I am using OpenCL (Xcode, Intel GPU), and I am trying to implement a kernel that calculates moving averages and deviations. I want to pass several double arrays of varying lengths to the kernel. Is this possible to implement, or do I need to pad smaller arrays with zeroes so all the arrays are the same size?
I am new to OpenCL and GPGPU, so please forgive my ignorance of any nomenclature.
You can pass to the kernel any buffer, the kernel does not need to use it all.
For example, if your kernel reduces a buffer you can query at run time the amount of work items (items to reduce) using get_global_size(0). And then call the kernel with the proper parameters.
An example (unoptimized):
__kernel reduce_step(__global float* data)
{
int id = get_global_id(0);
int size = get_global_size(0);
int size2 = size/2;
int size2p = (size+1)/2;
if(id<size2) //Only reduce up to size2, the odd element will remain in place
data[id] += data[id+size2p];
}
Then you can call it like this.
void reduce_me(std::vector<cl_float>& data){
size_t size = data.size();
//Copy to a buffer already created, equal or bigger size than data size
// ... TODO, check sizes of buffer or change the buffer set to the kernel args.
queue.enqueueWriteBuffer(buffer,CL_FALSE,0,sizeof(cl_float)*size,data.data());
//Reduce until 1024
while(size > 1024){
queue.enqueueNDRangeKernel(reduce_kernel,cl::NullRange,cl::NDRange(size),cl::NullRange);
size /= 2;
}
//Read out and trim
queue.enqueueReadBuffer(buffer,CL_TRUE,0,sizeof(cl_float)*size,data.data());
data.resize(size);
}
Related
I need to provide a huge circular buffer (a few GB) for the bus-mastering DMA PCIe device implemented in FPGA.
The buffers should not be reserved at the boot time. Therefore, the buffer may be not contiguous.
The device supports scatter-gather (SG) operation, but for performance reasons, the addresses and lengths of consecutive contiguous segments of the buffer are stored inside the FPGA.
Therefore, usage of standard 4KB pages is not acceptable (there would be up to 262144 segments for each 1GB of the buffer).
The right solution should allocate the buffer consisting of 2MB hugepages in the user space (reducing the maximum number of segments by factor of 512).
The virtual address of the buffer should be transferred to the kernel driver via ioctl. Then the addresses and the length of the segments should be calculated and written to the FPGA.
In theory, I could use get_user_pages to create the list of the pages, and then call sg_alloc_table_from_pages to obtain the SG list suitable to program the DMA engine in FPGA.
Unfortunately, in this approach I must prepare the intermediate list of page structures with length of 262144 pages per 1GB of the buffer. This list is stored in RAM, not in the FPGA, so it is less problematic, but anyway it would be good to avoid it.
In fact I don't need to keep the pages maped for the kernel, as the hugepages are protected against swapping out, and they are mapped for the user space application that will process the received data.
So what I'm looking for is a function sg_alloc_table_from_user_hugepages, that could take such a user-space address of the hugepages-based memory buffer, and transfer it directly into the right scatterlist, without performing unnecessary and memory-consuming mapping for the kernel.
Of course such a function should verify that the buffer indeed consists of hugepages.
I have found and read these posts: (A), (B), but couldn't find a good answer.
Is there any official method to do it in the current Linux kernel?
At the moment I have a very inefficient solution based on get_user_pages_fast:
int sgt_prepare(const char __user *buf, size_t count,
struct sg_table * sgt, struct page *** a_pages,
int * a_n_pages)
{
int res = 0;
int n_pages;
struct page ** pages = NULL;
const unsigned long offset = ((unsigned long)buf) & (PAGE_SIZE-1);
//Calculate number of pages
n_pages = (offset + count + PAGE_SIZE - 1) >> PAGE_SHIFT;
printk(KERN_ALERT "n_pages: %d",n_pages);
//Allocate the table for pages
pages = vzalloc(sizeof(* pages) * n_pages);
printk(KERN_ALERT "pages: %p",pages);
if(pages == NULL) {
res = -ENOMEM;
goto sglm_err1;
}
//Now pin the pages
res = get_user_pages_fast(((unsigned long)buf & PAGE_MASK), n_pages, 0, pages);
printk(KERN_ALERT "gupf: %d",res);
if(res < n_pages) {
int i;
for(i=0; i<res; i++)
put_page(pages[i]);
res = -ENOMEM;
goto sglm_err1;
}
//Now create the sg-list
res = sg_alloc_table_from_pages(sgt, pages, n_pages, offset, count, GFP_KERNEL);
printk(KERN_ALERT "satf: %d",res);
if(res < 0)
goto sglm_err2;
*a_pages = pages;
*a_n_pages = n_pages;
return res;
sglm_err2:
//Here we jump if we know that the pages are pinned
{
int i;
for(i=0; i<n_pages; i++)
put_page(pages[i]);
}
sglm_err1:
if(sgt) sg_free_table(sgt);
if(pages) kfree(pages);
* a_pages = NULL;
* a_n_pages = 0;
return res;
}
void sgt_destroy(struct sg_table * sgt, struct page ** pages, int n_pages)
{
int i;
//Free the sg list
if(sgt->sgl)
sg_free_table(sgt);
//Unpin pages
for(i=0; i < n_pages; i++) {
set_page_dirty(pages[i]);
put_page(pages[i]);
}
}
The sgt_prepare function builds the sg_table sgt structure that i can use to create the DMA mapping. I have verified that it contains the number of entries equal to the number of hugepages used.
Unfortunately, it requires that the list of the pages is created (allocated and returned via the a_pages pointer argument), and kept as long as the buffer is used.
Therefore, I really dislike that solution. Now I have 256 2MB hugepages used as a DMA buffer. It means that I have to create and keeep unnecessary 128*1024 page structures. I also waste 512 MB of kernel address space for unnecessary kernel mapping.
The interesting question is if the a_pages may be kept only temporarily (until the sg-list is created)? In theory it should be possible, as the pages are still locked...
I have encountered an interesting issue where a PERCPU_ARRAY created on one system with 2 processors creates an array with 2 per-CPU elements and on another system with 2 processors, an array with 128 per-CPU elements. The latter was rather unexpected to me!
The way I discovered this behavior is that a program that allocated an array for the number of CPUs (using get_nprocs_conf(3)) and then read in the PERCPU_ARRAY into it (using bpf_map_lookup_elem()) ended up writing past the end of the array and crashing.
I would like to find out what is the proper way to determine in a program that reads BPF maps the number of elements in a PERCPU_ARRAY used on a system.
Failing that, I think the second best approach is to pick a buffer for reading in that is "large enough." Here, the problem is similar: what is that number and is there way to learn it at runtime?
The question comes from reading the source of bpftool, which figures this out:
unsigned int get_possible_cpus(void)
{
int cpus = libbpf_num_possible_cpus();
if (cpus < 0) {
p_err("Can't get # of possible cpus: %s", strerror(-cpus));
exit(-1);
}
return cpus;
}
int libbpf_num_possible_cpus(void)
{
static const char *fcpu = "/sys/devices/system/cpu/possible";
static int cpus;
int err, n, i, tmp_cpus;
bool *mask;
/* ---8<--- snip */
}
So that's how they do it!
I am writing firmware for a data logging device. It reads data from sensors at 20 Hz and writes data to an SD card. However, the time to write data to SD card is not consistent (about 200-300 ms). Thus one solution is writing data to a buffer at a consistent rate (using a timer interrupt), and have a second thread that writes data to the SD card when the buffer is full.
Here is my naive implementation:
#define N 64
char buffer[N];
int count;
ISR() {
if (count < N) {
char a = analogRead(A0);
buffer[count] = a;
count = count + 1;
}
}
void loop() {
if (count == N) {
myFile.open("data.csv", FILE_WRITE);
int i = 0;
for (i = 0; i < N; i++) {
myFile.print(buffer[i]);
}
myFile.close();
count = 0;
}
}
The code has the following problems:
Writing data to the SD card is blocking reading when the buffer is full
It might have a race conditions.
What is the best way to solve this problem? Using a circular buffer, or double buffering? How do I ensure that a race condition does not happen?
You have rather answered your own question; you should use either double buffering or a circular buffer. Double-buffering is probably simpler to implement and appropriate for devices such as an SD card for which block-writes are generally more efficient.
Buffer length selection may need some consideration; generally you would make the buffer the same as the SD sector buffer size (typically 512 bytes), but that may not be practical, and with a sample rate as low as 20 sps, optimising SD write performance is perhaps not an issue.
Another consideration is that you need to match your sample rate to the file-system latency by selecting an appropriate buffer size. In this case the 64 sample buffer buffer will fill in a little more than three seconds, but the block write takes only up-to 300 ms - so you could use a much smaller buffer if required - 8 samples would be sufficient - although be careful, you may have observed latency of 300 ms, but it may be larger when specific boundaries are crossed in the physical flash memory - I have seen significant latency on some cards at 1 Mbyte boundaries - moreover card performance varies significantly between sizes and manufacturers.
An adaptation of your implementation with double-buffering is below. I have reduced the buffer length to 32 samples, but with double-buffering the total is unchanged at 64, but the write lag is reduced to 1.6 seconds.
// Double buffer and its management data/constants
static volatile char buffer[2][32];
static const int BUFFLEN = sizeof(buffer[0]);
static const unsigned char EMPTY = 0xff;
static volatile unsigned char inbuffer = 0;
static volatile unsigned char outbuffer = EMPTY;
ISR()
{
static int count = 0;
// Write to current buffer
char a = analogRead(A0);
buffer[inbuffer][count] = a;
count++ ;
// If buffer full...
if( count >= BUFFLEN )
{
// Signal to loop() that data available (not EMPTY)
outbuffer = inbuffer;
// Toggle input buffer
inbuffer = inbuffer == 0 ? 1 : 0;
count = 0;
}
}
void loop()
{
// If buffer available...
if( outbuffer != EMPTY )
{
// Write buffer
myFile.open("data.csv", FILE_WRITE);
for( int i = 0; i < BUFFLEN; i++)
{
myFile.print(buffer[outbuffer][i]);
}
myFile.close();
// Set the buffer to empty
outbuffer = EMPTY;
}
}
Note the use of volatile and unsigned char for the shared data. It is important that data shared between concurrent execution contexts is accessed explicitly and atomically; access to an int on 8-bit AVR based Arduino requires multiple machine instructions and the interrupt may occur part way through a read/write in loop() and cause an incorrect value to be read.
I would like to apply a reduce on this piece of my kernel code (1 dimensional data):
__local float sum = 0;
int i;
for(i = 0; i < length; i++)
sum += //some operation depending on i here;
Instead of having just 1 thread that performs this operation, I would like to have n threads (with n = length) and at the end having 1 thread to make the total sum.
In pseudo code, I would like to able to write something like this:
int i = get_global_id(0);
__local float sum = 0;
sum += //some operation depending on i here;
barrier(CLK_LOCAL_MEM_FENCE);
if(i == 0)
res = sum;
Is there a way?
I have a race condition on sum.
To get you started you could do something like the example below (see Scarpino). Here we also take advantage of vector processing by using the OpenCL float4 data type.
Keep in mind that the kernel below returns a number of partial sums: one for each local work group, back to the host. This means that you will have to carry out the final sum by adding up all the partial sums, back on the host. This is because (at least with OpenCL 1.2) there is no barrier function that synchronizes work-items in different work-groups.
If summing the partial sums on the host is undesirable, you can get around this by launching multiple kernels. This introduces some kernel-call overhead, but in some applications the extra penalty is acceptable or insignificant. To do this with the example below you will need to modify your host code to call the kernel repeatedly and then include logic to stop executing the kernel after the number of output vectors falls below the local size (details left to you or check the Scarpino reference).
EDIT: Added extra kernel argument for the output. Added dot product to sum over the float 4 vectors.
__kernel void reduction_vector(__global float4* data,__local float4* partial_sums, __global float* output)
{
int lid = get_local_id(0);
int group_size = get_local_size(0);
partial_sums[lid] = data[get_global_id(0)];
barrier(CLK_LOCAL_MEM_FENCE);
for(int i = group_size/2; i>0; i >>= 1) {
if(lid < i) {
partial_sums[lid] += partial_sums[lid + i];
}
barrier(CLK_LOCAL_MEM_FENCE);
}
if(lid == 0) {
output[get_group_id(0)] = dot(partial_sums[0], (float4)(1.0f));
}
}
I know this is a very old post, but from everything I've tried, the answer from Bruce doesn't work, and the one from Adam is inefficient due to both global memory use and kernel execution overhead.
The comment by Jordan on the answer from Bruce is correct that this algorithm breaks down in each iteration where the number of elements is not even. Yet it is essentially the same code as can be found in several search results.
I scratched my head on this for several days, partially hindered by the fact that my language of choice is not C/C++ based, and also it's tricky if not impossible to debug on the GPU. Eventually though, I found an answer which worked.
This is a combination of the answer by Bruce, and that from Adam. It copies the source from global memory into local, but then reduces by folding the top half onto the bottom repeatedly, until there is no data left.
The result is a buffer containing the same number of items as there are work-groups used (so that very large reductions can be broken down), which must be summed by the CPU, or else call from another kernel and do this last step on the GPU.
This part is a little over my head, but I believe, this code also avoids bank switching issues by reading from local memory essentially sequentially. ** Would love confirmation on that from anyone that knows.
Note: The global 'AOffset' parameter can be omitted from the source if your data begins at offset zero. Simply remove it from the kernel prototype and the fourth line of code where it's used as part of an array index...
__kernel void Sum(__global float * A, __global float *output, ulong AOffset, __local float * target ) {
const size_t globalId = get_global_id(0);
const size_t localId = get_local_id(0);
target[localId] = A[globalId+AOffset];
barrier(CLK_LOCAL_MEM_FENCE);
size_t blockSize = get_local_size(0);
size_t halfBlockSize = blockSize / 2;
while (halfBlockSize>0) {
if (localId<halfBlockSize) {
target[localId] += target[localId + halfBlockSize];
if ((halfBlockSize*2)<blockSize) { // uneven block division
if (localId==0) { // when localID==0
target[localId] += target[localId + (blockSize-1)];
}
}
}
barrier(CLK_LOCAL_MEM_FENCE);
blockSize = halfBlockSize;
halfBlockSize = blockSize / 2;
}
if (localId==0) {
output[get_group_id(0)] = target[0];
}
}
https://pastebin.com/xN4yQ28N
You can use new work_group_reduce_add() function for sum reduction inside single work group if you have support for OpenCL C 2.0 features
A simple and fast way to reduce data is by repeatedly folding the top half of the data into the bottom half.
For example, please use the following ridiculously simple CL code:
__kernel void foldKernel(__global float *arVal, int offset) {
int gid = get_global_id(0);
arVal[gid] = arVal[gid]+arVal[gid+offset];
}
With the following Java/JOCL host code (or port it to C++ etc):
int t = totalDataSize;
while (t > 1) {
int m = t / 2;
int n = (t + 1) / 2;
clSetKernelArg(kernelFold, 0, Sizeof.cl_mem, Pointer.to(arVal));
clSetKernelArg(kernelFold, 1, Sizeof.cl_int, Pointer.to(new int[]{n}));
cl_event evFold = new cl_event();
clEnqueueNDRangeKernel(commandQueue, kernelFold, 1, null, new long[]{m}, null, 0, null, evFold);
clWaitForEvents(1, new cl_event[]{evFold});
t = n;
}
The host code loops log2(n) times, so it finishes quickly even with huge arrays. The fiddle with "m" and "n" is to handle non-power-of-two arrays.
Easy for OpenCL to parallelize well for any GPU platform (i.e. fast).
Low memory, because it works in place
Works efficiently with non-power-of-two data sizes
Flexible, e.g. you can change kernel to do "min" instead of "+"
My PC has an AMD processor with an ATI 3200 GPU which doesn't support OpenCL. The rest of the codes all running by "Falling back to CPU itself".
I am converting one of the code from CUDA to OpenCL but stuck in some particular part for which there is no exact conversion code in OpenCL. since i have less experience in OpenCL I can't make out this, please suggest me some solution if any of you think will work,
The CUDA code is,
size_t pitch = 0;
cudaError error = cudaMallocPitch((void**)&gpu_data, (size_t*)&pitch,
instances->cols * sizeof(float), instances->rows);
for( int i = 0; i < instances->rows; i++ ){
error = cudaMemcpy((void*)(gpu_data + (pitch/sizeof(float))*i),
(void*)(instances->data + (instances->cols*i)),
instances->cols * sizeof(float) ,cudaMemcpyHostToDevice);
If I remove the pitch value from the above I end up with an problem which doesn't write to the device memory "gpu_data".
Somebody please convert this code to OpenCL and reply. I have converted it to OpenCL, but its not working and the data is not written to "gpu_data". My converted OpenCL code is
gpu_data = clCreateBuffer(context, CL_MEM_READ_WRITE, ((instances->cols)*(instances->rows))*sizeof(float), NULL, &ret);
for( int i = 0; i < instances->rows; i++ ){
ret = clEnqueueWriteBuffer(command_queue, gpu_data, CL_TRUE, 0, ((instances->cols)*(instances->rows))*sizeof(float),(void*)(instances->data + (instances->cols*i)) , 0, NULL, NULL);
Sometimes it runs well for this code and gets stuck in the reading part i.e.
ret = clEnqueueReadBuffer(command_queue, gpu_data, CL_TRUE, 0,sizeof( float ) * instances->cols* 1 , instances->data, 0, NULL, NULL);
overhere. And it gives error like
Unhandled exception at 0x10001098 in CL_kmeans.exe: 0xC000001D: Illegal Instruction.
when break is pressed , it gives:
No symbols are loaded for any call stack frame. The source code cannot be displayed.
while debugging. In the call stack it is displaying:
OCL8CA9.tmp.dll!10001098()
[Frames below may be incorrect and/or missing, no symbols loaded for OCL8CA9.tmp.dll]
amdocl.dll!5c39de16()
I really dont know what it means. someone please help me to rid of this problem.
First of all, in the CUDA code you're doing a horribly inefficient thing to copy the data. The CUDA runtime has the function cudaMemcpy2D that does exactly what you are trying to do by looping over different rows.
What cudaMallocPitch does is to compute an optimal pitch (= distance in byte between rows in a 2D array) such that each new row begins at an address that is optimal for coalescing, and then allocates a memory area as large as pitch times the number of rows you specify. You can emulate the same thing in OpenCL by first computing the optimal pitch and then doing the allocation of the correct size.
The optimal pitch is computed by (1) getting the base address alignment preference for your card (CL_DEVICE_MEM_BASE_ADDR_ALIGN property with clGetDeviceInfo: note that the returned value is in bits, so you have to divide by 8 to get it in bytes); let's call this base (2) find the largest multiple of base that is no less than your natural data pitch (sizeof(type) times number of columns); this will be your pitch.
You then allocate pitch times number of rows bytes, and pass the pitch information to kernels.
Also, when copying data from the host to the device and converesely, you want to use clEnqueue{Read,Write}BufferRect, that are specifically designed to copy 2D data (they are the counterparts to cudaMemcpy2D).