I've watched Mike Acton's talks about DOD a few times now to better understand it (it is not an easy subject to me). I'm referring to CppCon 2014: Mike Acton "Data-Oriented Design and C++"
and GDC 2015: How to Write Code the Compiler Can Actually Optimize.
But in both talks he presents some calculations that I'm confused with:
This shows that FooUpdateIn takes 12 bytes, but if you stack 32 of them you will get 6 fully packed cache lines. Same goes for FooUpdateOut, it takes 4 bytes and 32 of them gives you 2 fully packed cache lines.
In the UpdateFoos function, you can do ~5.33 loops per each cache line (assuming that count is indeed 32), then he proceeds by assuming that all the math done takes about 40 cycles which means that each cache line would take about 213.33 cycles.
Now here's where I'm confused, isn't he forgetting about reads and writes? Even though he has 2 fully packed data structures they are in different memory spaces.
In my head this is what's happening:
Read in[0].m_Velocity[0] (which would take about 200 cycles based on his previous slides)
Since in[0].m_Velocity[1] and in[0].m_Foo are in the same cache line as in[0].m_Velocity[0] their access is free
Do all the calculation
Write the result to out[0].m_Foo - Here is what I don't know what happens, I assume that it would discard the previous cache line (fetched in 1.) and load the new one to write the result
Read in[1].m_Velocity[0] which would discard again another cache line (fetched in 4.) (which would take again about 200 cycles)
...
So jumping from in and out the calculations goes from ~5.33 loops/cache line to 0.5 loops/cache line which would do 20 cycles per cache line.
Could someone explain why wasn't he concerned about reads/writes? Or what is wrong in my thinking?
Thank you.
If we assume L1 cache is 64KB and one cache line is 64 bytes then there are total 1000 cache lines. So, in step 4 write to the result out[0].m_Foo will not discard the data cache in step 2 as they both are in different memory locations. This is the reason why he is using separate structure for updating out m_Foo instead directly mutating it in inplace like in his first implementation. He is just talking till point of calculation value. Updating value/writing value will have same cost as in his first implementation. Also, processor can optimize loops quite well as it can do multiple calculations in parallel(not sequential as result of first loop and second loop are not dependent). I hope this helps
what is the best practice to access a changing 32bit register (like a counter) through a 16bit databus ?
I suppose i have to 'freeze' or copy the 32bit value on a read of the LSB until the MSB is also read and vise versa on a write to avoid data corruption if the LSB overflows to the MSB between the 2 accesses.
Is there a standard approach to this ?
As suggested in both the question and Morten's answer, a second register to hold the value at the time of the read of the first half is a common method. In some MCUs this register is common to multiple devices, meaning you need to either disable interrupts across the two accesses or ensure ISRs don't touch the extra register. Writes are similarly handled, frequently in the opposite order (write second word temporary storage, then write first word on device thus triggering the device to read the second word simultaneously).
There have also been cases where you just can't access the register atomically. In such cases, you might need to implement additional logic to figure out the true value. An example of such an algorithm assuming three reads take much less than 1<<15 counter ticks might be:
earlyMSB = highreg;
midLSB = lowreg;
lateMSB = highreg;
fullword = ((midLSB<0x8000 ? lateMSB : earlyMSB)<<16) | midLSB;
Other variants might use an overflow flag to signal the more significant word needs an increment (frequently used to implement that part of the counter in software).
There is no standard way, but an often used approach is to make read one address return the first 16 bits, while the remaining 16 bits are captured at the same time, and read later at another address.
Suppose I have a program that has an instruction to add two numbers and that operation takes 10 nanoseconds(constant, as enforced by the gate manufactures).
Now I have 3 different processors A, B and C(where A< B < C in terms of clock cycles). A's one clock cycle has 15 nanosec, B has 10 nanosec and C has 7 nanosec.
Firstly am I correct on my following assumptions-
1. Add operation takes 1 complete cycle of processor A(slow processor) and wastes rest of 5 ns of the cycle.
2. Add operation takes 1 complete cycle of processor B wasting no time.
3. Add operation takes 2 complete cycles(20 ns) of processor C(fast processor) wasting rest of the 20-14=7 ns.
If the above assumptions are correct then isn't this a contradiction to the regular assumption that processors with high clock cycles are faster. Here processor C which is the fastest actually takes 2 cycles and wastes 7ns whereas, the slower processor A takes just 1 cycle.
Processor C is fastest, no matter what. It takes 7 ns per cycle and therefore performs more cycles than A and B. It's not C's fault that the circuit is not fast enough. If you would implement the addition circuit in a way that it gives result in 1 ns, all processors will give the answer in 1 clock cycle (i.e. C will give you the answer in 7ns, B in 10ns and A in 15ns).
Firstly am I correct on my following assumptions-
1. Add operation takes 1 complete cycle of processor A(slow processor) and wastes rest of 5 ns of the cycle.
2. Add operation takes 1 complete cycle of processor B wasting no time.
3. Add operation takes 2 complete cycles(20 ns) of processor C(fast processor) wasting rest of the 20-7=13 ns.
No. It is because you are using incomplete data to express the time for an operation. Measure the time taken to finish an operation on a particular processor in clock cycles instead of nanoseconds as you are doing here. When you say ADD op takes 10 ns and you do not mention the processor on which you measured the time for the ADD op, the time measurement in ns is meaningless.
So when you say that ADD op takes 2 clock cycles on all three processors, then you have standardized the measurement. A standardized measurement can then be translated as:
Time taken by A for addition = 2 clock cycles * 15 ns per cycle = 30 ns
Time taken by B for addition = 2 clock cycles * 10 ns per cycle = 20 ns
Time taken by C for addition = 2 clock cycles * 07 ns per cycle = 14 ns
In case you haven't noticed, when you say:
A's one clock cycle has 15 nanosec, B has 10 nanosec and C has 7 nanosec.
which of the three processors is fastest?
Answer: C is fastest. It's one cycle is finished in 7ns. It implies that it finishes 109/7 (~= 1.4 * 108) cycles in one second, compared to B which finishes 109/10 (= 108) cycles in one second, compared to A which finishes only 109/15 (~= 0.6 * 108) cycles in one second.
What does a ADD instruction mean, does it purely mean only and only ADD(with operands available at the registers) or does it mean getting
the operands, decoding the instruction and then actually adding the
numbers.
Getting the operands is done by MOV op. If you are trying to compare how fast ADD op is being done, it should be compared by time to perform ADD op only. If you, on the other hand want to find out how fast addition of two numbers is being done, then it will involve more operations than simple ADD. However, if it's helpful, the list of all Original 8086/8088 instructions is available on Wikipedia too.
Based on the above context to what add actually means, how many cycles does add take, one or more than one.
It will depend on the processor because each processor may have the adder differently implemented. There are many ways to generate addition of two numbers. Quoting Wikipedia again - A full adder can be implemented in many different ways such as with a custom transistor-level circuit or composed of other gates.
Also, there may be pipelining in the instructions which can result in parallelizing of the addition of the numbers resulting in huge time savings.
Why is clock cycle a standard since it can vary with processor to processor. Shouldn't nanosec be the standard. Atleast its fixed.
Clock cycle along with the processor speed can be the standard if you want to tell the time taken by a processor to execute an instruction. Pick any two from:
Time to execute an instruction,
Processor Speed, and
Clock cycles needed for an instruction.
The third can be derived from it.
When you say the clock cycles taken by ADD is x and you know the processor speed is y MHz, you can calculate that the time to ADD is x / y. Also, you can mention the time to perform ADD as z ns and you know the processor speed is same y MHz as earlier, you can calculate the cycles needed to execute ADD as y * z.
I'm no expert BUT I'd say ...
the regular assumption that processors with high clock cycles are faster FOR THE VAST MAJORITY OF OPERATIONS
For example, a more intelligent processor might perform an "overhead task" that takes X ns. The "overhead task" might make it faster for repetitive operations but might actually cause it to take longer for a one-off operation such as adding 2 numbers.
Now, if the same processor performed that same operation 1 million times, it should be massively faster than the slower less intelligent processor.
Hope my thinking helps. Your feedback on my thoughts welcome.
Why would a faster processor take more cycles to do the same operation than a slower one?
Even more important: modern processors use Instruction pipelining, thus executing multiple operations in one clock cycle.
Also, I don't understand what you mean by 'wasting 5ns', the frequency determines the clock speed, thus the time it takes to execute 1 clock. Of course, cpu's can have to wait on I/O for example, but that holds for all cpu's.
Another important aspect of modern cpu's are the L1, L2 and L3 caches and the architecture of those caches in multicore systems. For example: if a register access takes 1 time unit, a L1 cache access will take around 2 while a normal memory access will take between 50 and 100 (and a harddisk access would take thousands..).
This is actually almost correct, except that on processor B taking 2 cycles means 14ns, so with 10ns being enough the next cycle starts 4ns after the result was already "stable" (though it is likely that you need some extra time if you chop it up, to latch the partial result). It's not that much of a contradiction, setting your frequency "too high" can require trade-offs like that. An other thing you might do it use more a different circuit or domino logic to get the actual latency of addition down to one cycle again. More likely, you wouldn't set addition at 2 cycles to begin with. It doesn't work out so well in this case, at least not for addition. You could do it, and yes, basically you will have to "round up" the time a circuit takes to an integer number of cycles. You can also see this in bitwise operations, which take less time than addition but nevertheless take a whole cycle. On machine C you could probably still fit bitwise operations in a single cycle, for some workloads it might even be worth splitting addition like that.
FWIW, Netburst (Pentium 4) had staggered adders, which computed the lower half in one "half-cycle" and the upper half in the next (and the flags in the third half cycle, in some sense giving the whole addition a latency of 1.5). It's not completely out of this world, though Netburst was over all, fairly mad - it had to do a lot of weird things to get the frequency up that high. But those half-cycles aren't very half (it wasn't, AFAIK, logic that advanced on every flank, it just used a clock multiplier), you could also see them as the real cycles that are just very fast, with most of the rest of the logic (except that crazy ALU) running at half speed.
Your broad point that 'a CPU will occasionally waste clock cycles' is valid. But overall in the real world, part of what makes a good CPU a good CPU is how it alleviates this problem.
Modern CPUs consist of a number of different components, none of whose operations will end up taking a constant time in practice. For example, an ADD instruction might 'burst' at 1 instruction per clock cycle if the data is immediately available to it... which in turn means something like 'if the CPU subcomponents required to fetch that data were immediately available prior to the instruction'. So depending on if e.g. another subcomponent had to wait for a cache fetch, the ADD may in practice take 2 or 3 cycles, say. A good CPU will attempt to re-order the incoming stream of instructions to maximise the availability of subcomponents at the right time.
So you could well have the situation where a particular series of instructions is 'suboptimal' on one processor compared to another. And the overall performance of a processor is certainly not just about raw clock speed: it is as much about the clever logic that goes around taking a stream of incoming instructions and working out which parts of which instructions to fire off to which subcomponents of the chip when.
But... I would posit that any modern chip contains such logic. Both a 2GHz and a 3GHz processor will regularly "waste" clock cycles because (to put it simply) a "fast" instruction executed on one subcomponent of the CPU has to wait for the result of the output from another "slower" subcomponent. But overall, you will still expect the 3GHz processor to "execute real code faster".
First, if the 10ns time to perform the addition does not include the pipeline overhead (clock skew and latch delay), then Processor B cannot complete an addition (with these overheads) in one 10ns clock cycle, but Processor A can and Processor C can still probably do it in two cycles.
Second, if the addition itself is pipelined (or other functional units are available), then a subsequent non-dependent operation can begin executing in the next cycle. (If the addition was width-pipelined/staggered (as mentioned in harold's answer) then even dependent additions, logical operations and left shifts could be started after only one cycle. However, if the exercise is constraining addition timing, it presumably also prohibits other optimizations to simplify the exercise.) If dependent operations are not especially common, then the faster clock of Processor C will result in higher performance. (E.g., if a dependence stall occurred every fourth cycle, then, ignoring other effects, Processor C can complete four instructions every five 7ns cycles (35 ns; the first three instruction overlap in execution) compared to 40ns for Processor B (assuming the add timing included pipelining overhead).) (Note: Your assumption 3 is incorrect, two cycles for Processor C would be 14ns.)
Third, the extra time in a clock cycle can be used to support more complex operations (e.g., preshifting one operand by a small immediate value and even adding three numbers — a carry-save adder has relatively little delay), to steal work from other pipeline stages (potentially reducing the number of pipeline stages, which generally reduces branch misprediction penalties), or to reduce area or power by using simpler logic. In addition, the extra time might be used to support a larger (or more associative) cache with fixed latency in cycles, reducing miss rates. Such factors can compensate for the "waste" of 5ns in Processor A.
Even for scalar (single issue per cycle) pipelines clock speed is not the single determinant of performance. Design choices become even more complex when power, manufacturing cost (related to yield, adjusted according to sellable bins, and area), time-to-market (and its variability/predictability), workload diversity, and more advanced architectural and microarchitectural techniques are considered.
The incorrect assumption that clock frequency determines performance even has a name: the Megahertz myth.
Suppose my kernel takes 4 (or 3, or 2) unrelated float or double args, or that I want to access 4 separate floats from global memory. Will this cause 4 separate global memory accesses? Is accessing a single vector of 4 floats or doubles faster than accessing 4 separate ones? If so, am I better off packing them into a single vector and then, say, using #defines to reference the individual members?
If this does increase the performance, do I have to do it myself, or might the compiler be smart enough to automatically convert 4 separate float reads into a single vector for me? Is this what "auto-vectorization" is? I've seen auto-vectorization mentioned in a few documents, without detailed explanation of exactly what it does, except that it seems to be an optional performance optimization for CPUs only, not GPUs.
Using vectors depends on kernel itself. If you need all four values at same time (for example: at start of kernel, at start of loop), it's better to pack them, because they will be assigned during one read (Values in single vector are stored sequential).
On the other hand, when you need only some of the values, you can speed up execution by reading only what you need.
Another case is when you read them one by one, each reading divided by some computation (i.e. give GPU some time to fetch data).
Basically, this data read segments, behaves like buffer. If you have enough instances, number of reads is same (in optional cause) and what really counts is how well are these reads used.
Compiler often unpack these structures so only speedup is, that you have all variables nicely stored, so when you read, you fill them all up with one read and rest of buffer is used for another instance.
As example, I will use 128 bits wide bus and 4 floats (32 bits).
(32b * 4) / 128b = 1 instance/read
For scalar data types, there are N reads (N = number of variables), each read filling one variable in each instance up to the number of fetched variables.
32b / 128b = 4 instance/read
So in my examples, if you have 4 instances, there will always be at least 4 reads no matter what and only thing, you can do with this is cover fetching time by some computation, if it's even possible.
This is an exercise from this text book (page 77):
Exercise 48 (External memory stacks and queues). Design a stack data structure that needs O(1/B) I/Os per operation in the I/O model
from Section 2.2. It suffices to keep two blocks in internal memory.
What can happen in a naive implementation with only one block in
memory? Adapt your data structure to implement FIFOs, again using two
blocks of internal buffer memory. Implement deques using four buffer
blocks.
I don't want the code. Can anyone explain me what the question needs, and how can i do operations in O(1/B)?
As the book goes, quoting Section 2.2 on page 27:
External Memory: <...> There are special I/O operations that transfer B consecutive words between slow and fast memory. For
example, the external memory could be a hard disk, M would then be the
main memory size and B would be a block size that is a good compromise
between low latency and high bandwidth. On current technology, M = 1
GByte and B = 1 MByte are realistic values. One I/O step would then be
around 10ms which is 107 clock cycles of a 1GHz machine. With another
setting of the parameters M and B, we could model the smaller access
time difference between a hardware cache and main memory.
So, doing things in O(1/B) most likely means, in other words, using a constant number of these I/O operations for each B stack/queue operations.