I have an OpenCL Kernel that throws an OutOfResources exception when run.
Note: I am using Cloo for C#
I created a minimum working example of my problem and the kernel now looks like this:
__kernel void MinBug
(
__global float * img,
__global float * background,
__global int * tau
)
{
int neighbourhoodSize = tau[0];
const int x = get_global_id(0);
const int y = get_global_id(1);
for (int i = -neighbourhoodSize; i <= neighbourhoodSize; i++)
{
for (int j = -neighbourhoodSize; j <= neighbourhoodSize; j++)
{
//...
}
}
}
For my original program, this runs fine when tau is small (ie: 2, 10, 15), but once tau gets to be around 27, this sometimes throws an exception. The minimum working example I created does not have this problem until tau gets near 300.
The specific error that I get in my C# program is
Cloo.OutOfResourcesComputeException: 'OpenCL error code detected:
OutOfResources.'
This always happens on the very next line after calling the Kernel.Execute() method.
What concept am I missing?
Thanks to Huseyin for his advice on installing the correct runtime.
I also needed to select the correct platform in the code.
On my computer I currently have three platforms.
Two of them seem to be associated with the CPU (intel i7).
And one seems to be the GPU (NVidia gtx 660 ti)
I tried explicitly running on my GPU and it ran out of juice. As you can see from the error message above.
When I specified the CPU
CLCalc.InitCL(Cloo.ComputeDeviceTypes.Cpu, 1);
It ran much better. Who'd have thought, my CPU seems to have more grunt than the GPU. Maybe its a simplistic metric. Its also worth noting my CPU supports a later version of OpenCL than the GPU.
Related
I am trying to learn how to use the Sparse Coding algorithm with the mlpack library. When I call Encode() on my instance of mlpack::sparse_coding:SparseCoding, I get the error
[WARN] There are 63 inactive atoms. They will be reinitialized randomly.
error: solve(): solution not found
Is it simply that the algorithm cannot learn a latent representation of the data. Or perhaps it is my usage? The relevant section follows
EDIT: One line was modified to fix an unrelated error, but the original error remains.
double* Application::GetSparseCodes(arma::mat* trainingExample, int atomCount)
{
double* latentRep = new double[atomCount];
mlpack::sparse_coding::SparseCoding<mlpack::sparse_coding::DataDependentRandomInitializer> sc(*trainingExample, Utils::ATOM_COUNT, 1.0);
sc.Encode(Utils::MAX_ITERATIONS);
arma::mat& latentRepMat = sc.Codes();
for (int i = 0; i < atomCount; i++)
latentRep[i] = latentRepMat.at(i, 0);
return latentRep;
}
Some relevant parameters
const static int IMAGE_WIDTH = 20;
const static int IMAGE_HEIGHT = 20;
const static int PIXEL_COUNT = IMAGE_WIDTH * IMAGE_HEIGHT;
const static int ATOM_COUNT = 64;
const static int MAX_ITERATIONS = 100000;
This could be one of a handful of issues but given the description it's a little difficult to tell which of these it is (or if it is something else entirely). However, these three ideas should provide a good place to start:
Matrices in mlpack are column-major. That means each observation should represent a column. If you use mlpack::data::Load() to load, e.g., a CSV file (which are generally one row per observation), it will automatically transpose the dataset. SparseCoding will act oddly if you pass it transposed data. See also http://www.mlpack.org/doxygen.php?doc=matrices.html.
If there are 63 inactive atoms, then only one atom is actually active (given that ATOM_COUNT is 64). This means that the algorithm has found that the best way to represent the dictionary (at a given step) uses only one atom. This could happen if the matrix you are passing consists of all zeros.
mlpack will provide verbose output, which may also be helpful for debugging. Usually this is used by using mlpack's CLI class to parse command-line input, but you can enable verbose output with mlpack::Log::Info.ignoreInput = false. You may obtain a lot of output that way, but it will give a better look at what is going on...
The mlpack project has its own mailing list where you may be likely to get a quicker or more comprehensive response, by the way.
I have a weird problem, so I thought I would ask and see if someone more experienced than me could see a solution.
I am writing a program with CUDA C/C++, and I have some constant integers that specify various things, like coordinates of the bounds of the calculation, etc.. Currently I just have those things in global device memory. They are accessed by every thread in every kernel call, and so I figured that if they are in global memory, then they never are being cached or broadcast (right?). And so these little integers are taking up a lot (relatively) of overhead, and have a lot of 'read redundancy.'
So I declare in a header:
__constant__ int* number;
I include that header, and, when I do memory stuff, I do:
cutilSafeCall( cudaMemcpyToSymbol(number, &(some_host_int), sizeof(int) );
I pass number into all my kernel's then:
__global__ void magical_kernel(int* number, ...){
//and I access 'number' like this
int data_thingy = big_array[ *number ];
}
My code crashes. With number in global memory, it is just fine. I have determined that it crashes sometime upon accessing number within the kernel. This means that either I am accessing or allocating it wrong. If it holds the wrong value, it will also cause a crash, because it is used to index into arrays.
To conclude, I will ask a few questions. First, what am I doing wrong? As a bonus: is there a better way than constant memory to accomplish this task - I don't know the value of number at compile time, so a simple #define won't work. Will constant memory even speed the code up at all, or has it been cached and broadcasted all along? Could I somehow put the data in shared memory for each threadblock and have it remain in shared memory through multiple kernel calls?
There are several problems here:
You have declared number as a pointer, but never assigned it a value which is valid address in GPU memory
You have a variable scope onflict: the argument variable int * number defined in magic_kernel is not the same variable as the __constant__ int * variable defined as compilation unit scope.
The first argument of the cudaMemcpyToSymbol call is almost certainly incorrect.
If you don't understand why either of the first two point are true, you have some revision to do on pointers and scope in C++.
Based on your response to a now deleted answer, I suspect what you are actually trying to do is this:
__constant__ int number;
__global__ void magical_kernel(...){
int data_thingy = big_array[ number ];
}
cudaMemcpyToSymbol("number", &(some_host_int), sizeof(int));
i.e. number is intended to be an integer in constant memory, not a pointer, and not a kernel argument.
EDIT: here is an exmaple which shows this in action:
#include <cstdio>
__constant__ int number;
__global__ void magical_kernel(int * out)
{
out[threadIdx.x] = number;
}
int main()
{
const int value = 314159;
const size_t sz = size_t(32) * sizeof(int);
cudaMemcpyToSymbol("number", &value, sizeof(int));
int * _out, * out;
out = (int *)malloc(sz);
cudaMalloc((void **)&_out, sz);
magical_kernel<<<1,32>>>(_out);
cudaMemcpy(out, _out, sz, cudaMemcpyDeviceToHost);
for(int i=0; i<32; i++)
fprintf(stdout, "%d %d\n", i, out[i]);
return 0;
}
You should be able to run this yourself and confirm it works as advertised.
I'm running the OpenCL kernel below with a two-dimensional global work size of 1000000 x 100 and a local work size of 1 x 100.
__kernel void myKernel(
const int length,
const int height,
and a bunch of other parameters) {
//declare some local arrays to be shared by all 100 work item in this group
__local float LP [length];
__local float LT [height];
__local int bitErrors = 0;
__local bool failed = false;
//here come my actual computations which utilize the space in LP and LT
}
This however refuses to compile, since the parameters length and height are not known at compile time. But it is not clear to my at all how to do this correctly. Should I use pointers with memalloc? How to handle this in a way that the memory is only allocated once for the entire workgroup and not once per work item?
All that I need is 2 arrays of floats, 1 int and 1 boolean that are shared among the entire workgroup (so all 100 work items). But I fail to find any method that does this correctly...
It's relatively simple, you can pass the local arrays as arguments to your kernel:
kernel void myKernel(const int length, const int height, local float* LP,
local float* LT, a bunch of other parameters)
You then set the kernelargument with a value of NULL and a size equal to the size you want to allocate for the argument (in byte). Therefore it should be:
clSetKernelArg(kernel, 2, length * sizeof(cl_float), NULL);
clSetKernelArg(kernel, 3, height* sizeof(cl_float), NULL);
local memory is always shared by the workgroup (as opposed to private), so I think the bool and int should be fine, but if not you can always pass those as arguments too.
Not really related to your problem (and not necessarily relevant, since I do not know what hardware you plan to run this on), but at least gpus don't particulary like workingsizes which are not a multiple of a particular power of two (I think it was 32 for nvidia, 64 for amd), meaning that will probably create workgroups with 128 items, of which the last 28 are basically wasted. So if you are running opencl on gpu it might help performance if you directly use workgroups of size 128 (and change the global work size appropriately)
As a side note: I never understood why everyone uses the underscore variant for kernel, local and global, seems much uglier to me.
You could also declare your arrays like this:
__local float LP[LENGTH];
And pass the LENGTH as a define in your kernel compile.
int lp_size = 128; // this is an example; could be dynamically calculated
char compileArgs[64];
sprintf(compileArgs, "-DLENGTH=%d", lp_size);
clBuildProgram(program, 0, NULL, compileArgs, NULL, NULL);
You do not have to allocate all your local memory outside the kernel, especially when it is a simple variable instead of a array.
The reason that your code cannot compile is that OpenCL does not support local memory initialization. This is specified in the document(https://www.khronos.org/registry/cl/sdk/1.1/docs/man/xhtml/local.html). It is also not feasible in CUDA(Is there a way of setting default value for shared memory array?)
ps:The answer from Grizzly is good enough and it would be better if I can post it as a comment, but I am restricted by the reputation policy. Sorry.
I m currently working on a painting app on ios.
I use a directly draw into a NSMutableData buffer and apply blending with my brush like this:
- (void) combineColorDestination:(unsigned char*) dest source:(unsigned char*) src
{
const unsigned char sra = ((unsigned char *)src)[3];
const float oneminusalpha = 1.0f - (sra / 255.f);
int d[4];
for (int i=0;i<4;i++)
{
d[i] = oneminusalpha * ((unsigned char *)dest)[i] + ((unsigned char *)src)[i];
if (d[i]>255)
d[i] = 255;
((unsigned char *)dest)[i] = (unsigned char)d[i];
}
}
Any suggestions for optimisations ?
I previously tried to use neon , but i ve got a bug I wasnt able to fix (the bordering pixels was buggy)
I was iterating pixels 2 by 2 like this :
uint8x8_t va = vld1_u8(dest);
uint8x8_t vb = vld1_u8(src);
uint8x8_t res = vqadd_u8(va,vb);
vst1_u8(dest, res);
Suggestions? Alright. Note that these are valid whichever multimedia manipulation you are doing and is hardly restricted to your case.
First, before you even do NEON, you should change your code to have one function that changes a bunch of pixels (at least a row, a rectangle if you can) at once, instead of a function (or method - even worse) that changes one pixel and is called a bunch of times: somehow I doubt the brush is only 1x1 pixel.
Second, except for the column loop (and eventual row loop), there should be no branch (that is, flow control structures). No for (i=0;i<4;i++); just write the code for the four channels in sequence (use a macro if necessary). No if (d[i]>255); express that as an alternative: dest[i] = (temp>255?255:temp); at the very least, if not replacing it by a more efficient way to do saturation (tricks using subtractions, shifts, and masks exist).
Third, avoid any conversion between floating-point and integer; this is always valid advice, but float->int conversions are particularly devastating on ARM. Since you're manipulating integers, this means foregoing floating-point here.
And once you've done that, surprise, besides making your code faster you have in fact done the preparation work for NEON: NEON is only remotely useful if you process a bunch of pixels at once, if there is no branch, and if you don't convert between floating-point and integer all over the place. So only then will we talk about NEON, if it is even necessary at this point.
I am using CUDA SDK 3.1 on MS VS2005 with GPU GTX465 1 GB. I have such a kernel function:
__global__ void CRT_GPU_2(float *A, float *X, float *Y, float *Z, float *pIntensity, float *firstTime, float *pointsNumber)
{
int holo_x = blockIdx.x*20 + threadIdx.x;
int holo_y = blockIdx.y*20 + threadIdx.y;
float k=2.0f*3.14f/0.000000054f;
if (firstTime[0]==1.0f)
{
pIntensity[holo_x+holo_y*MAX_FINAL_X]=0.0f;
}
for (int i=0; i<pointsNumber[0]; i++)
{
pIntensity[holo_x+holo_y*MAX_FINAL_X]=pIntensity[holo_x+holo_y*MAX_FINAL_X]+A[i]*cosf(k*sqrtf(pow(holo_x-X[i],2.0f)+pow(holo_y-Y[i],2.0f)+pow(Z[i],2.0f)));
}
__syncthreads();
}
and this is function which calls kernel function:
extern "C" void go2(float *pDATA, float *X, float *Y, float *Z, float *pIntensity, float *firstTime, float *pointsNumber)
{
dim3 blockGridRows(MAX_FINAL_X/20,MAX_FINAL_Y/20);
dim3 threadBlockRows(20, 20);
CRT_GPU_2<<<blockGridRows, threadBlockRows>>>(pDATA, X, Y, Z, pIntensity,firstTime, pointsNumber);
CUT_CHECK_ERROR("multiplyNumbersGPU() execution failed\n");
CUDA_SAFE_CALL( cudaThreadSynchronize() );
}
I am loading in loop all the paramteres to this function (for example 4096 elements for each parameter in one loop iteration). In total I want to make this kernel for 32768 elements for each parameter after all loop iterations.
The MAX_FINAL_X is 1920 and MAX_FINAL_Y is 1080.
When I am starting alghoritm first iteration goes very fast and after one or two iteration more I get information about CUDA timeout error. I used this alghoritm on GPU gtx260 and it was doing better as far as I remember...
Could You help me.. maybe I am doing some mistake according to new Fermi arch in this algorithm?
It will be better to call
CUT_CHECK_ERROR after
cudaThreadSynchronize(). Because
kernel run asynchronous and you must
wait for kernel ending to know about
errors... Maybe in second iteration you receive an error
from first kernel usage.
Be sure
that you have some valid number in the most interesting variable
pointsNumber[0] (it might cause a
long internal loop).
You could also
improve speed of your kernel
function:
Use better blocks. Threads configuration 20x20 will cause very slow memory usage (see Programming Guide and Best Practices). Try to use blocks 16x16.
Do not use pow(..., 2.0) function. It's faster to use SQR macro (#define SQR(x) (x)*(x))
You don't use shared mem, so __syncthreads() is not required.
PS: You could also pass value parameters to CUDA functions, not only pointers. Speed will be the same.
PPS: please improve code's readability... Now you must edit six places to change block configuration... Inside the kernel you could use blockDim variable and you could use constants in go2 function.
You could also use bool firstTime - it will be MUCH better then float.
Is your GPU connected to a display? If so, I believe the default is that kernel execution will be aborted after 5 seconds. You can check whether kernel execution will timeout by using cudaGetDeviceProperties - see reference page
In kernel's cycle you write in the same array, from which you read - for global memory usage it is the worst, because warps from different blocks wait for each other.