The questions stems from reading the Spectre attack paper. If I understand it correctly the attack stems from the possibility of CPU heuristics speculatively executing (the wrong) branch of code.
Consider the example (in C):
int arr[42];
if (i < 42) {
int j = arr[i];
}
If I understand the paper correctly, the int j = arr[i] can be (in certain circumstances) speculatively executed even when i >= 42. My question is - when I access array outside of its bounds my program would often crash (segmentation fault on Linux, "The program performed an illegal operation" error on Windows).
Why does speculative execution not cause programs to crash in case of array out of bound access?
The key point is that in modern CPUs the verb executing doesn't mean what you think it means.
To execute an instruction is the act of computing its output and side effects if any.
However, this doesn't change the program state.
This seems hard to grasp at first but it's really nothing exotic.
The CPU has a quite big internal memory made by all its registers, most of this memory is not visible to the programmer, the part that is it is known as the architectural state.
The architectural state (AS) is what is documented in the CPU manuals and what is altered by a sequence of instructions (a program, for example).
Since altering the AS can only happen with the semantics given in the ISA (the manuals) and the ISA specify a serial semantics (instructions are completed one after the other in the program order) this doesn't allow parallelism.
However, a modern CPU has a lot of resources (known as execution unit) that can do their work independently.
To exploit all these resources the front-end of the CPU (the part that is responsible for reading instructions from the memory hierarchy and feeding them to the execution units) is able to reach, decode and output more that one instruction per cycle.
The boundary between the front-end and the back-end (where the execution units lie) is not really dealing with instructions anymore (but with uops) but that's an x86 CISC nuisance.
So now the CPU is given 4/6 uops to "execute" at a time but if the ISA is serial, what it could possibly do other than queuing these uops?
Well, the front-end is made so that these uops don't operate on the AS but on a shadow state (SS, my terminology here), their operands are renamed, made of part of the big invisible memory of the CPU.
Altering the in parallel or out-of-order is fine as it is not the AS.
This is what execution is: altering the SS.
Does it really worth it? Afterall it is the AS that matters.
Well, transferring the SS to the AS is really fast compared to execution, so it's worth it.
It is a matter of "renaming back" (inverting the previous renaming) and it is called retiring of the instructions.
Actually, retiring is a bit more than that.
Since execution doesn't affect the AS, the side effect should also not affect it.
These side effects include exceptions but speculatively handling an exception is too cumbersome (it needs to coordinate a lot of resources) so exception handling is delayed until retirement.
This also has the advantage of having the correct AS at the moment the exception is handled and the advantage of raising an exception only when it must actually be.
The point of speculative execution is to bet, the CPU bets that the instructions sequence doesn't generate any exception (including page fault) and thus execute it with most checks off (I cannot exclude, off the top of my head, that some check is not made regardless) thereby gaining a lot of advantage.
When it's time to retire those instructions the bets are checked and if any fails, the SS is discarded.
That's why speculatively execution doesn't crash your program.
What Spectre relies on is the fact that speculatively execution does indeed alter the AS in some sense: the caches are not invalidated (again for performance reasons, the SS is simply not copied into the AS when a bet is off) and timing attacks are possible.
This could be corrected in a number of ways, including performing a basic privilege check when reading from the TLB (after all only privileges 0 and 3 are used, so the logic is simple) or adding a bit to the cache lines to mark them speculative (treated as invalid by non speculative code).
Related
What is the difference between the issue queue and lsq queue for
memory instructions? Do memory instructions pass through both queues, or do they only pass
through the lsq queue.
If they pass through both queues what is their order?
I'm assuming you use the arm-like nomenclature here so the issue queue is what Intel calls RS (reservation station) and by issue you mean sending a uop ready for execution.
The answer is that memory instructions need to pass both. All instructions need to be issued (except the ones that can be eliminated without execution, for example register moves, zero idioms, nops, etc..). Let's rephrase - all instructions that need to go through an ALU need to go through the issue process first. Memory instructions will simply use that step to calculate their addresses.
This is true for loads, for stores there is usually an internal split into store-address and store-data, so the store-address will behave like a load in that sense and calculate its address during that step.
There is usually a dedicated execution port for that and dedicated execution units because the address calculation usually follows one of few specific addressing modes (each architecture has a different set of these), but aside from that the execution needs to follow the same rules like any other operation in the CPU - it needs to have its sources ready and read from the register file or bypassed from an in flight operation, it needs to get arbitrated when the execution port is free and prioritized by the same aging rules, so it makes sense that it uses the common path.
Once the memory operation has finished execution, it will be sent to the LSU (load-store unit, or the DCU, data-cache unit on Intel) and perform the actual memory access using the generated address. The LSU pipe will take care of the address translation, TLB lookups, the page walk if needed (though this is sometimes done in a dedicated unit), the address range and property checks, the cache lookup (if cacheable) and sending a miss to the next cache level or memory if needed. It may also trigger prefetches as part of the process.
For a load, when the LSU pipe has completed (which may require multiple passes and wakeups if the data was not available in the L1), the LSU will signal the issue queue again in order to wakeup anyone who was depended on the result.
For a store, store-address may fetch the line to the cache in advance as an optimization but the actual next step is usually to wakeup after retirement (since stores may not be dispatched to memory while speculative, unless you have some tricks to handle that).
It's also worth to mention that some CPUs try to optimize loads that can forward the data directly from prior stores instead of fetching it from the cache/memory. This can include forwarding (very common) or memory renaming (less common). The former is usually handled by the LSU internally, but the latter can be done much earlier and without the LSU (though the LSU pipe is usually still activated to validate the result).
Alright, so I know that if a particular conditional branch has a condition that takes time to compute (memory access, for instance), the CPU assumes a condition result and speculatively executes along that path. However, what would happen if, along that path, yet another slow conditional branch pops up (assuming, of course, that the first condition hasn't been resolved yet and the CPU can't just commit the changes)? Does the CPU just speculate inside the speculation? What happens if the last condition is mispredicted but the first wasn't? Does it just rollback all the way?
I'm talking about something like this:
if (value_in_memory == y){
// computations
if (another_val_memory == x){
//computations
}
}
Speculative execution is the regular state of execution, not a special mode that an out of order CPU enters when it sees a branch and then leaves when the branch is no longer in flight.
This is easier to see if you consider that it's not just branches that can fault, but many instructions, including those that access memory, have restrictions on their input values, etc. So any substantial out of order execution implies constant speculation, and CPUs are built around that idea.
So "nested branches" doesn't end up being special in that sense.
Now, modern CPUs have a variety of methods for quick branch misprediction recovery, faster than recovery from other types of faults1. For example they may snapshot the state of the register mapping at some branches, to allow recovery to start before the branch is at the head of the reorder buffer. Since it is not always feasible to snapshot at all branches, there might be complicated heuristics involved to decide where to take snapshots.
I mention this last part because it is one way in which nested branches might matter: when there are lots of branches in flight, you might hit some microarchitectural limits related to the tracking of these branches for recovery purposes. For more details, you can look through patents for "branch order buffer" (for Intel techniques, but there are no doubt others).
1 The basic recovery method is keep executing until the faulting instruction is the next to retire, and then throw away all younger instructions. In the context of branch mispredictions, this means you could actually suffer two or more mispredictions only the oldest of which actually takes effect: e.g., a younger branch mispredicts, and while executing up to that branch (at which point recovery can occur), another mispredict occurs, so the younger one ends up getting discarded.
(Maybe not a complete answer, but I had some of this written when #BeeOnRope posted an answer. Posting this anyway for some more links and technical details in case anyone's curious.)
Everything is always speculative until it reaches retirement and becomes non-speculative, definitely happened, part of the architectural state.
e.g. any load might fault with a bad address, any div might trap on divide by zero. See also Out-of-order execution vs. speculative execution That and What exactly happens when a skylake CPU mispredicts a branch? mention that branch mispredicts are handled specially, because they're expected to be frequent. Fast-recovery can start before a mis-predicted branch reaches retirement, unlike the behaviour for a faulting load for example. (That's part of why Meltdown is exploitable.)
So even "regular" instructions are executed speculatively before being commited, and the only distinction between them is a human-made distinction, not computer-made? I presume, then, that the CPU stores multiple, possible rollback points? For instance if I have load instructions that may lead to page faults or simply use stale values, inside a conditional branch, the CPU identifies such instructions and scenarios and saves a state for each of them? I feel like I misunderstood because this may lead to a lot of storing register states and complicated dependencies.
The retirement state is always consistent so you can always roll back to there and discard all in-flight work, e.g. if an external interrupt arrives you want to handle it without waiting for a chain of a dozen cache miss loads to all execute. When an interrupt occurs, what happens to instructions in the pipeline?
This tracking basically happens for free or is something you need to do anyway to be able to detect which instruction faulted, not just that there was a problem somewhere. (This is called "precise exceptions")
The real distinction humans can usefully make is speculation that has a real chance of being wrong during execution of non-error cases. If your code gets a bad pointer, it doesn't really matter how it performs; it's going to page-fault and that's going to be very slow compared to local OoO exec details.
You're talking about a modern out-of-order (OoO) execution (not just fetch) CPU, like modern Intel or AMD x86, high-end ARM, MIPS r10000, etc.
The front-end is in-order (with speculation down predicted paths), and so is commit (aka retirement) from the out-of-order back-end into non-speculative retirement state. (aka known-good architectural state).
The CPU uses two major structures to track instructions (or on x86, uops = parts of instructions) in the back-end. The last stage of the front-end (after fetch / decode) allocates/renames instructions and adds them into both of these structures at once.
RS = Reservation Station = scheduler: not-yet-executed instructions, waiting for an execution unit. The RS tracks dependencies and sends the oldest-ready uops to execution units that are ready.
ROB = ReOrder Buffer: not-yet-retired instructions. Instructions enter and leave in-order so it can just be a circular buffer.
Includes a flag to mark each entry as executed or not, set once the RS has sent it to an execution unit which reports success. The oldest instructions in the ROB that all have their done-executing bit set can "retire".
Also includes a flag which indicates "fault if this reaches retirement". This avoids spending time handling page faults from load instruction on the wrong path of execution (that might well have pointers into an unmapped page), for example. Either in the shadow of a branch mispredict, or just after another instruction (in program order) that should have faulted first but OoO exec got to it later.
(I'm also leaving out register-renaming onto a large physical register file.
That's the "rename" part. Allocate includes choosing which execution port an instruction will use, and reserving a load or store buffer entry for memory instructions.)
(There's also a store-buffer; stores don't write directly to L1d cache, they write to the store buffer. This makes it possible to speculatively execute stores and still roll back without them becoming visible to other cores. It also decouples cache-miss stores from execution. Once a store instruction retires, the store-buffer entry "graduates" and is eligible to commit to L1d cache, once MESI gets exclusive access to the cache line, and once memory-ordering rules are satisfied.)
Execution units detect whether an instruction should fault, or was mis-speculated and should roll back, but don't necessarily act on that until the instruction reaches retirement.
In-order retirement is the step that recovers program-order after OoO exec, including the case of exceptions of mis-speculation.
Terminology: Intel calls it "issue" when instructions are sent from the front-end into the ROB + RS. Other computer architecture people often call that "dispatch".
Sending uops from the RS to execution units is called "dispatch" by Intel, "issue" by other people.
According to Operating System Concepts book, Resource-Allocation-Graph Algorithm can prevent deadlocks as follow:
If we have the following allocation graph
https://www.cs.uic.edu/~jbell/CourseNotes/OperatingSystems/images/Chapter7/7_07_DeadlockAvoidance.jpg
And P1 tried to allocate resource R2, the system prevents it and makes it wait, because that will lead to an unsafe state.
My question is as shown from the graph, P2 is waiting for P1 to release R1, and P1 is now waiting to allocate R2 and that leads to a deadlock. How this algorithm can prevent this type of deadlocks ?
I don't have a copy of your book, but I suspect a typo. The idea is to return an error (EDEADLOCK) to the resource allocation request that would complete the cycle; thus detecting pending deadlock rather than actively avoiding it. It is still up to the process with the failed request to take some corrective action, like dropping all its resources and trying to re-acquire them.
If you replace resources with semaphore or mutex, it should be clear that waiting isn't going to help anything.
To actively avoid deadlock, you pretty much need to either use semaphore sets -- that is acquire all the locks that a particular code path will need in one place (see system V semaphores) -- or arrange your code to use a particular ordering of locks. An example of the latter is to allocate locks by increasing address, thus all actors will attempt the allocation in the same order. Neither is practical for finely grained general purpose code, but possible for transaction processing applications.
Given the small program shown below (handcrafted to look the same from a sequential consistency / TSO perspective), and assuming it's being run by a superscalar out-of-order x86 cpu:
Load A <-- A in main memory
Load B <-- B is in L2
Store C, 123 <-- C is L1
I have a few questions:
Assuming a big enough instruction-window, will the three instructions be fetched, decoded, executed at the same time? I assume not, as that would break execution in program order.
The 2nd load is going to take longer to fetch A from memory than B. Will the later have to wait until the first is fully executed? Will the fetching of B only start after Load A is fully executed? or until when does it have to wait?
Why would the store have to wait for the loads? If yes, will the instruction just wait to be committed in the store buffer until the loads finish or after decoding it will have to sit and wait for the loads?
Thanks
Terminology: "instruction-window" normally means out-of-order execution window, over which the CPU can find ILP. i.e. ROB or RS size. See Understanding the impact of lfence on a loop with two long dependency chains, for increasing lengths
The term for how many instructions can go through the pipeline in a single cycle is pipeline width. e.g. Skylake is 4-wide superscalar out-of-order. (Parts of its pipeline, like decode, uop-cache fetch, and retirement, are wider than 4 uops, but issue/rename is the narrowest point.)
Terminology: "wait to be committed in the store buffer" store data + address gets written into the store buffer when a store executes. It commits from the store buffer to L1d at any point after retirement, when it's known to be non-speculative.
(In program order, to maintain the TSO memory model of no store reordering. A store buffer allows stores to execute inside this core out of order but still commit to L1d (and become globally visible) in-order. Executing a store = writing address + data to the store buffer.)
Can a speculatively executed CPU branch contain opcodes that access RAM?
Also what is a store buffer? and
Size of store buffers on Intel hardware? What exactly is a store buffer?
The front-end is irrelevant. 3 consecutive instructions might well be fetched in the same 16-byte fetch block, and might go through pre-decode and decode in the same cycle as a group. And (also or instead) issue into the out-of-order back-end as part of a group of 3 or 4 uops. IDK why you think any of that would cause any potential problem.
The front end (from fetch to issue/rename) processes instructions in program order. Processing simultaneously doesn't put later instructions before earlier ones, it puts them at the same time. And more importantly, it preserves the information of what program order is; that's not lost or discarded because it matters for instructions that depend on the previous one1!
There are queues between most pipeline stages, so (for example on Intel Sandybridge) instructions that pre-decode as part of a group of up-to-6 instructions might not hit the decoders as part of the same group of up-to-4 (or more with macro-fusion). See https://www.realworldtech.com/sandy-bridge/3/ for fetch, and the next page for decode. (And the uop cache.)
Executing (dispatching uops to execution ports from the out-of-order scheduler) is where ordering matters. The out-of-order scheduler has to avoid breaking single threaded code.2
Usually issue/rename is far ahead of execution, unless you're bottlenecked on the front-end. So there's normally no reason to expect that uops that issued together will execute together. (For the sake of argument, let's assume that the 2 loads you show do get dispatched for execution in the same cycle, regardless of how they got there via the front-end.)
But anyway, there's no problem here starting both loads and the store the same time. The uop scheduler doesn't know whether a load will hit or miss in L1d. It just sends 2 load uops to the load execution units in a cycle, and a store-address + store-data uop to those ports.
[load ordering]
This is the tricky part.
As I explained in an answer + comments on your last question, modern x86 CPUs will speculatively use the L2 hit result from Load B for later instructions, even though the memory model requires that this load happens after Load A.
But if no other cores write to cache line B before Load A completes, then nothing can tell the difference. The Memory-Order Buffer takes care of detecting invalidations of cache lines that were loaded from before earlier loads complete, and doing a memory-order mis-speculation pipeline flush (rollback to retirement state) in the rare case that allowing load re-ordering could change the result.
Why would the store have to wait for the loads?
It won't, unless the store-address depends on a load value. The uop scheduler will dispatch the store-address and store-data uops to execution units when their inputs are ready.
It's after the loads in program order, and the store buffer will make it even farther after the loads as far as global memory order is concerned. The store buffer won't commit the store data to L1d (making it globally visible) until after the store has retired. Since it's after the loads, they'll have also retired.
(Retirement is in-order to allow precise exceptions, and to make sure no previous instructions took an exception or were a mispredicted branch. In-order retirement allows us to say for sure that an instruction is non-speculative after it retires.)
So yes, this mechanism does ensure that the store can't commit to L1d until after both loads have taken data from memory (via L1d cache which provides a coherent view of memory to all cores). So this prevents LoadStore reordering (of earlier loads with later stores).
I'm not sure if any weakly-ordered OoO CPUs do LoadStore reordering. It is possible on in-order CPUs when a cache-miss load comes before a cache-hit store, and the CPU uses scoreboarding to avoid stalling until the load data is actually read from a register, if it still isn't ready. (LoadStore is a weird one: see also Jeff Preshing's Memory Barriers Are Like Source Control Operations). Maybe some OoO exec CPUs can also track cache-miss stores post retirement when they're known to be definitely happening, but the data just still hasn't arrived yet. x86 doesn't do this because it would violate the TSO memory model.
Footnote 1: There are some architectures (typically VLIW) where bundles of simultaneous instructions are part of the architecture in a way that's visible to software. So if software can't fill all 3 slots with instructions that can execute simultaneously, it has to fill them with NOPs. It might even be allowed to swap 2 registers with a bundle that contained mov r0, r1 and mov r1, r0, depending on whether the ISA allows instructions in the same bundle to read and write the same registers.
But x86 is not like that: superscalar out-of-order execution must always preserve the illusion of running instructions one at a time in program order. The cardinal rule of OoO exec is: don't break single-threaded code.
Anything that would violate this can only be done with checking for hazards, or speculatively with rollback upon detection of mistakes.
Footnote 2: (continued from footnote 1)
You can fetch / decode / issue two back-to-back inc eax instructions, but they can't execute in the same cycle because register renaming + the OoO scheduler has to detect that the 2nd one reads the output of the first.
I was reading cpu protection rings and system call working. But that leads me to a different question. What if I don't (i.e. a user program ) use kernel API calls (system calls), and if possible write everything in assembly and execute it. If the user program has some inconsistent code, the CPU must not execute them, or the system may crash. But at what point in time the CPU realizes that a particular instruction xyzw must not be executed? How the protection level plays the key role here? Does the underlying ISA have a predefined privilege level for each instruction?
Thank You.
If the user program has some inconsistent code, the CPU must not
execute them, or the system may crash. But at what point in time the
CPU realizes that a particular instruction xyzw must not be executed?
what does this mean?
if there is wrong stuff, say division by 0, the cpu will raise an exception while trying to execute it. this switches you to the kernel and the os decides what to do - typically kill the process. modulo cpu bugs this is what happens for all "inconsistent" instructions.
cpu raises exceptions and switches to the kernel all the time - page faults, first time fpu use and whatnot are the standard reasons.