store data, which is private and hasn't become globally visible / isn't part of the global memory order and cache coherency until it commits to L1D.
So this proposed "load queue" implementation mechanism for your idea is probably not feasible without tweaks: it would have to be checked by invalidation-requests from other cores, so that's another read-port needed in the MOB.
Any possible implementation would have the problem of needing to later commit to L1D like a store. I think it would be a significant burden not to be able to evict + allocate a new line when it arrived from off-core.
(Even allowing speculative eviction but not speculative replacement from conflicts leaves open a possible cache-timing attack. You'd prime all the lines and then do a load that would evict one from one set of lines or another, and find which line was evicted instead of which one was fetched using a similar cache-timing side channel. So using extra bits in L1D to find / evict lines loaded during recovery from mis-speculation wouldn't eliminate this side-channel.)
Footnote: all instructions are speculative. This question is worded well, but I think many people reading about OoO exec and thinking about Meltdown / Spectre fall into this trap of confusing speculative execution with mis-speculation.
Remember that all instructions are speculative when they're executed. It's not known to be correct speculation until retirement. Meltdown / Spectre depend on accessing secret data and using it during mis-speculation. But the basis of current OoO CPU designs is that you don't know whether you've speculated correctly or not; everything is speculative until retirement.
Any load or store could potentially fault, and so can some ALU instructions (e.g. floating point if exceptions are unmasked), so any performance cost that applies "only when executing speculatively" actually applies all the time. This is why stores can't commit from the store queue into L1D until after the store uops have retired from the out-of-order CPU core (with the store data in the store queue).
However, I think conditional and indirect branches are treated specially, because they're expected to mis-speculate some of the time, and optimizing recovery for them is important. Modern CPUs do better with branches than just rolling back to the current retirement state when a mispredict is detected, I think using a checkpoint buffer of some sort. So out-of-order execution for instructions before the branch can continue during recovery.
But loop and other branches are very common, so most code executes "speculatively" in this sense, too, with at least one branch-rollback checkpoint not yet verified as correct speculation. Most of the time it's correct speculation, so no rollback happens.
Recovery for mis-speculation of memory ordering or faulting loads is a full pipeline-nuke, rolling back to the retirement architectural state. So I think only branches consume the branch checkpoint microarchitectural resources.
Anyway, all of this is what makes Spectre so insidious: the CPU can't tell the difference between mis-speculation and correct speculation until after the fact. If it knew it was mis-speculating, it would initiate rollback instead of executing useless instructions / uops. Indirect branches are not rare, either (in user-space); every DLL or shared library function call uses one in normal executables on Windows and Linux.
I suspect the overhead from buffering and committing the buffer would render the specEx/caching useless?
This is purely speculative (no pun intended) - I would love to see someone with a lower level background weigh in this!
Norvig claims, that an mutex lock or unlock operation takes only a quarter of the time that is needed to do a fetch from memory.
This answer explains, that a mutex is
essentially a flag and a wait queue and that it would only take a few instructions to flip the flag on an uncontended mutex.
I assume, if a different CPU or core tries to lock that mutex, it needs to wait for
the cache line to be written back into the memory (if that didn't already happen) and its own memory read to get the state of the flag. Is that correct? What is the difference, if it is a different core compared to a different CPU?
So the numbers Norvig states are only for an uncontended mutex where the CPU or core trying the operation already has that flag in its cache and the cache line isn't dirty?
A typical PC runs a x86 CPU, Intel's CPUs can perform the locking entirely on the caches:
if the area of memory being locked during a LOCK operation is
cached in the processor that is performing the LOCK operation as write-back memory and is completely contained
in a cache line, the processor may not assert the LOCK# signal on the bus.
Instead, it will modify the memory location internally and allow it’s cache coherency mechanism to ensure that the operation is carried out atomically.
This
operation is called “cache locking.”
The cache coherency mechanism automatically prevents two or more processors that have cached the same area of memory from simultaneously modifying data in that area.
From Intel Software Developer Manual 3, Section 8.1.4
The cache coherence mechanism is a variation of the MESI protocol.
In such protocol before a CPU can write to a cached location, it must have the corresponding line in the Exclusive (E) state.
This means that only one CPU at a time has a given memory location in a dirty state.
When other CPUs want to read the same location, the owner CPU will delay such reads until the atomic operation is finished.
It then follows the coherence protocol to either forward, invalidate or write-back the line.
In the above scenario a lock can be performed faster than an uncached load.
Those times however are a bit off and surely outdated.
They are intended to give an order, along with an order of magnitude, among the typical operations.
The timing for an L1 hit is a bit odd, it isn't faster than the typical instruction execution (which by itself cannot be described with a single number).
The Intel optimization manual reports, for an old CPU like Sandy Bridge, an L1 access time of 4 cycles while there are a lot of instructions with a latency of 4 cycles of less.
I would take those numbers with a grain of salt, avoiding reasoning too much on them.
The lesson Norvig tried to teach us is: hardware is layered, the closer (from a topological point of view1) to the CPU, the faster.
So when parsing a file, a programmer should avoid moving data back and forth to a file, instead it should minimize the IO pressure.
The some applies when processing an array, locality will improve performance.
Note however that these are technically, micro-optimisations and the topic is not as simple as it appears.
1 In general divide the hardware in what is: inside the core (registers), inside the CPU (caches, possibly not the LLC), inside the socket (GPU, LLC), behind dedicated bus devices (memory, other CPUs), behind one generic bus (PCIe - internal devices like network cards), behind two or more buses (USB devices, disks) and in another computer entirely (servers).