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.
I think I have a fairly basic MIPS question but am still getting my head wrapped around how addressing works in mips.
My question is: What is the address of the register $t0?
I am looking at the following memory allocation diagram from the MIPs "green sheet"
I had two ideas:
The register $t0 has a register number of 8 so I'm wondering if it would have an address of 0x0000 0008 and be in the reserved portion of the memory block.
Or would it fall in the Static Data Section and have an address of 0x1000 0008?
I know that MARS and different assemblers might start the addressing differently as described in this related question:
How is the la instruction translated in MIPS?
I trying to understand what the "base" address is for register $t0 so I have a better understanding how offsets(base) work.
For example what the address of 8($t0) would be
Thanks for the help!
feature
Registers
Memory
count
very few
vast
speed
fast
slow
Named
yes
no
Addressable
no
yes
There are typically 32 or fewer registers for programmers to use. On a 32-bit machine we can address 2^32 different bytes of memory.
Registers are fast, while memory is slow, potentially taking dozens of cycles depending on cache features & conditions.
On a load-store machine, registers can be used directly in most instructions (by naming them in the machine code instruction), whereas memory access requires a separate load or store instruction. Computational instructions on such a machine typically allows naming up to 3 registers (e.g. target, source1, source2). Memory operands have to be brought into registers for computation (and sometimes moved back to memory).
Register can be named in instructions, but they do not have addresses and cannot be indexed. On MIPS no register can be found as alias at some address in memory. It is hard to put even a smallish array (e.g. array of 10) in registers because they have no addresses and cannot be indexed. Memory has numerical addresses, so we can rely on storing arrays and objects in a predictable pattern of addresses. (Memory generally doesn't have names, just addresses; however, there are usually special memory locations for working with I/O various devices, and, as you note memory is partitioned into sections that have start (and ending) addresses.)
To be clear, memory-based aliases have been designed into some processors of the past. The HP/1000 (circa 70s'-80's), for example, had 2 registers (A & B), and they had aliases at memory locations 0 and 1, respectively. However, this aliasing of CPU registers to memory is generally no longer done on modern processors.
For example what the address of 8($t0) would be
8($t0) refers to the memory address of (the contents of register $t0) + 8. With proper usage, the program fragment would $t0 would be using $t0 as a pointer, which is some variable that holds a memory address.
I worked with megafunctions to generate 32bit data memory in the fpga.but the output was addressed 32bit (4 bytes) at time , how to do 1 byte addressing ?
i have Altera Cyclone IV ep4ce6e22c8.
I'm designing a 32bit CPU in fpga ,
Nowadays every CPU address bus works in bytes. Thus to access your 32-bit wide memory you should NOT connect the LS 2 address bits. You can use the A[1:0] address bits to select a byte (or half word using A[1] only) from the memory when your read.
You still will need four byte write enable signals. This allows you to write word, half-words or bytes.
Have a look at existing CPU buses or existing connection standards like AHB or AXI.
Post edit:
but reading address 0001 , i get 0x05060708 but the desired value is 0x02030405.
What you are trying to do is read a word from a non-aligned address. There is no existing 32-bit wide memory that supports that. I suggest you have a look at how a 32-bit wide memory works.
The old Motorola 68020 architecture supported that. It requires a special memory controller which first reads the data from address 0 and then from address 4 and re-combines the data into a new 32-bit word.
With the cost of memory dropping and reducing CPU cycles becoming more important, no modern CPU supports that. They throw an exception: non-aligned memory access.
You have several choices:
Build a special memory controller which supports unaligned accesses.
Adjust your expectations.
I would go for the latter. In general it is based on the wrong idea how a memory works. As consolidation: You are not the first person on this website who thinks that is how you read words from memory.
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.