By definition, a thread is a path of execution within a process.
But during the implementation of a kernel, a thread_id or global_index is generated to access a memory location allocated. For instance, in the Matrix Multiplication code below, ROW and COL are generated to access matrix A and B sequential.
My doubt here is, index generated isn't pointing to a thread(by definition), instead, it is used to access the location of the data in the memory, then why do we refer to it as thread index or global thread index and why not memory index or something else?
__global__ void matrixMultiplicationKernel(float* A, float* B, float* C, int N) {
int ROW = blockIdx.y*blockDim.y+threadIdx.y;
int COL = blockIdx.x*blockDim.x+threadIdx.x;
float tmpSum = 0;
if (ROW < N && COL < N) {
// each thread computes one element of the block sub-matrix
for (int i = 0; i < N; i++) {
tmpSum += A[ROW * N + i] * B[i * N + COL];
}
}
C[ROW * N + COL] = tmpSum;
}
This question seems to be mostly about semantics, so let's start at Wikipedia
.... a thread of execution is the smallest sequence of programmed
instructions that can be managed independently by a scheduler ....
That is pretty much describes exactly what s thread in CUDA is -- the kernel is the sequence of instructions, and the scheduler is the warp/thread scheduler in each streaming multiprocessor on the GPU.
The code in your question is calculating the unique ID of the thread in the kernel launch, as it is abstracted in the CUDA programming/execution model. It has no intrinsic relationship to memory layouts, only to the unique ID in the kernel launch. The fact it is being used to ensure that each parallel operation is being performed on a different memory location is programming technique and nothing more.
Thread ID seems like a logical moniker to me, but to paraphrase Miles Davis when he was asked what the name of the jam his band just played at the Isle of Wight festival in 1970: "call it whatever you want".
Related
Recently I have faced a problem when the program running the memory keep increasing, and when program is closed the memory would restore normal level. Obviously, it's a memory leak. After some work, I have located the code responsible, but I don't know why? The program's work flow is simple:
first use lidar api to get point cloud and image data;
then transport to next tbb flow graph to process these data;
finally use open3d api to visualzie them.
In the first step, the lidar itself's api use asio to asynchronously invoke some callback function to transport data, so I create some tbb concurrent_queue to store these data, and a align function to match cloud and image with timestamp. The problem is in the align function. In the function, I create a vector<shared_ptr<open3d::..::PointCloud>> and use iterator to store point cloud elements. However, I found when the function complete, the shared_ptr use count don't reduce . Similar but simpler example code like this:
std::pair<std::shared_ptr<int>, int> helper() {
auto a = std::make_shared<int>(90);
auto c = 100;
std::vector<std::pair<std::shared_ptr<int>, int>> container;
container.reserve(5);
auto iter = container.begin();
for (int i = 0; i < 3; i++) {
*iter = std::make_pair(a, c);
iter++;
}
return *(iter-1);
}
int main() {
auto b = helper();
std::cout << "shared_ptr use count: " << std::get<0>(b).use_count() << std::endl;
return 0;
}
Ubuntu 20.04 + gcc 9.4, the print result is shared_ptr use count: 4.
Why the vector can't be auto destroyed when function is completed? Hope someone kindly explain this problem.
Thanks #Retired Ninja! The root of the problem is vector.reserve just reserve capacity not physical space. So the vector space after reserve is 0. The following iterator operation is assumed point to some undifined memory. While the result can be transport to main function with no value error, the shared_ptr use count can't reduce to 1 after function call.
To solve the problem, One can just modify the reserve to resize, which can change physical space of the vector and iterator point to defined memory space. Or avoid use iterator, just use push_back and return back().
I have a problem where I need to do a linear interpolation on some data as it is acquired from a sensor (it's technically position data, but the nature of the data doesn't really matter). I'm doing this now in matlab, but since I will eventually migrate this code to other languages, I want to keep the code as simple as possible and not use any complicated matlab-specific/built-in functions.
My implementation initially seems OK, but when checking my work against matlab's built-in interp1 function, it seems my implementation isn't perfect, and I have no idea why. Below is the code I'm using on a dataset already fully collected, but as I loop through the data, I act as if I only have the current sample and the previous sample, which mirrors the problem I will eventually face.
%make some dummy data
np = 109; %number of data points for x and y
x_data = linspace(3,98,np) + (normrnd(0.4,0.2,[1,np]));
y_data = normrnd(2.5, 1.5, [1,np]);
%define the query points the data will be interpolated over
qp = [1:100];
kk=2; %indexes through the data
cc = 1; %indexes through the query points
qpi = qp(cc); %qpi is the current query point in the loop
y_interp = qp*nan; %this will hold our solution
while kk<=length(x_data)
kk = kk+1; %update the data counter
%perform online interpolation
if cc<length(qp)-1
if qpi>=y_data(kk-1) %the query point, of course, has to be in-between the current value and the next value of x_data
y_interp(cc) = myInterp(x_data(kk-1), x_data(kk), y_data(kk-1), y_data(kk), qpi);
end
if qpi>x_data(kk), %if the current query point is already larger than the current sample, update the sample
kk = kk+1;
else %otherwise, update the query point to ensure its in between the samples for the next iteration
cc = cc + 1;
qpi = qp(cc);
%It is possible that if the change in x_data is greater than the resolution of the query
%points, an update like the above wont work. In this case, we must lag the data
if qpi<x_data(kk),
kk=kk-1;
end
end
end
end
%get the correct interpolation
y_interp_correct = interp1(x_data, y_data, qp);
%plot both solutions to show the difference
figure;
plot(y_interp,'displayname','manual-solution'); hold on;
plot(y_interp_correct,'k--','displayname','matlab solution');
leg1 = legend('show');
set(leg1,'Location','Best');
ylabel('interpolated points');
xlabel('query points');
Note that the "myInterp" function is as follows:
function yi = myInterp(x1, x2, y1, y2, qp)
%linearly interpolate the function value y(x) over the query point qp
yi = y1 + (qp-x1) * ( (y2-y1)/(x2-x1) );
end
And here is the plot showing that my implementation isn't correct :-(
Can anyone help me find where the mistake is? And why? I suspect it has something to do with ensuring that the query point is in-between the previous and current x-samples, but I'm not sure.
The problem in your code is that you at times call myInterp with a value of qpi that is outside of the bounds x_data(kk-1) and x_data(kk). This leads to invalid extrapolation results.
Your logic of looping over kk rather than cc is very confusing to me. I would write a simple for loop over cc, which are the points at which you want to interpolate. For each of these points, advance kk, if necessary, such that qp(cc) is in between x_data(kk) and x_data(kk+1) (you can use kk-1 and kk instead if you prefer, just initialize kk=2 to ensure that kk-1 exists, I just find starting at kk=1 more intuitive).
To simplify the logic here, I'm limiting the values in qp to be inside the limits of x_data, so that we don't need to test to ensure that x_data(kk+1) exists, nor that x_data(1)<pq(cc). You can add those tests in if you wish.
Here's my code:
qp = [ceil(x_data(1)+0.1):floor(x_data(end)-0.1)];
y_interp = qp*nan; % this will hold our solution
kk=1; % indexes through the data
for cc=1:numel(qp)
% advance kk to where we can interpolate
% (this loop is guaranteed to not index out of bounds because x_data(end)>qp(end),
% but needs to be adjusted if this is not ensured prior to the loop)
while x_data(kk+1) < qp(cc)
kk = kk + 1;
end
% perform online interpolation
y_interp(cc) = myInterp(x_data(kk), x_data(kk+1), y_data(kk), y_data(kk+1), qp(cc));
end
As you can see, the logic is a lot simpler this way. The result is identical to y_interp_correct. The inner while x_data... loop serves the same purpose as your outer while loop, and would be the place where you read your data from wherever it's coming from.
I have encountered a strange behavior of the torch.mm function in Lua/Torch. Here is a simple program that demonstrates the problem.
iteration = 0;
a = torch.Tensor(2, 2);
b = torch.Tensor(2, 2);
prod = torch.Tensor(2,2);
a:zero();
b:zero();
repeat
prod = torch.mm(a,b);
ent = prod[{2,1}];
iteration = iteration + 1;
until ent ~= ent
print ("error at iteration " .. iteration);
print (prod);
The program consists of one loop, in which the program multiplies two zero 2x2 matrices and tests if entry ent of the product matrix is equal to nan. It seems that the program should run forever since the product should always be equal to 0, and hence ent should be 0. However, the program prints:
error at iteration 548
0.000000 0.000000
nan nan
[torch.DoubleTensor of size 2x2]
Why is this happening?
Update:
The problem disappears if I replace prod = torch.mm(a,b) with torch.mm(prod,a,b), which suggests that something is wrong with the memory allocation.
My version of Torch was compiled without BLAS & LAPACK libraries. After I recompiled torch with OpenBLAS, the problem disappeared. However, I am still interested in its cause.
The part of code that auto-generates the Lua wrapper for torch.mm can be found here.
When you write prod = torch.mm(a,b) within your loop it corresponds to the following C code behind the scenes (generated by this wrapper thanks to cwrap):
/* this is the tensor that will hold the results */
arg1 = THDoubleTensor_new();
THDoubleTensor_resize2d(arg1, arg5->size[0], arg6->size[1]);
arg3 = arg1;
/* .... */
luaT_pushudata(L, arg1, "torch.DoubleTensor");
/* effective matrix multiplication operation that will fill arg1 */
THDoubleTensor_addmm(arg1,arg2,arg3,arg4,arg5,arg6);
So:
a new result tensor is created and resized with the proper dimensions,
but this new tensor is NOT initialized, i.e. there is no calloc or explicit fill here so it points to junk memory and could contain NaN-s,
this tensor is pushed on the stack so as to be available on the Lua side as the return value.
The last point means that this returned tensor is different from the initial prod one (i.e. within the loop, prod shadows the initial value).
On the other hand calling torch.mm(prod,a,b) does use your initial prod tensor to store the results (behind the scenes there is no need to create a dedicated tensor in that case). Since in your code snippet you do not initialize / fill it with given values it could also contain junk.
In both cases the core operation is a gemm multiplication like C = beta * C + alpha * A * B, with beta=0 and alpha=1. The naive implementation looks like that:
real *a_ = a;
for(i = 0; i < m; i++)
{
real *b_ = b;
for(j = 0; j < n; j++)
{
real sum = 0;
for(l = 0; l < k; l++)
sum += a_[l*lda]*b_[l];
b_ += ldb;
/*
* WARNING: beta*c[j*ldc+i] could give NaN even if beta=0
* if the other operand c[j*ldc+i] is NaN!
*/
c[j*ldc+i] = beta*c[j*ldc+i]+alpha*sum;
}
a_++;
}
Comments are mine.
So:
with torch.mm(a,b): at each iteration, a new result tensor is created without being initialized (it could contain NaN-s). So every iteration presents a risk of returning NaN-s (see above warning),
with torch.mm(prod,a,b): there is the same risk since you do not initialized the prod tensor. BUT: this risk only exists at the first iteration of the repeat / until loop since right after prod is filled with 0-s and re-used for the subsequent iterations.
So this is why you do not observe a problem here (it is less frequent).
In case 1: this should be improved at the Torch level, i.e. make sure the wrapper initializes the output (e.g. with THDoubleTensor_fill(arg1, 0);).
In case 2: you should initialize prod initially and use the torch.mm(prod,a,b) construct to avoid any NaN problem.
--
EDIT: this problem is now fixed (see this pull request).
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 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.