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Closed 10 years ago.
Using CUDA, I want to allocate memory for different arrays, one for each GPU from a different function than main(), but I must have missed something in regard to pointer arithmetic. Here's what I thought,
void InitThisMemory(int***, int N, int Nout, size_t* pitch, int height, int width); // This function's purpose is to initialize A and the pitch
int main(void){
int** A;
int N = 10;
int NOut = 2;
int height = 2, width = 2;
size_t pitch;
InitThisMemory(&A, N, NOut, &pitch, height, width);
return 0;
}
InitThisMemory(int ***A, int N, int Nout, size_t* pitch, int height, int width){
int i;
*A = (int**)malloc(Nout * sizeof(int*));
for(i = 0;i < Nout;i++){
cudaSetDevice(i);
cudaMallocPitch((void**)&(*A[i]), &(*pitch), width, height);
}
}
Disclaimer: Not my actual code but this should reproduce the error. Let me know if I missed an allocation of a variable somewhere.
Why do I think that the problem is in the arithmetic? Simply because this works pretty well if Nout = 1 (which means that I am using only one device).
Any ideas?
Your bug, I think, is writing (void**)&(*A[i]) instead of (void **) (&(*A)[i]), but I recommend you refactor as follows:
use a local int ** variable to hold the malloc() return value;
use that local in your call to cudaMallocPitch();
pass back the malloc() return value only if all cudaMallocPitch() calls succeed.
If you do these things, then it will be simpler to write correct cleanup code in the event that one of the cudaMallocPitch() calls fails, and you needn't propagate the passback unless everything has succeeded.
Related
I have a question that I found many threads in, but none did explicitly answer my question.
I am trying to have a multidimensional array inside the kernel of the GPU using thrust. Flattening would be difficult, as all the dimensions are non-homogeneous and I go up to 4D. Now I know I cannot have device_vectors of device_vectors, for whichever underlying reason (explanation would be welcome), so I tried going the way over raw-pointers.
My reasoning is, a raw pointer points onto memory on the GPU, why else would I be able to access it from within the kernel. So I should technically be able to have a device_vector, which holds raw pointers, all pointers that should be accessible from within the GPU. This way I constructed the following code:
thrust::device_vector<Vector3r*> d_fluidmodelParticlePositions(nModels);
thrust::device_vector<unsigned int***> d_allFluidNeighborParticles(nModels);
thrust::device_vector<unsigned int**> d_nFluidNeighborsCrossFluids(nModels);
for(unsigned int fluidModelIndex = 0; fluidModelIndex < nModels; fluidModelIndex++)
{
FluidModel *model = sim->getFluidModelFromPointSet(fluidModelIndex);
const unsigned int numParticles = model->numActiveParticles();
thrust::device_vector<Vector3r> d_neighborPositions(model->getPositions().begin(), model->getPositions().end());
d_fluidmodelParticlePositions[fluidModelIndex] = CudaHelper::GetPointer(d_neighborPositions);
thrust::device_vector<unsigned int**> d_fluidNeighborIndexes(nModels);
thrust::device_vector<unsigned int*> d_nNeighborsFluid(nModels);
for(unsigned int pid = 0; pid < nModels; pid++)
{
FluidModel *fm_neighbor = sim->getFluidModelFromPointSet(pid);
thrust::device_vector<unsigned int> d_nNeighbors(numParticles);
thrust::device_vector<unsigned int*> d_neighborIndexesArray(numParticles);
for(unsigned int i = 0; i < numParticles; i++)
{
const unsigned int nNeighbors = sim->numberOfNeighbors(fluidModelIndex, pid, i);
d_nNeighbors[i] = nNeighbors;
thrust::device_vector<unsigned int> d_neighborIndexes(nNeighbors);
for(unsigned int j = 0; j < nNeighbors; j++)
{
d_neighborIndexes[j] = sim->getNeighbor(fluidModelIndex, pid, i, j);
}
d_neighborIndexesArray[i] = CudaHelper::GetPointer(d_neighborIndexes);
}
d_fluidNeighborIndexes[pid] = CudaHelper::GetPointer(d_neighborIndexesArray);
d_nNeighborsFluid[pid] = CudaHelper::GetPointer(d_nNeighbors);
}
d_allFluidNeighborParticles[fluidModelIndex] = CudaHelper::GetPointer(d_fluidNeighborIndexes);
d_nFluidNeighborsCrossFluids[fluidModelIndex] = CudaHelper::GetPointer(d_nNeighborsFluid);
}
Now the compiler won't complain, but accessing for example d_nFluidNeighborsCrossFluids from within the kernel will work, but return wrong values. I access it like this (again, from within a kernel):
d_nFluidNeighborsCrossFluids[iterator1][iterator2][iterator3];
// Note: out of bounds indexing guaranteed to not happen, indexing is definitely right
The question is, why does it return wrong values? The logic behind it should work in my opinion, since my indexing is correct and the pointers should be valid addresses from within the kernel.
Thank you already for your time and have a great day.
EDIT:
Here is a minimal reproducable example. For some reason the values appear right despite of having the same structure as my code, but cuda-memcheck reveals some errors. Uncommenting the two commented lines leads me to my main problem I am trying to solve. What does the cuda-memcheck here tell me?
/* Part of this example has been taken from code of Robert Crovella
in a comment below */
#include <thrust/device_vector.h>
#include <stdio.h>
template<typename T>
static T* GetPointer(thrust::device_vector<T> &vector)
{
return thrust::raw_pointer_cast(vector.data());
}
__global__
void k(unsigned int ***nFluidNeighborsCrossFluids, unsigned int ****allFluidNeighborParticles){
const unsigned int i = blockIdx.x*blockDim.x + threadIdx.x;
if(i > 49)
return;
printf("i: %d nNeighbors: %d\n", i, nFluidNeighborsCrossFluids[0][0][i]);
//for(int j = 0; j < nFluidNeighborsCrossFluids[0][0][i]; j++)
// printf("i: %d j: %d neighbors: %d\n", i, j, allFluidNeighborParticles[0][0][i][j]);
}
int main(){
const unsigned int nModels = 2;
const int numParticles = 50;
thrust::device_vector<unsigned int**> d_nFluidNeighborsCrossFluids(nModels);
thrust::device_vector<unsigned int***> d_allFluidNeighborParticles(nModels);
for(unsigned int fluidModelIndex = 0; fluidModelIndex < nModels; fluidModelIndex++)
{
thrust::device_vector<unsigned int*> d_nNeighborsFluid(nModels);
thrust::device_vector<unsigned int**> d_fluidNeighborIndexes(nModels);
for(unsigned int pid = 0; pid < nModels; pid++)
{
thrust::device_vector<unsigned int> d_nNeighbors(numParticles);
thrust::device_vector<unsigned int*> d_neighborIndexesArray(numParticles);
for(unsigned int i = 0; i < numParticles; i++)
{
const unsigned int nNeighbors = i;
d_nNeighbors[i] = nNeighbors;
thrust::device_vector<unsigned int> d_neighborIndexes(nNeighbors);
for(unsigned int j = 0; j < nNeighbors; j++)
{
d_neighborIndexes[j] = i + j;
}
d_neighborIndexesArray[i] = GetPointer(d_neighborIndexes);
}
d_nNeighborsFluid[pid] = GetPointer(d_nNeighbors);
d_fluidNeighborIndexes[pid] = GetPointer(d_neighborIndexesArray);
}
d_nFluidNeighborsCrossFluids[fluidModelIndex] = GetPointer(d_nNeighborsFluid);
d_allFluidNeighborParticles[fluidModelIndex] = GetPointer(d_fluidNeighborIndexes);
}
k<<<256, 256>>>(GetPointer(d_nFluidNeighborsCrossFluids), GetPointer(d_allFluidNeighborParticles));
if (cudaGetLastError() != cudaSuccess)
printf("Sync kernel error: %s\n", cudaGetErrorString(cudaGetLastError()));
cudaDeviceSynchronize();
}
A device_vector is a class definition. That class has various methods and operators associated with it. The thing that allows you to do this:
d_nFluidNeighborsCrossFluids[...]...;
is a square-bracket operator. That operator is a host operator (only). It is not usable in device code. Issues like this give rise to the general statements that "thrust::device_vector is not usable in device code." The device_vector object itself is generally not usable. However the data it contains is usable in device code, if you attempt to access it via a raw pointer.
Here is an example of a thrust device vector that contains an array of pointers to the data contained in other device vectors. That data is usable in device code, as long as you don't attempt to make use of the thrust::device_vector object itself:
$ cat t1509.cu
#include <thrust/device_vector.h>
#include <stdio.h>
template <typename T>
__global__ void k(T **data){
printf("the first element of vector 1 is: %d\n", (int)(data[0][0]));
printf("the first element of vector 2 is: %d\n", (int)(data[1][0]));
printf("the first element of vector 3 is: %d\n", (int)(data[2][0]));
}
int main(){
thrust::device_vector<int> vector_1(1,1);
thrust::device_vector<int> vector_2(1,2);
thrust::device_vector<int> vector_3(1,3);
thrust::device_vector<int *> pointer_vector(3);
pointer_vector[0] = thrust::raw_pointer_cast(vector_1.data());
pointer_vector[1] = thrust::raw_pointer_cast(vector_2.data());
pointer_vector[2] = thrust::raw_pointer_cast(vector_3.data());
k<<<1,1>>>(thrust::raw_pointer_cast(pointer_vector.data()));
cudaDeviceSynchronize();
}
$ nvcc -o t1509 t1509.cu
$ cuda-memcheck ./t1509
========= CUDA-MEMCHECK
the first element of vector 1 is: 1
the first element of vector 2 is: 2
the first element of vector 3 is: 3
========= ERROR SUMMARY: 0 errors
$
EDIT: In the mcve you have now posted, you point out that an ordinary run of the code appears to give correct results, but when you use cuda-memcheck, errors are reported. You have a general design problem that will cause this.
In C++, when an object is defined within a curly-braces region:
{
{
Object A;
// object A is in-scope here
}
// object A is out-of-scope here
}
// object A is out of scope here
k<<<...>>>(anything that points to something in object A); // is illegal
and you exit that region, the object defined within the region is now out of scope. For objects with constructors/destructors, this usually means the destructor of the object will be called when it goes out-of-scope. For a thrust::device_vector (or std::vector) this will deallocate any underlying storage associated with that vector. That does not necessarily "erase" any data, but attempts to use that data are illegal and would be considered UB (undefined behavior) in C++.
When you establish pointers to such data inside an in-scope region, and then go out-of-scope, those pointers no longer point to anything that would be legal to access, so attempts to dereference the pointer would be illegal/UB. Your code is doing this. Yes, it does appear to give the correct answer, because nothing is actually erased on deallocation, but the code design is illegal, and cuda-memcheck will highlight that.
I suppose one fix would be to pull all this stuff out of the inner curly-braces, and put it at main scope, just like the d_nFluidNeighborsCrossFluids device_vector is. But you might also want to rethink your general data organization strategy and flatten your data.
You should really provide a minimal, complete, verifiable/reproducible example; yours is neither minimal, nor complete, nor verifiable.
I will, however, answer your side-question:
I know I cannot have device_vectors of device_vectors, for whichever underlying reason (explanation would be welcome)
While a device_vector regards a bunch of data on the GPU, it's a host-side data structure - otherwise you would not have been able to use it in host-side code. On the host side, what it holds should be something like: The capacity, the size in elements, the device-side pointer to the actual data, and maybe more information. This is similar to how an std::vector variable may refer to data that's on the heap, but if you create the variable locally the fields I mentioned above will exist on the stack.
Now, those fields of the device vector that are located in host memory are not generally accessible from the device-side. In device-side code you would typically use the raw pointer to the device-side data the device_vector manages.
Also, note that if you have a thrust::device_vector<T> v, each use of operator[] means a bunch of separate CUDA calls to copy data to or from the device (unless there's some caching going on under the hoold). So you really want to avoid using square-brackets with this structure.
Finally, remember that pointer-chasing can be a performance killer, especially on a GPU. You might want to consider massaging your data structure somewhat in order to make it amenable to flattening.
I'm working with an 3D array of dimension xdim=49, ydim=1024 and zdim=64. my DEVICE_MAX_WORK_ITEM_SIZES is only 512/512/512. If I declare my
size_t global_work_size = {xdim, ydim, zdim}; and launch an 3D kernel,
I'm getting wrong results since my ydim > 512. If all my dimensions are below 512, I'm getting the expected results. Please let me know if there's an alternative for this?
CL_DEVICE_MAX_WORK_ITEM_SIZES only limits the size of work groups, not the global work item size (yea, it's a terrible name for the constant). You are much more tightly restricted by CL_DEVICE_MAX_WORK_GROUP_SIZE which is the total number of items allowed in a work group (you'd typically hit this far sooner than CL_DEVICE_MAX_WORK_ITEM_SIZES because of multiplication.
So go ahead an launch your global work size of 49, 1024, 64. It should work. If it's not, you're using get_local_id instead of get_global_id or have some other bug. We regularly launch 2D kernels with 4096 x 4096 global work size.
See also Questions about global and local work size
If you don't use shared local memory, you don't need to worry about local work group sizes. In fact, you can pass NULL instead of a pointer to an array of sizes for local_work_size and let the runtime pick something (it helps if your global dimensions are easily divisible by small numbers).
Assuming the dimensions you provided are the size of your data, you can decrease the global work size by making each GPU thread calculate more data. What I mean is, every thread in your case will do one calculation and if you change your kernels to do let's say 2 calculations in y dimension, than you could cut the number of threads you are firing into half. The global_work_size decides how many threads in each direction you are executing. Let me give you an example:
Let's assume you have an array you want to do some calculations with and the array size you have is 2048. If you write your kernel in the following way, you are going to need 2048 as the global_work_size:
__kernel void calc (__global int *A, __global int *B)
{
int i = get_global_id(0);
B[i] = A[i] * 5;
}
The global work size in this case will be:
size_t global_work_size = {2048, 1, 1};
However, if you change your kernel into the following kernel, you can lower your global work size as well: ()
__kernel void new_calc (__global int *A, __global int *B)
{
int i = get_global_id(0);
for (int ind = 0; ind < 8; ind++)
B[i*8 + ind] = A[i*8 + ind] * 5;
}
Then this way, you can use global size as:
size_t global_work_size = {256, 1, 1};
Also with the second kernel, each of your threads will execute more work, resulting in more utilisation.
I'm new to OpenCL and in order to get a better grasp of a few concepts I contrived a simple example of a geometric progression as follows (emphasis on contrived):
An array of N values and N coefficients (whose values could be
anything, but in the example they all are the same) are allocated.
M steps are performed in sequence where each value in the values array
is multiplied by its corresponding coefficient in the coefficients
array and assigned as the new value in the values array. Each step needs to fully complete before the next step can complete. I know this part is a bit contrived, but this is a requirement I want to enforce to help my understanding of OpenCL.
I'm only interested in the values in the values array after the final step has completed.
Here is the very simple OpenCL kernel (MultiplyVectors.cl):
__kernel void MultiplyVectors (__global float4* x, __global float4* y, __global float4* result)
{
int i = get_global_id(0);
result[i] = x[i] * y[i];
}
And here is the host program (main.cpp):
#include <CL/cl.hpp>
#include <vector>
#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
int main ()
{
auto context = cl::Context (CL_DEVICE_TYPE_GPU);
auto *sourceFile = fopen("MultiplyVectors.cl", "r");
if (sourceFile == nullptr)
{
perror("Couldn't open the source file");
return 1;
}
fseek(sourceFile, 0, SEEK_END);
const auto sourceSize = ftell(sourceFile);
auto *sourceBuffer = new char [sourceSize + 1];
sourceBuffer[sourceSize] = '\0';
rewind(sourceFile);
fread(sourceBuffer, sizeof(char), sourceSize, sourceFile);
fclose(sourceFile);
auto program = cl::Program (context, cl::Program::Sources {std::make_pair (sourceBuffer, sourceSize + 1)});
delete[] sourceBuffer;
const auto devices = context.getInfo<CL_CONTEXT_DEVICES> ();
program.build (devices);
auto kernel = cl::Kernel (program, "MultiplyVectors");
const size_t vectorSize = 1024;
float coeffs[vectorSize] {};
for (size_t i = 0; i < vectorSize; ++i)
{
coeffs[i] = 1.000001;
}
auto coeffsBuffer = cl::Buffer (context, CL_MEM_READ_ONLY | CL_MEM_COPY_HOST_PTR, sizeof (coeffs), coeffs);
float values[vectorSize] {};
for (size_t i = 0; i < vectorSize; ++i)
{
values[i] = static_cast<float> (i);
}
auto valuesBuffer = cl::Buffer (context, CL_MEM_READ_WRITE | CL_MEM_COPY_HOST_PTR, sizeof (values), values);
kernel.setArg (0, coeffsBuffer);
kernel.setArg (1, valuesBuffer);
kernel.setArg (2, valuesBuffer);
auto commandQueue = cl::CommandQueue (context, devices[0]);
for (size_t i = 0; i < 1000000; ++i)
{
commandQueue.enqueueNDRangeKernel (kernel, cl::NDRange (0), cl::NDRange (vectorSize / 4), cl::NullRange);
}
printf ("All kernels enqueued. Waiting to read buffer after last kernel...");
commandQueue.enqueueReadBuffer (valuesBuffer, CL_TRUE, 0, sizeof (values), values);
return 0;
}
What I'm basically asking is for advice on how to best optimize this OpenCL program to run on a GPU. I have the following questions based on my limited OpenCL experience to get the conversation going:
Could I be handling the buffers better? I'd like to minimize any
unnecessary ferrying of data between the host and the GPU.
What's the optimal work group configuration (in general at least, I
know this can very by GPU)? I'm not actually sharing any data
between work items and it doesn't seem like I'd benefit from work
groups much here, but just in case.
Should I be allocating and loading anything into local memory for a
work group (if that would at all makes sense)?
I'm currently enqueing one kernel for each step, which will create a
work item for each 4 floats to take advantage of a hypothetical GPU with a SIMD
width of 128 bits. I'm attempting to enqueue all of this
asynchronously (although I'm noticing the Nvidia implementation I have
seems to block each enqueue until the kernel is complete) at once
and then wait on the final one to complete. Is there a whole better
approach to this that I'm missing?
Is there a design that would allow for only one call to
enqueueNDRangeKernel (instead of one call per step) while
maintaining the ability for each step to be efficiently processed in
parallel?
Obviously I know that the example problem I'm solving can be done in much better ways, but I wanted to have as simple of an example as possible that illustrated a vector of values being operated on in a series of steps where each step has to be completed fully before the next. Any help and pointers on how to best go about this would be greatly appreciated.
Thanks!
I am trying to write a mutex for OpenCL. The idea is for every single individual work item to be able to proceed atomically. Currently, I believe the problem may be that thread warps are unable to proceed when one thread in a warp gets the lock.
My current simple kernel below, for summing numbers. "numbers" is an array of floats as input. "sum" is a one element array for the result, and "semaphore" is a one element array for holding the semaphore. I based it heavily off the example here.
void acquire(__global int* semaphore) {
int occupied;
do {
occupied = atom_xchg(semaphore, 1);
} while (occupied>0);
}
void release(__global int* semaphore) {
atom_xchg(semaphore, 0); //the previous value, which is returned, is ignored
}
__kernel void test_kernel(__global float* numbers, __global float* sum, __global int* semaphore) {
int i = get_global_id(0);
acquire(semaphore);
*sum += numbers[i];
release(semaphore);
}
I am calling the kernel effectively like:
int numof_dimensions = 1;
size_t offset_global[1] = {0};
size_t size_global[1] = {4000}; //the length of the numbers array
size_t* size_local = NULL;
clEnqueueNDRangeKernel(command_queue, kernel, numof_dimensions,offset_global,size_global,size_local, 0,NULL, NULL);
As above, when running, the graphics card hangs, and the driver restarts itself. How can I fix it so that it doesn't?
What you are trying to do is not possible because of the GPU execution model, where all threads on a "processor" share the instruction pointer, even in branches. Here is a post that explains the problem in detail: http://vansa.ic.cz/author/admin/.
BTW, the example code that you found has the exact same problem and would never work.
The answer to this might seem obvious in retrospect, but it's not unless you thought of it.
Basically, the GPU's prediction of the ideal local group size (size of a thread warp) is greater than 1, and so thread warps lock up. To fix it, you just need to specify it to be 1 (i.e. "size_t size_local[1] = {1};"). Doing this produces a correct result.
This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.
Closed 10 years ago.
I haven't got good results with an iterative equation solving.
I am using a 2D array with "size_y" rows with "size_x" elements for each row.
The problem is that the code only does one iteration because the error cumulative is equal to zero. This cumulative error is computed in the kernel code for each cell of the array.
Here are 2 parts of sources files of this solving :
kernel code :
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#define min(a,b) a <= b ? a : b
// kernel code
const char *source =
"__kernel void line_compute(__global double diagx, __global double diagy,\
__global double weightx, __global double weighty, __global int size_x,\
__global double* tab_process, __global double* tab_new, __global double* r) {\
const unsigned int iy = get_global_id(0);\
const unsigned int ix = get_global_id(1);\
/* do computation */\
tab_process[iy*size_x+ix] = weighty *( tab_new[(iy-1)*size_x+ix] +\
tab_new[(i+1)*size_x+ix] + tab_new[iy*size_x+ix]*diagy)+\
weightx *( tab_new[iy*size_x+(ix-1)] + tab_new[iy*size_x+(ix+1)] + tab_new[iy*size_x+ix]*diagx) ; \
r[iy*size_x+ix] = 0;\
rk = tab_new[iy*size_x+ix] - tab_process[iy*size_x+ix];\
r[iy*size_x+ix] =r[iy*size_x+ix]+ rk * rk;\
tab_new[iy*size_x+ix] = tab_process[iy*size_x+ix]\
}";
At the execution, the cumulative error that I print with :
result = 0.0;
for(i=1;i<=size_x*size_y;i++)
{ result = result + r[i];
printf("r[%d]=%20.18f\n",i,r[i]);
}
printf("result=%f\n",result);
*error=result;
is equal to zero. That's why the code only does one iteration.
I don't understand where the probem is.
If anyone could see what's wrong.
Please, when posting questions to Stack Overflow, isolate only the relevant sections of the code (and format it properly). This one is too much for anyone to look at.
Besides prematurely updating the tab_new at the end of the kernel (you should do it only once after all threads have finished since neighbouring values are dependent), you have a syntax error in the kernel source:
tab_process[iy*size_x+ix] = weighty *( tab_new[(iy-1)*size_x+ix] +\
>>> tab_new[(i+1)*size_x+ix] <<< + tab_new[iy*size_x+ix]*diagy)+\
weightx *( tab_new[iy*size_x+(ix-1)] + tab_new[iy*size_x+(ix+1)] + tab_new[iy*size_x+ix]*diagx) ; \
You have mistakenly written i instead of iy. So the program would most likely not compile in clCreateProgramWithSource. Because you don't check the return code in ret, you miss that fact and then the following clCreateKernel and clEnqueueNDRangeKernel are also failing. With no kernel being executed the value of r_mem_obj remains the same as its initial value - all zeros, because it is a copy of r which, as a freshly allocated heap memory, is also all zeros (newly commited after a read fault pages on Linux are CoW mapped to a special all-zeros page in the kernel). Summing up all zeros gives zero.