Array vs long vectors in Verilog - arrays

I am writing a Verilog code that needs to hold many data in a memory-like structure.
I have implemented them using both an array of vectors and a really long single vector.
Although I don't think there is any difference internally, which, if there is, is a better way of storing data?
I'm actually writing a code that will synthesize onto a board, so any practical advice from those who've had a lot of experience with FPGA will help.
For example, I could store 32x1024 data using
reg[31:0] temp_storage [0:1023]
or
reg[32767:0] temp_storage
The array method is much easier for the programmer to manage, but is there any disadvtange from the perspective of the hardware?
Would it actually be the same if I declared everything one by one?
reg[31:0] temp_storage0001
reg[31:0] temp_stoarge0002
.
.
.
reg[31:0] temp_storage1024
Thank you.

There is a big difference between these two formats.
Format one:
reg[31:0] temp_storage [0:1023];
This is can be mapped* on a memory block and as such will use significant less FPGA registers/recources. But there is penalty: you can read or write maximum two entries at a time.
(Two entries if you use dual ported memory, one if you use single ported memory. All embedded block rams in FPGAs are dual ported these days)
Format two:
reg[31:0] temp_storage0001
reg[31:0] temp_stoarge0002
In this case every temp_storage... is stored in a separate set of 32 registers. You can access as many as you like simultaneous reading and writing (or as much until you run out of FPGA gates). Thus the flexibility is much larger but this will use up you FPGA gates/LUTS much faster.
* As #B.Go says: check your FPGA documentation how exactly you get this mapped onto memory, or infer memory macro/IP directly.
What do you exactly mean by maximum of two entries: temp_storage[0] <= t_data; temp_storage[1] <= t_data.
reg[31:0] temp_storage [0:1023];
The above definition is for a memory which has 1024 entries each 32 bits wide. You can select two entries of the 1024 and you can read or you can write each entry. (Your dual-ported memory normally has two address buses, two read data ports and two write data ports and often also two clocks, on per port).
You would normally access this memory using something like:
always #( posedge clk)
begin
if (write_enable_0)
temp_storage[address_0] <= write_data_0;
else
read_data_0 <= temp_storage[address_0];
if (write_enable_1)
temp_storage[address_1] <= write_data_1;
else
read_data_1 <= temp_storage[address_1];
end
Find the 'memory' section /application note of your FPGA family which will tell you how to do thus and also the pitfalls (e.g. writing and reading at the same time from the same location)

Related

C: Most efficient way to store variables where every bit matters

To start off: this might be a duplicate, but i can't seem to find a definitive answer on this question after having searched for it on google.
For a project i am designing a script that makes 2 ATMega328p chips communicate. At this moment i'm testing the best speed to do this, but my goal is to achieve really high baudrates. I have plenty of experience with making code efficient, but not with the memory management part. The problem:
I want to store a multiple of 8 bits (ex.: 48 bits). My first thought was to use an array of length 6 and type uint8_t, but I don't know how efficient arrays are compared to other types. Some people say pointers are more efficient and others say it doesn't matter, but I cant find a definitive answer on what the case is for really small amounts of memory. last quesion: I know the size of the sent bits will never be bigger than 64 bits, so would it matter if i just Always jused uint64_t?
Edit:
to clarify: My goal is to minimize the storage size, not the transmission size
Edit2:
What i meant by having a varying size: The size is determined on compile time, not while running the program.
The ATmega328p is a 8bit processor. All of its instructions are 8bit. Nothing will be faster than simply having an uint8_t array.
What you can do is, when you compile, look at your .lss file, it will show you the assmebly code, then you can look up the AVR instruction set and see the clock cycles each one will take. I think you will find using a uint64_t will just add unncessary overhead unless you are very careful with the way you are putting the bytes into it.
If the length of your packages might vary, the most efficient approach would be to compress the package before communication.
For example the first 3 bits of each package, could determine the size of that package.
The compressed packages are communicated faster, and use up less memory space.

How can we allocate memory of order 10^15 in C

I need to allocate memory of order of 10^15 to store integers which can be of long long type.
If i use an array and declare something like
long long a[1000000000000000];
that's never going to work. So how can i allocate such a huge amount of memory.
Really large arrays generally aren't a job for memory, more one for disk. 1015 array elements at 64 bits apiece is (I think) 8 petabytes. You can pick up 8G memory slices for about $15 at the moment so, even if your machine could handle that much memory or address space, you'd be outlaying about $15 million dollars.
In addition, with upcoming DDR4 being clocked up to about 4GT/s (giga-transfers), even if each transfer was a 64-bit value, it would still take about one million seconds just to initialise that array to zero. Do you really want to be waiting around for eleven and a half days before your code even starts doing anything useful?
And, even if you go the disk route, that's quite a bit. At (roughly) $50 per TB, you're still looking at $400,000 and you'll possibly have to provide your own software for managing those 8,000 disks somehow. And I'm not even going to contemplate figuring out how long it would take to initialise the array on disk.
You may want to think about rephrasing your question to indicate the actual problem rather than what you currently have, a proposed solution. It may be that you don't need that much storage at all.
For example, if you're talking about an array where many of the values are left at zero, a sparse array is one way to go.
You can't. You don't have all this memory, and you'll don't have it for a while. Simple.
EDIT: If you really want to work with data that does not fit into your RAM, you can use some library that work with mass storage data, like stxxl, but it will work a lot slower, and you have always disk size limits.
MPI is what you need, that's actually a small size for parallel computing problems the blue gene Q monster at Lawerence Livermore National Labs holds around 1.5 PB of ram. you need to use block decomposition to divide up your problem and viola!
the basic approach is dividing up the array into equal blocks or chunks among many processors
You need to uppgrade to a 64-bit system. Then get 64-bit-capable compiler then put a l at the end of 100000000000000000.
Have you heard of sparse matrix implementation? In one of the sparse matrices, you just use very little part of the matrix despite of the matrix being huge.
Here are some libraries for you.
Here is a basic info about sparse-matrices You dont actually use all of it. Just the needed few points.

Is it possible to create a float array of 10^13 elements in C?

I am writing a program in C to solve an optimisation problem, for which I need to create an array of type float with an order of 1013 elements. Is it practically possible to do so on a machine with 20GB memory.
A float in C occupies 4 bytes (assuming IEEE floating point arithmetic, which is pretty close to universal nowadays). That means 1013 elements are naïvely going to require 4×1013 bytes of space. That's quite a bit (40 TB, a.k.a. quite a lot of disk for a desktop system, and rather more than most people can afford when it comes to RAM) so you need to find another approach.
Is the data sparse (i.e., mostly zeroes)? If it is, you can try using a hash table or tree to store only the values which are anything else; if your data is sufficiently sparse, that'll let you fit everything in. Also be aware that processing 1013 elements will take a very long time. Even if you could process a billion items a second (very fast, even now) it would still take 104 seconds (several hours) and I'd be willing to bet that in any non-trivial situation you'll not be able to get anything near that speed. Can you find some way to make not just the data storage sparse but also the processing, so that you can leave that massive bulk of zeroes alone?
Of course, if the data is non-sparse then you're doomed. In that case, you might need to find a smaller, more tractable problem instead.
I suppose if you had a 64 bit machine with a lot of swap space, you could just declare an array of size 10^13 and it may work.
But for a data set of this size it becomes important to consider carefully the nature of the problem. Do you really need random access read and write operations for all 10^13 elements? Is the array at all sparse? Could you express this as a map/reduce problem? If so, sequential access to 10^13 elements is much more practical than random access.

Memory management for gauss elimination

A matrix is created in processor 0 and scattered to other processors. A matrix is a symmetric dense matrix. That's why it is initialized in processor 0.
A matrix is created in this way:
A=malloc(sizeof(double)*N*N);
for (i=0; i<N; i++)
for(j=0; j<N; j++)
A(i,j)=rand()%10; // The code will be changed.
A(i,j) is defined as:
#define A(i,j) A[i*N+j]
and N has to be 100,000 to test the algorithm.
The problem here is: if N=100,000 then the memory needed is approximately 76GB. What do you suggest to store the A matrix?
PS: Algorithm works very well when N<20.000 and the cluster is a distrubed memory system(2GB RAM per processor)
If you are doing this, as stated in comments, to do a scaling test, then Oli Charlesworth is completely right; anything you do is going to make this an apples-to-oranges comparison, because your node doesn't have 76GB to use. Which is fine; one of the big reasons to use MPI is to tackle problems that couldn't fit on one node. But by trying to shoehorn 76GB of data onto one processor, the comparison you're doing isn't going to make any sense. As mentioned by both Oli Charlesworth and caf, through various methods you can use disk instead of RAM, but then your 1 processor answer is going not going to be directly comparable to the fits-in-RAM numbers you get from larger number of nodes, so you're going to be going to a lot of work to get a number which won't actually mean anything.
If you want scaling results on this sort of problem, you either start with the lowest number of nodes that the problem does fit on, and take data at increasing numbers of processors, or you do weak scaling, rather than strong scaling tests -- you keep the work-per-processor constant while scaling up the number of processors, rather than the total work being constant.
Incidentally, however you do the measurements, you'll end up with better results if, as Oli Charlesworth suggests, you have each procesor generate its own data rather than have a serial bottleneck by having rank 0 do the generation of the matrix and then have all the processors receive their parts.
If you are programming on a POSIX system with sufficient virtual address space (which in practice will mean a 64 bit system), you can use mmap().
Either create an anonymous mapping of the required size (this will be swap-backed, which will mean you'll need at least 76GB of swap), or create a real file of the required size and map that.
The file-backed solution has the advantage that if your cluster has a shared file system, you don't need to explicitly transfer the matrix to each processor - you can simply msync() it after creating it, and then map the right region on each processor.
If you can switch to C++, you might look into STXXL, which is an STL implementation specifically designed for huge datasets, with transparent disk-backed support, etc.

Why is that data structures usually have a size of 2^n?

Is there a historical reason or something ? I've seen quite a few times something like char foo[256]; or #define BUF_SIZE 1024. Even I do mostly only use 2n sized buffers, mostly because I think it looks more elegant and that way I don't have to think of a specific number. But I'm not quite sure if that's the reason most people use them, more information would be appreciated.
There may be a number of reasons, although many people will as you say just do it out of habit.
One place where it is very useful is in the efficient implementation of circular buffers, especially on architectures where the % operator is expensive (those without a hardware divide - primarily 8 bit micro-controllers). By using a 2^n buffer in this case, the modulo, is simply a case of bit-masking the upper bits, or in the case of say a 256 byte buffer, simply using an 8-bit index and letting it wraparound.
In other cases alignment with page boundaries, caches etc. may provide opportunities for optimisation on some architectures - but that would be very architecture specific. But it may just be that such buffers provide the compiler with optimisation possibilities, so all other things being equal, why not?
Cache lines are usually some multiple of 2 (often 32 or 64). Data that is an integral multiple of that number would be able to fit into (and fully utilize) the corresponding number of cache lines. The more data you can pack into your cache, the better the performance.. so I think people who design their structures in that way are optimizing for that.
Another reason in addition to what everyone else has mentioned is, SSE instructions take multiple elements, and the number of elements input is always some power of two. Making the buffer a power of two guarantees you won't be reading unallocated memory. This only applies if you're actually using SSE instructions though.
I think in the end though, the overwhelming reason in most cases is that programmers like powers of two.
Hash Tables, Allocation by Pages
This really helps for hash tables, because you compute the index modulo the size, and if that size is a power of two, the modulus can be computed with a simple bitwise-and or & rather than using a much slower divide-class instruction implementing the % operator.
Looking at an old Intel i386 book, and is 2 cycles and div is 40 cycles. A disparity persists today due to the much greater fundamental complexity of division, even though the 1000x faster overall cycle times tend to hide the impact of even the slowest machine ops.
There was also a time when malloc overhead was occasionally avoided at great length. Allocation's available directly from the operating system would be (still are) a specific number of pages, and so a power of two would be likely to make the most use of the allocation granularity.
And, as others have noted, programmers like powers of two.
I can think of a few reasons off the top of my head:
2^n is a very common value in all of computer sizes. This is directly related to the way bits are represented in computers (2 possible values), which means variables tend to have ranges of values whose boundaries are 2^n.
Because of the point above, you'll often find the value 256 as the size of the buffer. This is because it is the largest number that can be stored in a byte. So, if you want to store a string together with a size of the string, then you'll be most efficient if you store it as: SIZE_BYTE+ARRAY, where the size byte tells you the size of the array. This means the array can be any size from 1 to 256.
Many other times, sizes are chosen based on physical things (for example, the size of the memory an operating system can choose from is related to the size of the registers of the CPU etc) and these are also going to be a specific amount of bits. Meaning, the amount of memory you can use will usually be some value of 2^n (for a 32bit system, 2^32).
There might be performance benefits/alignment issues for such values. Most processors can access a certain amount of bytes at a time, so even if you have a variable whose size is let's say) 20 bits, a 32 bit processor will still read 32 bits, no matter what. So it's often times more efficient to just make the variable 32 bits. Also, some processors require variables to be aligned to a certain amount of bytes (because they can't read memory from, for example, addresses in the memory that are odd). Of course, sometimes it's not about odd memory locations, but locations that are multiples of 4, or 6 of 8, etc. So in these cases, it's more efficient to just make buffers that will always be aligned.
Ok, those points came out a bit jumbled. Let me know if you need further explanation, especially point 4 which IMO is the most important.
Because of the simplicity (read also cost) of base 2 arithmetic in electronics: shift left (multiply by 2), shift right (divide by 2).
In the CPU domain, lots of constructs revolve around base 2 arithmetic. Busses (control & data) to access memory structure are often aligned on power 2. The cost of logic implementation in electronics (e.g. CPU) makes for arithmetics in base 2 compelling.
Of course, if we had analog computers, the story would be different.
FYI: the attributes of a system sitting at layer X is a direct consequence of the server layer attributes of the system sitting below i.e. layer < x. The reason I am stating this stems from some comments I received with regards to my posting.
E.g. the properties that can be manipulated at the "compiler" level are inherited & derived from the properties of the system below it i.e. the electronics in the CPU.
I was going to use the shift argument, but could think of a good reason to justify it.
One thing that is nice about a buffer that is a power of two is that circular buffer handling can use simple ands rather than divides:
#define BUFSIZE 1024
++index; // increment the index.
index &= BUFSIZE; // Make sure it stays in the buffer.
If it weren't a power of two, a divide would be necessary. In the olden days (and currently on small chips) that mattered.
It's also common for pagesizes to be powers of 2.
On linux I like to use getpagesize() when doing something like chunking a buffer and writing it to a socket or file descriptor.
It's makes a nice, round number in base 2. Just as 10, 100 or 1000000 are nice, round numbers in base 10.
If it wasn't a power of 2 (or something close such as 96=64+32 or 192=128+64), then you could wonder why there's the added precision. Not base 2 rounded size can come from external constraints or programmer ignorance. You'll want to know which one it is.
Other answers have pointed out a bunch of technical reasons as well that are valid in special cases. I won't repeat any of them here.
In hash tables, 2^n makes it easier to handle key collissions in a certain way. In general, when there is a key collission, you either make a substructure, e.g. a list, of all entries with the same hash value; or you find another free slot. You could just add 1 to the slot index until you find a free slot; but this strategy is not optimal, because it creates clusters of blocked places. A better strategy is to calculate a second hash number h2, so that gcd(n,h2)=1; then add h2 to the slot index until you find a free slot (with wrap around). If n is a power of 2, finding a h2 that fulfills gcd(n,h2)=1 is easy, every odd number will do.

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