I am writing an algorithm which all blocks are reading a same address. Such as we have a list=[1, 2, 3, 4], and all blocks are reading it and store it to their own shared memory...My test shows the more blocks reading it, the slower it will be...I guess no broadcast happen here? Any idea I can make it faster? Thank you!!!
I learnt from previous post that this can be broadcast in one wrap, seems can not happen in different wrap....(Actually in my case, the threads in one wrap are not reading a same location...)
Once list element is accessed by first warp of a SM unit, the second warp in same SM unit gets it from cache and broadcasts to all simt lanes. But another SM unit's warp may not have it in L1 cache so it fetches from L2 to L1 first.
It is similar in __constant__ memory but it requires same address to be accessed by all threads. Its latency is closer to register access. __constant__ memory is like instruction cache, you get more performance when all threads do same thing.
For example, if you have a Gaussian-filter that iterates over same coefficient-list of filter on all threads, it is better to use constant memory. Using shared memory does not have much advantage as the filter array is not scanned randomly. Shared memory is better when the filter array content is different per block or if it needs random access.
You can also combine constant memory and shared memory. Get half of list from constant memory, then the other half from shared memory. This should let 1024 threads hide latency of one memory type hidden behind the other.
If list is small enough, you can use registers directly (has to be compile-time known indices). But it increases register pressure and may decrease occupancy so be careful about this.
Some old cuda architectures (in case of fma operation) required one operand fetched from constant memory and the other operand from a register to achieve better performance in compute-bottlenecked algorithms.
In a test with 12000 floats as filter to be applied on all threads inputs, shared memory version with 128 threads-per-block completed work in 330 milliseconds while constant-memory version completed in 260 milliseconds and the L1 access performance was the real bottleneck in both versions so the real constant-memory performance is even better, as long as it is similar-index for all threads.
Related
Why is primary and cache memory divided into blocks?
Hi just got posed with this question, I haven't been able to find a detailed explanation corresponding to both primary memory and cache memory, if you have a solution It would be greatly appreciated :)
Thank you
Cache blocks exploit locality of reference based on two types of locality. Temporal locality, after you reference location x you are likely to access location x again shorty. Spatial locality, after you reference location x you are likely to access nearby locations, location x+1, ... shortly.
If you use a value at some distant data center x, you are likely to reuse that value and so it is copied geographically closer, 150ms. If you use a value on disk block x, you are likely to reuse disk block x and so it is kept in memory, 20 ms. If you use a value on memory page x, you are like to reuse memory page x and so the translation of its virtual address to its physical address is kept in the TLB cache. If you use a particular memory location x you are likely to reuse it and its neighbors and so it is kept in cache.
Cache memory is very small, L1D on an M1 is 192kB, and DRAM is very big, 8GB on an M1 Air. L1D cache is much faster than DRAM, maybe 5 cycles vs maybe 200 cycles. I wish this table was in cycles and included registers but it gives a useful overview of latencies:
https://gist.github.com/jboner/2841832
The moral of this is to pack data into aligned structures which fit. If you randomly access memory instead, you will miss in the cache, the TLB, the virtual page cache, ... and everything will be excruciatingly slow.
Most things in computer systems are divided into chunks of fixed sizes: bytes, words, cache blocks, pages.
One reason for this is that while hardware can do many things at once, it is hardware and thus, generally can only do what it was designed for. Making bytes of blocks of 8 bits, making words blocks of 4 bytes (32-bit systems) or 8 bytes (64-bit systems) is something that we can design hardware to do, and mostly in parallel.
Not using fixed-sized chunks or blocks, on the other hand, can make things much more difficult for the hardware, so, data structures like strings — an example of data that is highly variable in length — are usually handled with software loops.
Usually these fixed sizes are in powers of 2 (32, 64, etc..) — because division and modulus, which are very useful operations are easy to do in binary for powers of 2.
In summary, we must subdivide data into blocks because, we cannot treat all the data as one lump sum (hardware wise at least), and, treating all data as individual bits is also too cumbersome. So, we chunk or group data into blocks as appropriate for various levels of hardware to deal with in parallel.
We are taught that the abstraction of the RAM memory is a long array of bytes. And that for the CPU it takes the same amount of time to access any part of it. What is the device that has the ability to access any byte out of the 4 gigabytes (on my computer) in the same time? As this does not seem as a trivial task for me.
I have asked colleagues and my professors, but nobody can pinpoint to the how this task can be achieved with simple logic gates, and if it isn't just a tricky combination of logic gates, then what is it?
My personal guess is that you could achieve the access of any memory in O(log(n)) speed, where n would be the size of memory. Because each gate would split the memory in two and send you memory access instruction to the next split the memory in two gate. But that requires ALOT of gates. I can't come up with any other educated guess, and I don't even know the name of the device that I should look up in Google.
Please help my anguished curiosity, and thanks in advance.
edit<
This is what I learned!
quote from yours "the RAM can send the value from cell addressed X to some output pins", here is where everyone skip (again) the thing that is not trivial for me. The way that I see it, In order to build a gate that from 64 pins decides which byte out of 2^64 to get, each pin needs to split the overall possible range of memory into two. If bit at index 0 is 0 -> then the address is at memory 0-2^64/2, else address is at memory 2^64/2-2^64. And so on, However the amout of gates (lets call them) that the memory fetch will go through will be 64, (a constant). However the amount of gates needed is N, where N is the number of memory bytes there are.
Just because there is 64 pins, it doesn't mean that you can still decode it into a single fetch from a range of 2^64. Does 4 gigabytes memory come with a 4 gigabytes gates in the memory control???
now this can be improved, because as I read furiously more and more about how this memory is architectured, if you place the memory into a matrix with sqrt(N) rows and sqrt(N) columns, the amount of gates that a fetch memory will need to go through is O(log(sqrt(N)*2) and the amount of gates that will be required will be 2*sqrt(N), which is much better, and I think that its probably a trade secret.
/edit<
What the heck, I might as well make this an answer.
Yes, in the physical world, memory access cannot be constant time.
But it cannot even be logarithmic time. The O(log n) circuit you have in mind ultimately involves some sort of binary (or whatever) tree, and you cannot make a binary tree with constant-length wires in a 3D universe.
Whatever the "bits per unit volume" capacity of your technology is, storing n bits requires a sphere with radius O(n^(1/3)). Since information can only travel at the speed of light, accessing a bit at the other end of the sphere requires time O(n^(1/3)).
But even this is wrong. If you want to talk about actual limitations of our universe, our physics friends say the absolute maximum number of bits you can store in any sphere is proportional to the sphere's surface area, not its volume. So the actual radius of a minimal sphere containing n bits of information is O(sqrt(n)).
As I mentioned in my comment, all of this is pretty much moot. The models of computation we generally use to analyze algorithms assume constant-access-time RAM, which is close enough to the truth in practice and allows a fair comparison of competing algorithms. (Although good engineers working on high-performance code are very concerned about locality and the memory hierarchy...)
Let's say your RAM has 2^64 cells (places where it is possible to store a single value, let's say 8-bit). Then it needs 64 pins to address every cell with a different number. When at the input pins of your RAM there 'appears' a binary number X the RAM can send the value from cell addressed X to some output pins, and your CPU can get the value from there. In hardware the addressing can be done quite easily, for example by using multiple NAND gates (such 'addressing device' from some logic gates is called a decoder).
So it is all happening at the hardware-level, this is just direct addressing. If the CPU is able to provide 64 bits to 64 pins of your RAM it can address every single memory cell (as 64 bit is enough to represent any number up to 2^64 -1). The only reason why you do not get the value immediately is a kind of 'propagation time', so time it takes for the signal to go through all the logic gates in the circuit.
The component responsible for dealing with memory accesses is the memory controller. It is used by the CPU to read from and write to memory.
The access time is constant because memory words are truly layed out in a matrix form (thus, the "byte array" abstraction is very realistic), where you have rows and columns. To fetch a given memory position, the desired memory address is passed on to the controller, which then activates the right column.
From http://computer.howstuffworks.com/ram1.htm:
Memory cells are etched onto a silicon wafer in an array of columns
(bitlines) and rows (wordlines). The intersection of a bitline and
wordline constitutes the address of the memory cell.
So, basically, the answer to your question is: the memory controller figures it out. Of course that, given a memory address, the mapping to column and row must be easy to calculate in a constant time.
To fully understand this topic, I recommend you to read this guide on how memory works: http://computer.howstuffworks.com/ram.htm
There are so many concepts to master that it is difficult to explain it all in one answer.
I've been reading your comments and questions until I answered. I think you are on the right track, but there is some confusion here. The random access in which you are implying doesn't exist in the same way you think it does.
Reading, writing, and refreshing are done in a continuous cycle. A particular cell in memory is only read or written in a certain interval if a signal is detected to do so in that cycle. There is going to be support circuitry that includes "sense amplifiers to amplify the signal or charge detected on a memory cell."
Unless I am misunderstanding what you are implying, your confusion is in how easy it is to read/write to a cell. It's different dependent on chip design but there IS a minimum number of cycles it takes to read or write data to a cell.
These are my sources:
http://www.doc.ic.ac.uk/~dfg/hardware/HardwareLecture16.pdf
http://www.electronics.dit.ie/staff/tscarff/memory/dram_cycles.htm
http://www.ece.cmu.edu/~ece548/localcpy/dramop.pdf
To avoid a humungous answer, I left most of the detail out but all three of these will describe the process you are looking for.
I am implementing a spiking neural network using the CUDA library and am really unsure of how to proceed with regard to the following things:
Allocating memory (cudaMalloc) to many different arrays. Up until now, simply using cudaMalloc 'by hand' has sufficed, as I have not had to make more than 10 or so arrays. However, I now need to make pointers to, and allocate memory for thousands of arrays.
How to decide how much memory to allocate to each of those arrays. The arrays have a height of 3 (1 row for the postsynaptic neuron ids, 1 row for the number of the synapse on the postsynaptic neuron, and 1 row for the efficacy of that synapse), but they have an undetermined length which changes over time with the number of outgoing synapses.
I have heard that dynamic memory allocation in CUDA is very slow and so toyed with the idea of allocating the maximum memory required for each array, however the number of outgoing synapses per neuron varies from 100-10,000 and so I thought this was infeasible, since I have on the order of 1000 neurons.
If anyone could advise me on how to allocate memory to many arrays on the GPU, and/or how to code a fast dynamic memory allocation for the above tasks I would have more than greatly appreciative.
Thanks in advance!
If you really want to do this, you can call cudaMalloc as many times as you want; however, it's probably not a good idea. Instead, try to figure out how to lay out the memory so that neighboring threads in a block will access neighboring elements of RAM whenever possible.
The reason this is likely to be problematic is that threads execute in groups of 32 at a time (a warp). NVidia's memory controller is quite smart, so if neighboring threads ask for neighboring bytes of RAM, it coalesces those loads into a single request that can be efficiently executed. In contrast, if each thread in a warp is accessing a random memory location, the entire warp must wait till 32 memory requests are completed. Furthermore, reads and writes to the card's memory happen a whole cache line at a time, so if the threads don't use all the RAM that was read before it gets evicted from the cache, memory bandwidth is wasted. If you don't optimize for coherent memory access within thread blocks, expect a 10x to 100x slowdown.
(side note: The above discussion is still applicable with post-G80 cards; the first generation of CUDA hardware (G80) was even pickier. It also required aligned memory requests if the programmer wanted the coalescing behavior.)
I'm using a Tesla, and for the first time, I'm running low on CPU memory instead of GPU memory! Hence, I thought I could cut the size of my host memory by switching all integers to short (all my values are below 255).
However, I want my device memory to use integers, since the memory access is faster. So is there a way to copy my host memory (in short) to my device global memory (in int)? I guess this won't work:
short *buf_h = new short[100];
int *buf_d = NULL;
cudaMalloc((void **)&buf_d, 100*sizeof(int));
cudaMemcpy( buf_d, buf_h, 100*sizeof(short), cudaMemcpyHostToDevice );
Any ideas? Thanks!
There isn't really a way to do what you are asking directly. The CUDA API doesn't support "smart copying" with padding or alignment, or "deep copying" of nested pointers, or anything like that. Memory transfers require linear host and device memory, and alignment must be the same between source and destination memory.
Having said that, one approach to circumvent this restriction would be to copy the host short data to an allocation of short2 on the device. Your device code can retrieve a short2 containing two packed shorts, extract the value it needs and then cast the value to int. This will give the code 32 bit memory transactions per thread, allowing for memory coalescing, and (if you are using Fermi GPUs) good L1 cache hit rates, because adjacent threads within a block would be reading the same 32 bit word. On non Fermi GPUs, you could probably use a shared memory scheme to efficiently retrieve all the values for a block using coalesced reads.
Admittedly I don't get it. Say you have a memory with a memory word of length of 1 byte. Why can't you access a 4 byte long variable in a single memory access on an unaligned address(i.e. not divisible by 4), as it's the case with aligned addresses?
The memory subsystem on a modern processor is restricted to accessing memory at the granularity and alignment of its word size; this is the case for a number of reasons.
Speed
Modern processors have multiple levels of cache memory that data must be pulled through; supporting single-byte reads would make the memory subsystem throughput tightly bound to the execution unit throughput (aka cpu-bound); this is all reminiscent of how PIO mode was surpassed by DMA for many of the same reasons in hard drives.
The CPU always reads at its word size (4 bytes on a 32-bit processor), so when you do an unaligned address access — on a processor that supports it — the processor is going to read multiple words. The CPU will read each word of memory that your requested address straddles. This causes an amplification of up to 2X the number of memory transactions required to access the requested data.
Because of this, it can very easily be slower to read two bytes than four. For example, say you have a struct in memory that looks like this:
struct mystruct {
char c; // one byte
int i; // four bytes
short s; // two bytes
}
On a 32-bit processor it would most likely be aligned like shown here:
The processor can read each of these members in one transaction.
Say you had a packed version of the struct, maybe from the network where it was packed for transmission efficiency; it might look something like this:
Reading the first byte is going to be the same.
When you ask the processor to give you 16 bits from 0x0005 it will have to read a word from 0x0004 and shift left 1 byte to place it in a 16-bit register; some extra work, but most can handle that in one cycle.
When you ask for 32 bits from 0x0001 you'll get a 2X amplification. The processor will read from 0x0000 into the result register and shift left 1 byte, then read again from 0x0004 into a temporary register, shift right 3 bytes, then OR it with the result register.
Range
For any given address space, if the architecture can assume that the 2 LSBs are always 0 (e.g., 32-bit machines) then it can access 4 times more memory (the 2 saved bits can represent 4 distinct states), or the same amount of memory with 2 bits for something like flags. Taking the 2 LSBs off of an address would give you a 4-byte alignment; also referred to as a stride of 4 bytes. Each time an address is incremented it is effectively incrementing bit 2, not bit 0, i.e., the last 2 bits will always continue to be 00.
This can even affect the physical design of the system. If the address bus needs 2 fewer bits, there can be 2 fewer pins on the CPU, and 2 fewer traces on the circuit board.
Atomicity
The CPU can operate on an aligned word of memory atomically, meaning that no other instruction can interrupt that operation. This is critical to the correct operation of many lock-free data structures and other concurrency paradigms.
Conclusion
The memory system of a processor is quite a bit more complex and involved than described here; a discussion on how an x86 processor actually addresses memory can help (many processors work similarly).
There are many more benefits to adhering to memory alignment that you can read at this IBM article.
A computer's primary use is to transform data. Modern memory architectures and technologies have been optimized over decades to facilitate getting more data, in, out, and between more and faster execution units–in a highly reliable way.
Bonus: Caches
Another alignment-for-performance that I alluded to previously is alignment on cache lines which are (for example, on some CPUs) 64B.
For more info on how much performance can be gained by leveraging caches, take a look at Gallery of Processor Cache Effects; from this question on cache-line sizes
Understanding of cache lines can be important for certain types of program optimizations. For example, the alignment of data may determine whether an operation touches one or two cache lines. As we saw in the example above, this can easily mean that in the misaligned case, the operation will be twice slower.
It's a limitation of many underlying processors. It can usually be worked around by doing 4 inefficient single byte fetches rather than one efficient word fetch, but many language specifiers decided it would be easier just to outlaw them and force everything to be aligned.
There is much more information in this link that the OP discovered.
you can with some processors (the nehalem can do this), but previously all memory access was aligned on a 64-bit (or 32-bit) line, because the bus is 64 bits wide, you had to fetch 64 bit at a time, and it was significantly easier to fetch these in aligned 'chunks' of 64 bits.
So, if you wanted to get a single byte, you fetched the 64-bit chunk and then masked off the bits you didn't want. Easy and fast if your byte was at the right end, but if it was in the middle of that 64-bit chunk, you'd have to mask off the unwanted bits and then shift the data over to the right place. Worse, if you wanted a 2 byte variable, but that was split across 2 chunks, then that required double the required memory accesses.
So, as everyone thinks memory is cheap, they just made the compiler align the data on the processor's chunk sizes so your code runs faster and more efficiently at the cost of wasted memory.
Fundamentally, the reason is because the memory bus has some specific length that is much, much smaller than the memory size.
So, the CPU reads out of the on-chip L1 cache, which is often 32KB these days. But the memory bus that connects the L1 cache to the CPU will have the vastly smaller width of the cache line size. This will be on the order of 128 bits.
So:
262,144 bits - size of memory
128 bits - size of bus
Misaligned accesses will occasionally overlap two cache lines, and this will require an entirely new cache read in order to obtain the data. It might even miss all the way out to the DRAM.
Furthermore, some part of the CPU will have to stand on its head to put together a single object out of these two different cache lines which each have a piece of the data. On one line, it will be in the very high order bits, in the other, the very low order bits.
There will be dedicated hardware fully integrated into the pipeline that handles moving aligned objects onto the necessary bits of the CPU data bus, but such hardware may be lacking for misaligned objects, because it probably makes more sense to use those transistors for speeding up correctly optimized programs.
In any case, the second memory read that is sometimes necessary would slow down the pipeline no matter how much special-purpose hardware was (hypothetically and foolishly) dedicated to patching up misaligned memory operations.
#joshperry has given an excellent answer to this question. In addition to his answer, I have some numbers that show graphically the effects which were described, especially the 2X amplification. Here's a link to a Google spreadsheet showing what the effect of different word alignments look like.
In addition here's a link to a Github gist with the code for the test.
The test code is adapted from the article written by Jonathan Rentzsch which #joshperry referenced. The tests were run on a Macbook Pro with a quad-core 2.8 GHz Intel Core i7 64-bit processor and 16GB of RAM.
If you have a 32bit data bus, the address bus address lines connected to the memory will start from A2, so only 32bit aligned addresses can be accessed in a single bus cycle.
So if a word spans an address alignment boundary - i.e. A0 for 16/32 bit data or A1 for 32 bit data are not zero, two bus cycles are required to obtain the data.
Some architectures/instruction sets do not support unaligned access and will generate an exception on such attempts, so compiler generated unaligned access code requires not just additional bus cycles, but additional instructions, making it even less efficient.
If a system with byte-addressable memory has a 32-bit-wide memory bus, that means there are effectively four byte-wide memory systems which are all wired to read or write the same address. An aligned 32-bit read will require information stored in the same address in all four memory systems, so all systems can supply data simultaneously. An unaligned 32-bit read would require some memory systems to return data from one address, and some to return data from the next higher address. Although there are some memory systems that are optimized to be able to fulfill such requests (in addition to their address, they effectively have a "plus one" signal which causes them to use an address one higher than specified) such a feature adds considerable cost and complexity to a memory system; most commodity memory systems simply cannot return portions of different 32-bit words at the same time.
On PowerPC you can load an integer from an odd address with no problems.
Sparc and I86 and (I think) Itatnium raise hardware exceptions when you try this.
One 32 bit load vs four 8 bit loads isnt going to make a lot of difference on most modern processors. Whether the data is already in cache or not will have a far greater effect.