C/CUDA Program Output - c++

The following is a CUDA programming example which is basically C but with NVidia CUDA functions within. I've been trying to interpret this code example and figure out what it is trying to do. My question is this the program compiles just fine, but what arguments does it take? For example this CUDA program is being run in a linux emulator however upon running ./program it returns:
Usage: ./program number
Segmentation fault
What are the programs input arguments. Thank you.
#include <assert.h>
#include <stdio.h>
//#define N 100000
__host__ void saxpy_host(int length, float alpha, float * x, float * y)
{
for (int i = 0; i < length; ++i)
y[i] = alpha*x[i] + y[i];
}
__global__ void saxpy (int length, float alpha, float * x, float * y)
{
int i;
i = blockIdx.x*blockDim.x + threadIdx.x;
if (i < length) y[i] = alpha*x[i]+y[i];
__syncthreads();
}
int main(int argc, char* argv[]) {
if (argc != 2) {
printf("Usage: %s number\n", argv[0]);
return -1;
}
int N = atoi(argv[1]);
// host data
float alpha = 0.5;
float x[N], xback[N];
float y[N], yback[N];
int size;
int i;
int blocks;
// determining size
size = sizeof(float)*N;
// device data
float * dxp, * dyp;
// fill host data
for (i = 0; i < N; i++) {
x[i] = (float) (rand () % 128);
y[i] = (float) (rand () % 256);
}
// Allocating and Moving data to device
cudaMalloc((void**) &dxp, size);
cudaMalloc((void**) &dyp, size);
cudaMemcpy (dxp, x, size, cudaMemcpyHostToDevice);
cudaMemcpy (dyp, y, size, cudaMemcpyHostToDevice);
// size of thread blocks
blocks = (N + 31)/32;
saxpy <<< blocks, 32 >>> (N, alpha, dxp, dyp);
// bring back data
cudaMemcpy (xback, dxp, size, cudaMemcpyDeviceToHost);
cudaMemcpy (yback, dyp, size, cudaMemcpyDeviceToHost);
// Calculating host SAXPY
saxpy_host (N, alpha, (float *) &x, (float *) &y);
// checking computation on host matches computation on GPU
for (i = 0; i < N; i++) {
assert (yback[i] == y[i]) ;
//printf ("%i %f %f \n", i, yback[i], y[i]);
}
// free device data
cudaFree(dxp); cudaFree(dyp);
return 0;
}

int N = atoi(argv[1]);
The program takes a single integer as a command line argument. (Try calling it as ./program 5, for example.)
It then calculates a SAXPY (An old term originating from early BLAS implementations, but it stuck. It means "single (precision, aka float) real alpha x plus y".) with vectors of dimension N.

Related

Allocate 1 Dimension array with cudaMallocPitch and then copy to device with cudaMemcpy2D 3

I have read this post Allocate 2D array with cudaMallocPitch and copying with cudaMemcpy2D among many others including NVIDIA docs and I can't get cudaMallocPitch to work together with cudaMemcpy2D.
I need to copy a very big matrix in an array format (Matrix[width*height]) along with a simple array to perform Matrix * vector operations. It is not optional for me to use cudaMallocPitch in order to avoid conflicts and have a better performance.
So, I started by just trying to copy the matrix (vector in my case) to the device and check if it was correctly copied but my code does not print anything. If I use cudaMalloc and cudaMemcpy everything works fine. But I do not know what to do with cudaMallocPitch and cudaMemcpy2D.
What can I do to fix this?
#include <stdio.h>
__global__ void kernel(size_t mpitch, double * A, int N)
{
int idx = threadIdx.x + blockIdx.x * blockDim.x;
while (idx < N)
{
double e = *(double *)(((char *) A + idx * mpitch) + N);
printf("(%f)", e);
}
}
int main()
{
int N = 1500;
double * A = new double[N], * d_A;
size_t pitch;
for (int i = 0; i < N; ++i)
{
A[i] = i;
}
cudaMallocPitch(&d_A, &pitch, sizeof(double) * N, 1);
cudaMemcpy2D(d_A, pitch, A, N * sizeof(double), sizeof(double) * N, 1, cudaMemcpyHostToDevice);
unsigned int blocksize = 1024;
unsigned int nblocks = (N + blocksize - 1) / blocksize;
kernel <<<nblocks, blocksize>>>(pitch, d_A, N);
cudaFree(d_A);
delete [] A;
return 0;
}
Error checking can make a big difference in debugging. You should always use it before coming here.
It wasn't clear if you wanted a row or column vector i.e. a matrix of [1xN] or [Nx1]
I've added an explanation on Talomnies suggestion, but first the 'working slabs of code'
Here's [Nx1]
#include <cstdio>
#include <iostream>
#include <cuda.h>
using namespace std;
__global__ void kernel(size_t mpitch, double * A, int N)
{
int idx = threadIdx.x + blockIdx.x * blockDim.x;
if(idx>=N) return;
double e = *(double *)(((char *) A + idx * mpitch));
printf("(%f)", e);
}
int main()
{
int N = 15;
double * A = new double[N], * d_A;
size_t pitch;
for (int i = 0; i < N; ++i)
{
A[i] = i;
}
cudaError_t err = cudaMallocPitch(&d_A, &pitch, sizeof(double), N);
if(err!=cudaSuccess) cout<<"err0:"<<cudaGetErrorString(err)<<endl;
err = cudaMemcpy2D(d_A, pitch, A, sizeof(double), sizeof(double), N, cudaMemcpyHostToDevice);
if(err!=cudaSuccess) cout<<"err1:"<<cudaGetErrorString(err)<<endl;
unsigned int blocksize = 1024;
unsigned int nblocks = (N + blocksize - 1) / blocksize;
kernel <<<nblocks, blocksize>>>(pitch, d_A, N);
cudaDeviceSynchronize();
err = cudaGetLastError();
if(err!=cudaSuccess) cout<<"err2:"<<cudaGetErrorString(err)<<endl;
cudaFree(d_A);
delete [] A;
return 0;
}
[1xN]:
#include <cstdio>
#include <iostream>
#include <cuda.h>
using namespace std;
__global__ void kernel(size_t mpitch, double * A, int N)
{
int idx = threadIdx.x + blockIdx.x * blockDim.x;
if(idx>=N) return;
int row=0;//only one row
double *row_ptr = (double *)( (char *) (A + mpitch * row) );
double e = row_ptr[idx];
printf("(%f)", e);
}
int main()
{
int N = 15;
double * A = new double[N], * d_A;
size_t pitch;
for (int i = 0; i < N; ++i)
{
A[i] = i;
}
cudaError_t err = cudaMallocPitch(&d_A, &pitch, sizeof(double)*N, 1);
if(err!=cudaSuccess) cout<<"err0:"<<cudaGetErrorString(err)<<endl;
err = cudaMemcpy2D(d_A, pitch, A, sizeof(double)*N, sizeof(double)*N, 1, cudaMemcpyHostToDevice);
if(err!=cudaSuccess) cout<<"err1:"<<cudaGetErrorString(err)<<endl;
unsigned int blocksize = 1024;
unsigned int nblocks = (N + blocksize - 1) / blocksize;
kernel <<<nblocks, blocksize>>>(pitch, d_A, N);
cudaDeviceSynchronize();
err = cudaGetLastError();
if(err!=cudaSuccess) cout<<"err2:"<<cudaGetErrorString(err)<<endl;
cudaFree(d_A);
delete [] A;
return 0;
}
Explanation
Firslty, Error Handling:
Considering how easy error handling is in CUDA there isn't a good excuse not to put it in.
cudaError_t err = cudaMallocPitch(&d_A, &pitch, sizeof(double)*N, 1);
if(err!=cudaSuccess) cout<<"err0:"<<cudaGetErrorString(err)<<endl;
Second, you didn't specify if you wanted a column vector or a row vector. Since a row vector is simply a 1-D array in linear memory and you don't need pitched memory to do that, I will assume for this explanation that you meant a column vector.
The reoccurring problem you were having was "misaligned address" in the kernel. This indicates that the problem is book-keeping, so lets walk through the three major steps of handling an aligned 2D array (even though our arrays will be either a column or row vector).
Allocating:
Your allocation was written out as
cudaMallocPitch(&d_A, &pitch, sizeof(double) * N, 1);
This is correct for the row vector as the API is cudaMallocPitch(void*** pointer, size_t* pitch_return, size_t row_width_in_bytes, size_t count_of_rows) However if we would like to do a column vector correct call is
cudaMallocPitch(&d_A, &pitch, sizeof(double), N);
Accessing:
For accessing you were mixing up accessing a row, and accessing an element in the row.
double e = *(double *)(((char *) A + idx * mpitch) + N);
Once again stick to the documentation. The API documentation for cudaMallocPitch includes
T* pElement = (T*)((char*)BaseAddress + Row * pitch) + Column;
for us this translates into
int column=0;
double element=(double*) ((char*)A + idx * mpitch) + column;
I've used column = 0 for completeness since we do not have more than one column.
Copying:
cudaMemcpy2D(d_A, pitch, A, N * sizeof(double), sizeof(double) * N, 1, cudaMemcpyHostToDevice);
For this case this is correct. API for cudaMemcpy2D is
cudaMemcpy2D(void* destination, size_t pitch_from_mallocPitch, const void* source, size_t source_pitch_bytes, size_t src_width_in_bytes, size_t src_rows_count, enum type_of_xfer);

Fast calculation of many scalar products [closed]

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I have a program that calculates 1-10 million scalar products.
It looks like this. ts and A are arrays of about 1000-10000 3D points (each element is a 3x1 vector). For the moment, with ts.size() = 10,000 and A.size() = 1000, my code takes about 41ms. I have not done any parallelization so far. Will the calculations be much faster, for example, in CUDA? I have no such experience. Or is there any other way? Thanks.
for(int i = 0; i< ts.size(); i++){
for(int j = 0; j< A.size(); j++){
if(abs(scalarProduct(ts.at(i), A.at(j))) <epsilon){
score[i] +=1;
}
}
}
This is my implementation of the scalar product.
double scalarProduct(const Point &p1,const Point &p2)
{
return (p1.getX()*p2.getX() + p1.getY()*p2.getY() + p1.getZ()*p2.getZ()) ;
}
Could I use Lapack or Eigen instead, formulating the problem as matrix multiplication? I've done that in Matlab and it is only 5 times slower. Any speedup would be great. With OpenMP i guess I could be 4x faster.
This answer consists of two parts:
Accelerating the calculation of many independent scalar products;
Solving your specific problem.
PART 1
The problem of calculating a large number of independent scalar products is an embarassingly parallel problem. If you aim at accelerating only the mentioned scalar products, retaining the rest of the computation on the CPU, then I agree with Calvin that most of the time will be spent in device-> memory transaction of the large N*M resulting matrix. However, if you purge your timing from the mentioned transaction, accelerating the calculations will be worth. This is shown by the code below, tested on an Intel Xeon E5-2650 2.00 GHz, Eight core processor equipped with an NVIDIA Kepler K20c cards, and whose timing is the following:
CPU: 27ms; GPU (without D2H transaction): 0.08ms; GPU (with D2H transaction): 23ms
#include <stdio.h>
#include <time.h>
#define BLOCKSIZE_X 16
#define BLOCKSIZE_Y 16
/********************/
/* CUDA ERROR CHECK */
/********************/
#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, char *file, int line, bool abort=true)
{
if (code != cudaSuccess)
{
fprintf(stderr,"GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
if (abort) exit(code);
}
}
/*******************/
/* iDivUp FUNCTION */
/*******************/
int iDivUp(int a, int b) { return ((a % b) != 0) ? (a / b + 1) : (a / b); }
/*************************************************/
/* DEVICE FUNCTION PERFORMING THE SCALAR PRODUCT */
/*************************************************/
__host__ __device__ float scalarProduct(float p1x, float p1y, float p1z, float p2x, float p2y, float p2z)
{
return (p1x * p2x + p1y * p2y + p1z * p2z) ;
}
/*******************/
/* KERNEL FUNCTION */
/*******************/
__global__ void kernel(const float* __restrict__ p1x, const float* __restrict__ p1y, const float* __restrict__ p1z,
const float* __restrict__ p2x, const float* __restrict__ p2y, const float* __restrict__ p2z,
float* __restrict__ output, const int N, const int M) {
int idx = threadIdx.x + blockIdx.x * blockDim.x;
int idy = threadIdx.y + blockIdx.y * blockDim.y;
if ((idx < N) && (idy < M))
output[idy * N + idx] = scalarProduct(p1x[idx], p1y[idx], p1z[idx], p2x[idy], p2y[idy], p2z[idy]);
}
/********/
/* MAIN */
/********/
int main() {
const int N = 10000;
const int M = 1000;
// --- Host side allocations
float *Ax = (float*)malloc(N*sizeof(float));
float *Ay = (float*)malloc(N*sizeof(float));
float *Az = (float*)malloc(N*sizeof(float));
float *Bx = (float*)malloc(M*sizeof(float));
float *By = (float*)malloc(M*sizeof(float));
float *Bz = (float*)malloc(M*sizeof(float));
float *C = (float*)malloc(N*M*sizeof(float));
float *D = (float*)malloc(N*M*sizeof(float));
// --- Device side allocations
float *d_Ax; gpuErrchk(cudaMalloc((void**)&d_Ax, N*sizeof(float)));
float *d_Ay; gpuErrchk(cudaMalloc((void**)&d_Ay, N*sizeof(float)));
float *d_Az; gpuErrchk(cudaMalloc((void**)&d_Az, N*sizeof(float)));
float *d_Bx; gpuErrchk(cudaMalloc((void**)&d_Bx, M*sizeof(float)));
float *d_By; gpuErrchk(cudaMalloc((void**)&d_By, M*sizeof(float)));
float *d_Bz; gpuErrchk(cudaMalloc((void**)&d_Bz, M*sizeof(float)));
float *d_C; gpuErrchk(cudaMalloc((void**)&d_C, N*M*sizeof(float)));
// --- Initialization
srand(time(NULL));
for (int i=0; i<N; i++) {
Ax[i] = rand() / RAND_MAX;
Ay[i] = rand() / RAND_MAX;
Az[i] = rand() / RAND_MAX;
}
for (int i=0; i<M; i++) {
Bx[i] = rand() / RAND_MAX;
By[i] = rand() / RAND_MAX;
Bz[i] = rand() / RAND_MAX;
}
// --- Host side computations
double t1 = clock();
for (int i=0; i<N; i++)
for (int j=0; j<M; j++)
C[i*M + j] = scalarProduct(Ax[i], Ay[i], Az[i], Bx[j], By[j], Bz[j]);
double t2 = clock();
printf("CPU elapsed time: %3.4f ms \n", 1000.*((double)(t2-t1))/CLOCKS_PER_SEC);
// --- Device side computations
dim3 dimBlock(BLOCKSIZE_X, BLOCKSIZE_Y);
dim3 dimGrid(iDivUp(N, BLOCKSIZE_X), iDivUp(M, BLOCKSIZE_Y));
float time;
cudaEvent_t start, stop;
cudaEventCreate(&start);
cudaEventCreate(&stop);
cudaEventRecord(start, 0);
// --- Host to device memory transfers
gpuErrchk(cudaMemcpy(d_Ax, Ax, N*sizeof(float), cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_Ay, Ay, N*sizeof(float), cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_Az, Az, N*sizeof(float), cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_Bx, Bx, M*sizeof(float), cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_By, By, M*sizeof(float), cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_Bz, Bz, M*sizeof(float), cudaMemcpyHostToDevice));
// --- Computations
kernel<<<dimGrid, dimBlock>>>(d_Ax, d_Ay, d_Az, d_Bx, d_By, d_Bz, d_C, N, M);
gpuErrchk(cudaPeekAtLastError());
gpuErrchk(cudaDeviceSynchronize());
gpuErrchk(cudaMemcpy(D, d_C, N*M*sizeof(float), cudaMemcpyDeviceToHost));
cudaEventRecord(stop, 0);
cudaEventSynchronize(stop);
cudaEventElapsedTime(&time, start, stop);
printf("Elapsed time: %3.4f ms \n", time);
for (int i=0; i<N*M; i++) {
if (D[i] != C[i]) {
printf("Mismatch at i = %i; Host= %f, Device = %f\n", i, C[i], D[i]);
return 1;
}
}
printf("Results match!\n");
cudaDeviceReset();
return 0;
}
PART 2
For solving your specific problem, the CUDA will be worth, even by considering the D2H memory transaction (which is very cheap). This is confirmed by the code below, tested on the same system as above, and whose timing is the following:
CPU: 46ms; GPU (with D2H transaction): 0.31ms;
#include <stdio.h>
#include <time.h>
#define BLOCKSIZE_X 16
#define BLOCKSIZE_Y 16
/********************/
/* CUDA ERROR CHECK */
/********************/
#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, char *file, int line, bool abort=true)
{
if (code != cudaSuccess)
{
fprintf(stderr,"GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
if (abort) exit(code);
}
}
/*******************/
/* iDivUp FUNCTION */
/*******************/
int iDivUp(int a, int b) { return ((a % b) != 0) ? (a / b + 1) : (a / b); }
/*************************************************/
/* DEVICE FUNCTION PERFORMING THE SCALAR PRODUCT */
/*************************************************/
__host__ __device__ float scalarProduct(float p1x, float p1y, float p1z, float p2x, float p2y, float p2z)
{
return (p1x * p2x + p1y * p2y + p1z * p2z) ;
}
/*******************/
/* KERNEL FUNCTION */
/*******************/
__global__ void kernel(const float* __restrict__ p1x, const float* __restrict__ p1y, const float* __restrict__ p1z,
const float* __restrict__ p2x, const float* __restrict__ p2y, const float* __restrict__ p2z,
float* __restrict__ output, const int N, const int M) {
int idx = threadIdx.x + blockIdx.x * blockDim.x;
int idy = threadIdx.y + blockIdx.y * blockDim.y;
if ((idx < N) && (idy < M))
if(abs(scalarProduct(p1x[idx], p1y[idx], p1z[idx], p2x[idy], p2y[idy], p2z[idy])) < 0.01f)
output[idx] = 1.f;
else
output[idx] = 0.f;
}
/********/
/* MAIN */
/********/
int main() {
const int N = 10000;
const int M = 1000;
// --- Host side allocations
float *Ax = (float*)malloc(N*sizeof(float));
float *Ay = (float*)malloc(N*sizeof(float));
float *Az = (float*)malloc(N*sizeof(float));
float *Bx = (float*)malloc(M*sizeof(float));
float *By = (float*)malloc(M*sizeof(float));
float *Bz = (float*)malloc(M*sizeof(float));
float *C = (float*)malloc(N*sizeof(float));
float *D = (float*)malloc(N*sizeof(float));
// --- Device side allocations
float *d_Ax; gpuErrchk(cudaMalloc((void**)&d_Ax, N*sizeof(float)));
float *d_Ay; gpuErrchk(cudaMalloc((void**)&d_Ay, N*sizeof(float)));
float *d_Az; gpuErrchk(cudaMalloc((void**)&d_Az, N*sizeof(float)));
float *d_Bx; gpuErrchk(cudaMalloc((void**)&d_Bx, M*sizeof(float)));
float *d_By; gpuErrchk(cudaMalloc((void**)&d_By, M*sizeof(float)));
float *d_Bz; gpuErrchk(cudaMalloc((void**)&d_Bz, M*sizeof(float)));
float *d_C; gpuErrchk(cudaMalloc((void**)&d_C, N*sizeof(float)));
// --- Initialization
srand(time(NULL));
for (int i=0; i<N; i++) {
Ax[i] = rand() / RAND_MAX;
Ay[i] = rand() / RAND_MAX;
Az[i] = rand() / RAND_MAX;
}
for (int i=0; i<M; i++) {
Bx[i] = rand() / RAND_MAX;
By[i] = rand() / RAND_MAX;
Bz[i] = rand() / RAND_MAX;
}
// --- Host side computations
double t1 = clock();
for (int i=0; i<N; i++)
for (int j=0; j<M; j++)
if(abs(scalarProduct(Ax[i], Ay[i], Az[i], Bx[j], By[j], Bz[j])) < 0.01f)
C[i] = 1.f;
else
C[i] = 0.f;
double t2 = clock();
printf("CPU elapsed time: %3.4f ms \n", 1000.*((double)(t2-t1))/CLOCKS_PER_SEC);
// --- Device side computations
float time;
cudaEvent_t start, stop;
cudaEventCreate(&start);
cudaEventCreate(&stop);
cudaEventRecord(start, 0);
// --- Host to device memory transfers
gpuErrchk(cudaMemcpy(d_Ax, Ax, N*sizeof(float), cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_Ay, Ay, N*sizeof(float), cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_Az, Az, N*sizeof(float), cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_Bx, Bx, M*sizeof(float), cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_By, By, M*sizeof(float), cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_Bz, Bz, M*sizeof(float), cudaMemcpyHostToDevice));
// --- Computations
kernel<<<iDivUp(N, BLOCKSIZE_X), BLOCKSIZE_X>>>(d_Ax, d_Ay, d_Az, d_Bx, d_By, d_Bz, d_C, N, M);
gpuErrchk(cudaPeekAtLastError());
gpuErrchk(cudaDeviceSynchronize());
gpuErrchk(cudaMemcpy(D, d_C, N*sizeof(float), cudaMemcpyDeviceToHost));
cudaEventRecord(stop, 0);
cudaEventSynchronize(stop);
cudaEventElapsedTime(&time, start, stop);
printf("Elapsed time: %3.4f ms \n", time);
for (int i=0; i<N; i++) {
if (D[i] != C[i]) {
printf("Mismatch at i = %i; Host= %f, Device = %f\n", i, C[i], D[i]);
return 1;
}
}
printf("Results match!\n");
cudaDeviceReset();
return 0;
}
Instead of optimising for arithmetic, you should use better algorithm first.
In most practical situation ts and A are not totally random per each cycle, and you may somehow organise (sort) them spatially, and greatly reduce the need for calculating spatial metric.
Now if you insist to stick with current algorithm, you may enable compiler to emit SSE code, this should give some instant boost without any programming work.
Now since you have to ask this question, the chance that you may further squeeze cycles by manually code with compiler intrinsics, is relatively narrow.
About CUDA, for just 10 million dot product the overhead for CPU-RAM-DISPLAY RAM-GPU communication is significant and not worth all the trouble.
To parallelize this using MIMD with OpenMP you can do this:
#pragma omp parallel for
for(int i = 0; i< ts.size(); i++){
for(int j = 0; j< A.size(); j++){
if(abs(scalarProduct(ts.at(i), A.at(j))) <epsilon){
score[i] +=1;
}
}
}
You could also consider using SIMD. In that case you should change your data structure and store blocks of points equal to the SIMD width (4 for SSE with floats). Something like
class PointBlock4 {
float x[4];
float y[4];
float z[4];
//
}
Each block has four points. This is obviously more complicated but it's achievable. You could get a speed up as four as well. Combining SIMD and MIMD you could get a speedup of 16x (with four cores). But for large n your algorithm will become memory bound and not compute bound so you will achieve a much lower speedup. In fact your algorithm may already be memory bound so you might achieve much with SIMD or MIMD. I would test OpenMP first to see if you gain much.

MPI Gather Corrupting Arrays

I have written an MPI code in C++ for my Raspberry Pi cluster, which generates an image of the Mandelbrot Set. What happens is on each node (excluding the master, processor 0) part of the Mandelbrot Set is calculated, resulting in each node having a 2D array of ints that indicates whether each xy point is in the set.
It appears to work well on each node individually, but when all the arrays are gathered to the master using this command:
MPI_Gather(&inside, 1, MPI_INT, insideFull, 1, MPI_INT, 0, MPI_COMM_WORLD);
it corrupts the data, and the result is an array full of garbage.
(inside is the nodes' 2D arrays of part of the set. insideFull is also a 2D array but it holds the whole set)
Why would it be doing this?
(This led to me wondering if it corrupting because the master isn't sending its array to itself (or at least I don't want it to). So part of my question also is is there an MPI_Gather variant that doesn't send anything from the root process, just collects from everything else?)
Thanks
EDIT: here's the whole code. If anyone can suggest better ways of how I'm transferring the arrays, please say.
#include <mpi.h>
#include <stdio.h>
#include <stdlib.h>
// ONLY USE MULTIPLES OF THE NUMBER OF SLAVE PROCESSORS
#define ImageHeight 128
#define ImageWidth 128
double MinRe = -1.9;
double MaxRe = 0.5;
double MinIm = -1.2;
double MaxIm = MinIm + (MaxRe - MinRe)*ImageHeight / ImageWidth;
double Re_factor = (MaxRe - MinRe) / (ImageWidth - 1);
double Im_factor = (MaxIm - MinIm) / (ImageHeight - 1);
unsigned n;
unsigned MaxIterations = 50;
int red;
int green;
int blue;
// MPI variables ****
int processorNumber;
int processorRank;
//*******************//
int main(int argc, char** argv) {
// Initialise MPI
MPI_Init(NULL, NULL);
// Get the number of procesors
MPI_Comm_size(MPI_COMM_WORLD, &processorNumber);
// Get the rank of this processor
MPI_Comm_rank(MPI_COMM_WORLD, &processorRank);
// Get the name of this processor
char processorName[MPI_MAX_PROCESSOR_NAME];
int name_len;
MPI_Get_processor_name(processorName, &name_len);
// A barrier just to sync all the processors, make timing more accurate
MPI_Barrier(MPI_COMM_WORLD);
// Make an array that stores whether each point is in the Mandelbrot Set
int inside[ImageWidth / processorNumber][ImageHeight / processorNumber];
if(processorRank == 0) {
printf("Generating Mandelbrot Set\n");
}
// We don't want the master to process the Mandelbrot Set, only the slaves
if(processorRank != 0) {
// Determine which coordinates to test on each processor
int xMin = (ImageWidth / (processorNumber - 1)) * (processorRank - 1);
int xMax = ((ImageWidth / (processorNumber - 1)) * (processorRank - 1)) - 1;
int yMin = (ImageHeight / (processorNumber - 1)) * (processorRank - 1);
int yMax = ((ImageHeight / (processorNumber - 1)) * (processorRank - 1)) - 1;
// Check each value to see if it's in the Mandelbrot Set
for (int y = yMin; y <= yMax; y++) {
double c_im = MaxIm - y *Im_factor;
for (int x = xMin; x <= xMax; x++) {
double c_re = MinRe + x*Re_factor;
double Z_re = c_re, Z_im = c_im;
int isInside = 1;
for (n = 0; n <= MaxIterations; ++n) {
double Z_re2 = Z_re * Z_re, Z_im2 = Z_im * Z_im;
if (Z_re2 + Z_im2 > 10) {
isInside = 0;
break;
}
Z_im = 2 * Z_re * Z_im + c_im;
Z_re = Z_re2 - Z_im2 + c_re;
}
if (isInside == 1) {
inside[x][y] = 1;
}
else{
inside[x][y] = 0;
}
}
}
}
// Wait for all processors to finish computing
MPI_Barrier(MPI_COMM_WORLD);
int insideFull[ImageWidth][ImageHeight];
if(processorRank == 0) {
printf("Sending parts of set to master\n");
}
// Send all the arrays to the master
MPI_Gather(&inside[0][0], 1, MPI_INT, &insideFull[0][0], 1, MPI_INT, 0, MPI_COMM_WORLD);
// Output the data to an image
if(processorRank == 0) {
printf("Generating image\n");
FILE * image = fopen("mandelbrot_set.ppm", "wb");
fprintf(image, "P6 %d %d 255\n", ImageHeight, ImageWidth);
for(int y = 0; y < ImageHeight; y++) {
for(int x = 0; x < ImageWidth; x++) {
if(insideFull[x][y]) {
putc(0, image);
putc(0, image);
putc(255, image);
}
else {
putc(0, image);
putc(0, image);
putc(0, image);
}
// Just to see what values return, no actual purpose
printf("%d, %d, %d\n", x, y, insideFull[x][y]);
}
}
fclose(image);
printf("Complete\n");
}
MPI_Barrier(MPI_COMM_WORLD);
// Finalise MPI
MPI_Finalize();
}
You call MPI_Gether with the following parameters:
const void* sendbuf : &inside[0][0] Starting address of send buffer
int sendcount : 1 Number of elements in send buffer
const MPI::Datatype& sendtype : MPI_INT Datatype of send buffer elements
void* recvbuf : &insideFull[0][0]
int recvcount : 1 Number of elements for any single receive
const MPI::Datatype& recvtype : MPI_INT Datatype of recvbuffer elements
int root : 0 Rank of receiving process
MPI_Comm comm : MPI_COMM_WORLD Communicator (handle).
Sending/receiving only one element is not sufficient. Instead of 1 use
(ImageWidth / processorNumber)*(ImageHeight / processorNumber)
Then think about the different memory layout of your source and target 2D arrays:
int inside[ImageWidth / processorNumber][ImageHeight / processorNumber];
vs.
int insideFull[ImageWidth][ImageHeight];
As the copy is a memory bloc copy, and not an intelligent 2D array copy, all your source integers will be transfered contiguously to the target adress, regardless of the different size of the lines.
I'd recommend to send the data fisrt into an array of the same size as the source, and then in the receiving process, to copy the elements to the right lines & columns in the full array, for example with a small function like:
// assemble2d():
// copys a source int sarr[sli][sco] to a destination int darr[dli][sli]
// using an offset to starting at darr[doffli][doffco].
// The elements that are out of bounds are ignored. Negative offset possible.
void assemble2D(int*darr, int dli, int dco, int*sarr, int sli, int sco, int doffli=0, int doffco=0)
{
for (int i = 0; i < sli; i++)
for (int j = 0; j < sco; j++)
if ((i + doffli >= 0) && (j + doffco>=0) && (i + doffli<dli) && (j + doffco<dco))
darr[(i+doffli)*dli + j+doffco] = sarr[i*sli+j];
}

Passing values from host to device and vice versa Cuda [duplicate]

I'm new to cuda. I want to add up two 2d array into a third array.
I use following code:
cudaMallocPitch((void**)&device_a, &pitch, 2*sizeof(int),2);
cudaMallocPitch((void**)&device_b, &pitch, 2*sizeof(int),2);
cudaMallocPitch((void**)&device_c, &pitch, 2*sizeof(int),2);
now my problem is that i dont want to use these array as flattened 2-d array
all in my kernel code i want to di is use two for loop & put the result in the third array like
__global__ void add(int *dev_a ,int *dev_b,int* dec_c)
{
for i=0;i<2;i++)
{
for j=0;j<2;j++)
{
dev_c[i][j]=dev_a[i][j]+dev_b[i][j];
}
}
}
How i can do this in CUDA?
please tell me how to use 2-d array in this way ?
What should be the kernel call for using 2d-array ?
If possible, please explain using code samples.
The short answer is, you can't. The cudaMallocPitch()function does exactly what its name implies, it allocates pitched linear memory, where the pitch is chosen to be optimal for the GPU memory controller and texture hardware.
If you wanted to use arrays of pointers in the kernel, the kernel code would have to look like this:
__global___ void add(int *dev_a[] ,int *dev_b[], int* dec_c[])
{
for i=0;i<2;i++) {
for j=0;j<2;j++) {
dev_c[i][j]=dev_a[i][j]+dev_b[i][j];
}
}
}
and then you would need nested cudaMalloc calls on the host side to construct the array of pointers and copy it to device memory. For your rather trivial 2x2 example, the code to allocate a single array would look like this:
int ** h_a = (int **)malloc(2 * sizeof(int *));
cudaMalloc((void**)&h_a[0], 2*sizeof(int));
cudaMalloc((void**)&h_a[1], 2*sizeof(int));
int **d_a;
cudaMalloc((void ***)&d_a, 2 * sizeof(int *));
cudaMemcpy(d_a, h_a, 2*sizeof(int *), cudaMemcpyHostToDevice);
Which would leave the allocated device array of pointers in d_a, and you would pass that to your kernel.
For code complexity and performance reasons, you really don't want to do that, using arrays of pointers in CUDA code is both harder and slower than the alternative using linear memory.
To show what folly using arrays of pointers is in CUDA, here is a complete working example of your sample problem which combines the two ideas above:
#include <cstdio>
__global__ void add(int * dev_a[], int * dev_b[], int * dev_c[])
{
for(int i=0;i<2;i++)
{
for(int j=0;j<2;j++)
{
dev_c[i][j]=dev_a[i][j]+dev_b[i][j];
}
}
}
inline void GPUassert(cudaError_t code, char * file, int line, bool Abort=true)
{
if (code != 0) {
fprintf(stderr, "GPUassert: %s %s %d\n", cudaGetErrorString(code),file,line);
if (Abort) exit(code);
}
}
#define GPUerrchk(ans) { GPUassert((ans), __FILE__, __LINE__); }
int main(void)
{
const int aa[2][2]={{1,2},{3,4}};
const int bb[2][2]={{5,6},{7,8}};
int cc[2][2];
int ** h_a = (int **)malloc(2 * sizeof(int *));
for(int i=0; i<2;i++){
GPUerrchk(cudaMalloc((void**)&h_a[i], 2*sizeof(int)));
GPUerrchk(cudaMemcpy(h_a[i], &aa[i][0], 2*sizeof(int), cudaMemcpyHostToDevice));
}
int **d_a;
GPUerrchk(cudaMalloc((void ***)&d_a, 2 * sizeof(int *)));
GPUerrchk(cudaMemcpy(d_a, h_a, 2*sizeof(int *), cudaMemcpyHostToDevice));
int ** h_b = (int **)malloc(2 * sizeof(int *));
for(int i=0; i<2;i++){
GPUerrchk(cudaMalloc((void**)&h_b[i], 2*sizeof(int)));
GPUerrchk(cudaMemcpy(h_b[i], &bb[i][0], 2*sizeof(int), cudaMemcpyHostToDevice));
}
int ** d_b;
GPUerrchk(cudaMalloc((void ***)&d_b, 2 * sizeof(int *)));
GPUerrchk(cudaMemcpy(d_b, h_b, 2*sizeof(int *), cudaMemcpyHostToDevice));
int ** h_c = (int **)malloc(2 * sizeof(int *));
for(int i=0; i<2;i++){
GPUerrchk(cudaMalloc((void**)&h_c[i], 2*sizeof(int)));
}
int ** d_c;
GPUerrchk(cudaMalloc((void ***)&d_c, 2 * sizeof(int *)));
GPUerrchk(cudaMemcpy(d_c, h_c, 2*sizeof(int *), cudaMemcpyHostToDevice));
add<<<1,1>>>(d_a,d_b,d_c);
GPUerrchk(cudaPeekAtLastError());
for(int i=0; i<2;i++){
GPUerrchk(cudaMemcpy(&cc[i][0], h_c[i], 2*sizeof(int), cudaMemcpyDeviceToHost));
}
for(int i=0;i<2;i++) {
for(int j=0;j<2;j++) {
printf("(%d,%d):%d\n",i,j,cc[i][j]);
}
}
return cudaThreadExit();
}
I recommend you study it until you understand what it does, and why it is such a poor idea compared to using linear memory.
You don't need to use for loops inside the device. Try this code.
#include <stdio.h>
#include <cuda.h>
#include <stdlib.h>
#include <time.h>
#define N 800
__global__ void matrixAdd(float* A, float* B, float* C){
int i = threadIdx.x;
int j = blockIdx.x;
C[N*j+i] = A[N*j+i] + B[N*j+i];
}
int main (void) {
clock_t start = clock();
float a[N][N], b[N][N], c[N][N];
float *dev_a, *dev_b, *dev_c;
cudaMalloc((void **)&dev_a, N * N * sizeof(float));
cudaMalloc((void **)&dev_b, N * N * sizeof(float));
cudaMalloc((void **)&dev_c, N * N * sizeof(float));
for (int i = 0; i < N; i++){
for (int j = 0; j < N; j++){
a[i][j] = rand() % 10;
b[i][j] = rand() % 10;
}
}
cudaMemcpy(dev_a, a, N * N * sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(dev_b, b, N * N * sizeof(float), cudaMemcpyHostToDevice);
matrixAdd <<<N,N>>> (dev_a, dev_b, dev_c);
cudaMemcpy(c, dev_c, N * N * sizeof(float), cudaMemcpyDeviceToHost);
for (int i = 0; i < N; i++){
for (int j = 0; j < N; j++){
printf("[%d, %d ]= %f + %f = %f\n",i,j, a[i][j], b[i][j], c[i][j]);
}
}
printf("Time elapsed: %f\n", ((double)clock() - start) / CLOCKS_PER_SEC);
cudaFree(dev_a);
cudaFree(dev_b);
cudaFree(dev_c);
return 0;
}

Copying from cuda 3D memory to linear memory: copied data is not where I expected

Here is my issue:
I have a 3D array of float3 on my device:
int size[3] = {416,464,512};
cudaExtent extent = make_cudaExtent(size[0]*sizeof(float3),size[1],size[2]);
cudaPitchedPtr renderedVolume;
int ret = cudaMalloc3D(&renderedVolume, extent);
size_t pitch = renderedVolume.pitch; //pitch = 5,120
size_t slicePitch = pitch * size[1]; //slicePitch = 2,375,680
Then I work with it and make it full of outstanding data.
After that I wish to copy it on a 1D linear memory on my host:
float *host_memory = (float*)malloc(size[0]*size[1]*size[2]*sizeof(float3));
cudaMemcpy3DParms p = {0};
p.srcPtr = renderedVolume;
p.dstPtr = make_cudaPitchedPtr(host_memory,size[0]*sizeof(float3),size[0],size[1]);
p.extent = make_cudaExtent(size[0]*sizeof(float3),size[1],size[2]);
p.srcPos = make_cudaPos(0,0,0);
p.dstPos = make_cudaPos(0,0,0);
p.kind=cudaMemcpyDeviceToHost;
cudaMemcpy3D(&p);
I am comparing the result in host_memory with the data I initially wrote tu renderedVolume (my_data) and with the data I read in my 3Dmemory, slice by slice:
float* test1 = (float*)malloc(size[0]*size[1]*sizeof(float3));
cudaMemcpy(test1, myData, size[0]*size[1]*sizeof(float3) , cudaMemcpyDeviceToHost);
float* test2 = (float*)malloc(size[0]*size[1]*sizeof(float3));
cudaMemcpy(test2,(char*)renderedVolume.ptr + slicePitch * i,size[0]*size[1]*sizeof(float3), cudaMemcpyDeviceToHost);
Problem:
The first slice (i=0) is ok, I have the same data in host_memory, test1 and test2.
In the second slice, I have the same data in test1 and test2. However, I should find this data in host_memory+579072 (=number of float per slice, also heigth*pitch of the destination pitched pointer) and I find it in host_memory+577504. It is off by 1568 bytes, which corresponds to nothing that I am aware of, and this is why I would very much appreciate if any of you have an idea of what the problem might be in my code ?
This is a late answer provided to remove this question from the unanswered list.
Below, I'm providing a full code showing how to allocate 3D memory by cudaMalloc3D, moving a host allocated 1D memory to 3D device memory by cudaMemcpy3D, performing some operations on the 3D device data by the test_kernel_3D __global__ function and moving the 3D result data back to 1D host memory, again by cudaMemcpy3D.
The __global__ function test_kernel_3D squares each element of the 3D device memory. In particular, each thread of a 2D grid takes care of performing a for loop along the "depth" dimension.
#include<stdio.h>
#include<cuda.h>
#include<cuda_runtime.h>
#include<device_launch_parameters.h>
#include<conio.h>
#define BLOCKSIZE_x 16
#define BLOCKSIZE_y 16
#define N 128
#define M 64
#define W 16
/*****************/
/* CUDA MEMCHECK */
/*****************/
#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, char *file, int line, bool abort=true)
{
if (code != cudaSuccess)
{
fprintf(stderr,"GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
if (abort) { getch(); exit(code); }
}
}
/*******************/
/* iDivUp FUNCTION */
/*******************/
int iDivUp(int a, int b){ return ((a % b) != 0) ? (a / b + 1) : (a / b); }
/******************/
/* TEST KERNEL 3D */
/******************/
__global__ void test_kernel_3D(cudaPitchedPtr devPitchedPtr)
{
int tidx = blockIdx.x*blockDim.x+threadIdx.x;
int tidy = blockIdx.y*blockDim.y+threadIdx.y;
char* devPtr = (char*) devPitchedPtr.ptr;
size_t pitch = devPitchedPtr.pitch;
size_t slicePitch = pitch * N;
for (int w = 0; w < W; w++) {
char* slice = devPtr + w * slicePitch;
float* row = (float*)(slice + tidy * pitch);
row[tidx] = row[tidx] * row[tidx];
}
}
/********/
/* MAIN */
/********/
int main()
{
float a[N][M][W];
for (int i=0; i<N; i++)
for (int j=0; j<M; j++)
for (int w=0; w<W; w++) {
a[i][j][w] = 3.f;
//printf("row %i column %i depth %i value %f \n",i,j,w,a[i][j][w]);
}
// --- 3D pitched allocation and host->device memcopy
cudaExtent extent = make_cudaExtent(M * sizeof(float), N, W);
cudaPitchedPtr devPitchedPtr;
gpuErrchk(cudaMalloc3D(&devPitchedPtr, extent));
cudaMemcpy3DParms p = { 0 };
p.srcPtr.ptr = a;
p.srcPtr.pitch = M * sizeof(float);
p.srcPtr.xsize = M;
p.srcPtr.ysize = N;
p.dstPtr.ptr = devPitchedPtr.ptr;
p.dstPtr.pitch = devPitchedPtr.pitch;
p.dstPtr.xsize = M;
p.dstPtr.ysize = N;
p.extent.width = M * sizeof(float);
p.extent.height = N;
p.extent.depth = W;
p.kind = cudaMemcpyHostToDevice;
gpuErrchk(cudaMemcpy3D(&p));
dim3 GridSize(iDivUp(M,BLOCKSIZE_x),iDivUp(N,BLOCKSIZE_y));
dim3 BlockSize(BLOCKSIZE_y,BLOCKSIZE_x);
test_kernel_3D<<<GridSize,BlockSize>>>(devPitchedPtr);
gpuErrchk(cudaPeekAtLastError());
gpuErrchk(cudaDeviceSynchronize());
p.srcPtr.ptr = devPitchedPtr.ptr;
p.srcPtr.pitch = devPitchedPtr.pitch;
p.dstPtr.ptr = a;
p.dstPtr.pitch = M * sizeof(float);
p.kind = cudaMemcpyDeviceToHost;
gpuErrchk(cudaMemcpy3D(&p));
for (int i=0; i<N; i++)
for (int j=0; j<M; j++)
for (int w=0; w<W; w++)
printf("row %i column %i depth %i value %f\n",i,j,w,a[i][j][w]);
getch();
return 0;
}