cuda calc distance of two points - cuda

Here I want to calculate the distance of each two points, and decide if they are neighbours. here is my simple code in cuda.
__global__ void calcNeighbors(const DataPoint* points,
const float doubleRadius, bool* neighbors) {
int tid = threadIdx.x + blockIdx.x * blockDim.x;
float dis = 0.0f;
while (tid < N) {
DataPoint p1 = points[tid];
for (int i=0; i<N; i++) {
DataPoint p2 = points[i];
dis = 0;
dis += (p1.pfDimens[0]-p2.pfDimens[0]) * (p1.pfDimens[0]-p2.pfDimens[0]) +
(p1.pfDimens[1]-p2.pfDimens[1]) * (p1.pfDimens[1]-p2.pfDimens[1]) +
(p1.pfDimens[2]-p2.pfDimens[2]) * (p1.pfDimens[2]-p2.pfDimens[2]);
if (dis <= doubleRadius) {
neighbors[tid*N+i] = true;
} else {
neighbors[tid*N+i] = false;
}
}
tid += blockDim.x * gridDim.x;
}
}
The DataPoint is a struct is
typedef struct DataPoint {
float pfDimens[3];
} DataPoint;
so here i want to reduce the time, How can i do? I have tried to use memory coalesing and share memory, but i didn't get a good speed up?
===============use share memory==============
__global__ void calcNeighbors2(const DataPoint* points,
const float doubleRadius, bool* neighbors) {
__shared__ DataPoint sharedpoints[threadsPerBlock];
int start = blockIdx.x * blockDim.x;
int len = start+threadIdx.x;
if (len < N) {
sharedpoints[threadIdx.x] = points[len];
}
len = imin(N, blockDim.x + start);
__syncthreads();
int tid = threadIdx.x;
float dis;
while (tid < N) {
DataPoint p1 = points[tid];
for (int i=start; i<len; i++) {
dis = 0;
dis += (p1.pfDimens[0]-sharedpoints[i-start].pfDimens[0]) * (p1.pfDimens[0]-sharedpoints[i-start].pfDimens[0]) +
(p1.pfDimens[1]-sharedpoints[i-start].pfDimens[1]) * (p1.pfDimens[1]-sharedpoints[i-start].pfDimens[1]) +
(p1.pfDimens[2]-sharedpoints[i-start].pfDimens[2]) * (p1.pfDimens[2]-sharedpoints[i-start].pfDimens[2]);
if (dis <= doubleRadius) {
neighbors[i*N+tid] = true;
} else {
neighbors[i*N+tid] = false;
}
}
tid += blockDim.x;
}
}
Here i changed the neighbors[tid*N+i] to neighbors[i*N+tid], it give me amlost 8x speed up on Tesla K10.G2.8GB. But when i use share memory to store some points, it is no use?

There are at least 4 ideas, some of which have already been stated in the comments:
Transform your point distance storage from AoS format:
struct DataPoint {
float pfDimens[3];
};
to SoA format:
struct DataPoint {
float pfDimens_x[NPTS];
float pfDimens_y[NPTS];
float pfDimens_z[NPTS];
};
this will enable full coalescing on loading of the data. In fact, to help with point 4 below, I would just switch to using 3 bare arrays, rather than a structure.
reduce the computation to (slightly less than) half:
for (int i=N-1; i>tid; i--) {
then, either in the thread code itself, or in the host, you can populate the other "half" of the output matrix by copying data.
Transpose the storage in your output matrix, so that you can write a storage operation like this:
neighbors[i*N+tid] = true;
which will nicely coalesce, as opposed to this:
neighbors[tid*N+i] = true;
which will not.
Since your input point data is read only, mark the kernel parameter appropriately:
const float * __restrict__ points_x, const float * __restrict__ points_y, const float * __restrict__ points_z
in some cases, and on some GPUs, this will often lead to a speed-up due to use of the read-only cache. If you really want to get aggressive with caching, and your data array is small enough (4K or less float points), you could put a copy of the point data in global memory as well as a copy in __constant__ memory, and load the "uniform" load you are doing here through constant memory:
DataPoint p2 = c_points[i];
thus you could perform the coalesced load through the read-only cache, the uniform load through the constant cache, and the coalesced store going to ordinary global memory.
On a K40c, on linux/CUDA 7, for N = 4096, the net effect of these changes appears to be about a 3.5x speedup, at the kernel level:
$ cat t749.cu
#include <stdio.h>
#define N 4096
// if N is 16K/3 or less, we can use constant
#define USE_CONSTANT
#define THRESH 0.2f
#define nTPB 256
#define nBLK (N/nTPB+1)
#define cudaCheckErrors(msg) \
do { \
cudaError_t __err = cudaGetLastError(); \
if (__err != cudaSuccess) { \
fprintf(stderr, "Fatal error: %s (%s at %s:%d)\n", \
msg, cudaGetErrorString(__err), \
__FILE__, __LINE__); \
fprintf(stderr, "*** FAILED - ABORTING\n"); \
exit(1); \
} \
} while (0)
#include <time.h>
#include <sys/time.h>
#define USECPSEC 1000000ULL
unsigned long long dtime_usec(unsigned long long start){
timeval tv;
gettimeofday(&tv, 0);
return ((tv.tv_sec*USECPSEC)+tv.tv_usec)-start;
}
struct DataPoint {
float pfDimens[3];
};
__global__ void calcNeighbors(const DataPoint* points,
const float doubleRadius, bool* neighbors) {
int tid = threadIdx.x + blockIdx.x * blockDim.x;
float dis = 0.0f;
while (tid < N) {
DataPoint p1 = points[tid];
for (int i=0; i<N; i++) {
DataPoint p2 = points[i];
dis = 0;
dis += (p1.pfDimens[0]-p2.pfDimens[0]) * (p1.pfDimens[0]-p2.pfDimens[0]) +
(p1.pfDimens[1]-p2.pfDimens[1]) * (p1.pfDimens[1]-p2.pfDimens[1]) +
(p1.pfDimens[2]-p2.pfDimens[2]) * (p1.pfDimens[2]-p2.pfDimens[2]);
if (dis <= doubleRadius) {
neighbors[tid*N+i] = true;
} else {
neighbors[tid*N+i] = false;
}
}
tid += blockDim.x * gridDim.x;
}
}
#ifdef USE_CONSTANT
__constant__ float cpx[N];
__constant__ float cpy[N];
__constant__ float cpz[N];
#endif
__global__ void calcNeighbors2(const float * __restrict__ pts_x, const float * __restrict__ pts_y, const float * __restrict__ pts_z, const float doubleRadius, bool * __restrict__ neighbors) {
int tid = threadIdx.x+blockDim.x*blockIdx.x;
while (tid < N) {
float p1x = pts_x[tid];
float p1y = pts_y[tid];
float p1z = pts_z[tid];
for (int i = N-1; i > tid; i--){
float p2x, p2y, p2z;
#ifdef USE_CONSTANT
p2x = cpx[i];
p2y = cpy[i];
p2z = cpz[i];
#else
p2x = pts_x[i];
p2y = pts_y[i];
p2z = pts_z[i];
#endif
float dis = ((p1x-p2x)*(p1x-p2x)) + ((p1y-p2y)*(p1y-p2y)) + ((p1z-p2z)*(p1z-p2z));
neighbors[i*N+tid] = (dis <= doubleRadius);
}
tid += blockDim.x * gridDim.x;
}
}
int main(){
float *dx, *dy, *dz, *hx, *hy, *hz;
DataPoint *dp, *hp;
bool *dn, *hn1, *hn2;
hx =(float *)malloc(N*sizeof(float));
hy =(float *)malloc(N*sizeof(float));
hz =(float *)malloc(N*sizeof(float));
hp =(DataPoint *)malloc(N*sizeof(DataPoint));
hn1=(bool *)malloc(N*N*sizeof(bool));
hn2=(bool *)malloc(N*N*sizeof(bool));
cudaMalloc(&dx, N*sizeof(float));
cudaMalloc(&dy, N*sizeof(float));
cudaMalloc(&dz, N*sizeof(float));
cudaMalloc(&dp, N*sizeof(DataPoint));
cudaMalloc(&dn, N*N*sizeof(bool));
for (int i =0; i < N; i++){
hx[i] = rand()/(float)RAND_MAX;
hy[i] = rand()/(float)RAND_MAX;
hz[i] = rand()/(float)RAND_MAX;
hp[i].pfDimens[0] = hx[i];
hp[i].pfDimens[1] = hy[i];
hp[i].pfDimens[2] = hz[i];}
cudaMemcpy(dx, hx, N*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(dy, hy, N*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(dz, hz, N*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(dp, hp, N*sizeof(DataPoint), cudaMemcpyHostToDevice);
// warm-up
calcNeighbors<<<nBLK, nTPB>>>(dp, THRESH, dn);
cudaDeviceSynchronize();
cudaMemset(dn, 0, N*N*sizeof(bool));
unsigned long long t1 = dtime_usec(0);
calcNeighbors<<<nBLK, nTPB>>>(dp, THRESH, dn);
cudaDeviceSynchronize();
cudaCheckErrors("kernel 1 error");
t1 = dtime_usec(t1);
cudaMemcpy(hn1, dn, N*N*sizeof(bool), cudaMemcpyDeviceToHost);
// warm-up
calcNeighbors2<<<nBLK, nTPB>>>(dx, dy, dz, THRESH, dn);
cudaDeviceSynchronize();
cudaMemset(dn, 0, N*N*sizeof(bool));
unsigned long long t2 = dtime_usec(0);
calcNeighbors2<<<nBLK, nTPB>>>(dx, dy, dz, THRESH, dn);
cudaDeviceSynchronize();
cudaCheckErrors("kernel 2 error");
t2 = dtime_usec(t2);
cudaMemcpy(hn2, dn, N*N*sizeof(bool), cudaMemcpyDeviceToHost);
cudaCheckErrors("some error");
printf("t1: %fs, t2: %fs\n", t1/(float)USECPSEC, t2/(float)USECPSEC);
// results validation
for (int i = 0; i < N; i++)
for (int j = i+1; j < N; j++)
if (hn1[i*N+j] != hn2[j*N+i]) {printf("mismatch at %d, %d, was: %d, should be: %d\n", i, j, hn2[j*N+i], hn1[i*N+j]); return 1;}
return 0;
}
$ nvcc -arch=sm_35 -o t749 t749.cu
$ ./t749
t1: 0.004903s, t2: 0.001395s
$
In the case of K40c, the limited number of blocks being launched above (16) is a significant impediment to performance, due to latency. If we comment out the USE_CONSTANT define, and change N to 16384, we observe an even higher speedup with the improved kernel:
$ ./t749
t1: 0.267107s, t2: 0.008209s
$
the resultant ~48 blocks being enough to approximately "fill" the K40c which has 15 SMs.
EDIT: now that you've posted a shared memory kernel, I added it to my test case as calcNeighbors3 and compared it's timing performance (as t3). It is almost as fast as my kernel, and it seems to provide the correct result (matches your original kernel) so I'm not sure what your concerns are.
Here's the updated code and test case:
$ cat t749.cu
#include <stdio.h>
#include <math.h>
#define imin(X,Y) ((X)<(Y))?(X):(Y)
#define N 32768
// if N is 16K/3 or less, we can use constant
// #define USE_CONSTANT
#define THRESH 0.2f
#define nTPB 256
#define nBLK (N/nTPB+1)
#define cudaCheckErrors(msg) \
do { \
cudaError_t __err = cudaGetLastError(); \
if (__err != cudaSuccess) { \
fprintf(stderr, "Fatal error: %s (%s at %s:%d)\n", \
msg, cudaGetErrorString(__err), \
__FILE__, __LINE__); \
fprintf(stderr, "*** FAILED - ABORTING\n"); \
exit(1); \
} \
} while (0)
#include <time.h>
#include <sys/time.h>
#define USECPSEC 1000000ULL
unsigned long long dtime_usec(unsigned long long start){
timeval tv;
gettimeofday(&tv, 0);
return ((tv.tv_sec*USECPSEC)+tv.tv_usec)-start;
}
struct DataPoint {
float pfDimens[3];
};
__global__ void calcNeighbors(const DataPoint* points,
const float doubleRadius, bool* neighbors) {
int tid = threadIdx.x + blockIdx.x * blockDim.x;
float dis = 0.0f;
while (tid < N) {
DataPoint p1 = points[tid];
for (int i=0; i<N; i++) {
DataPoint p2 = points[i];
dis = 0;
dis += (p1.pfDimens[0]-p2.pfDimens[0]) * (p1.pfDimens[0]-p2.pfDimens[0]) +
(p1.pfDimens[1]-p2.pfDimens[1]) * (p1.pfDimens[1]-p2.pfDimens[1]) +
(p1.pfDimens[2]-p2.pfDimens[2]) * (p1.pfDimens[2]-p2.pfDimens[2]);
if (dis <= doubleRadius) {
neighbors[tid*N+i] = true;
} else {
neighbors[tid*N+i] = false;
}
}
tid += blockDim.x * gridDim.x;
}
}
#ifdef USE_CONSTANT
__constant__ float cpx[N];
__constant__ float cpy[N];
__constant__ float cpz[N];
#endif
__global__ void calcNeighbors2(const float * __restrict__ pts_x, const float * __restrict__ pts_y, const float * __restrict__ pts_z, const float doubleRadius, bool * __restrict__ neighbors) {
int tid = threadIdx.x+blockDim.x*blockIdx.x;
while (tid < N) {
float p1x = pts_x[tid];
float p1y = pts_y[tid];
float p1z = pts_z[tid];
for (int i = N-1; i > tid; i--){
float p2x, p2y, p2z;
#ifdef USE_CONSTANT
p2x = cpx[i];
p2y = cpy[i];
p2z = cpz[i];
#else
p2x = pts_x[i];
p2y = pts_y[i];
p2z = pts_z[i];
#endif
float dis = ((p1x-p2x)*(p1x-p2x)) + ((p1y-p2y)*(p1y-p2y)) + ((p1z-p2z)*(p1z-p2z));
neighbors[i*N+tid] = (dis <= doubleRadius);
}
tid += blockDim.x * gridDim.x;
}
}
__global__ void calcNeighbors3(const DataPoint* points,
const float doubleRadius, bool* neighbors) {
__shared__ DataPoint sharedpoints[nTPB];
int start = blockIdx.x * blockDim.x;
int len = start+threadIdx.x;
if (len < N) {
sharedpoints[threadIdx.x] = points[len];
}
len = imin(N, blockDim.x + start);
__syncthreads();
int tid = threadIdx.x;
float dis;
while (tid < N) {
DataPoint p1 = points[tid];
for (int i=start; i<len; i++) {
dis = 0;
dis += (p1.pfDimens[0]-sharedpoints[i-start].pfDimens[0]) * (p1.pfDimens[0]-sharedpoints[i-start].pfDimens[0]) +
(p1.pfDimens[1]-sharedpoints[i-start].pfDimens[1]) * (p1.pfDimens[1]-sharedpoints[i-start].pfDimens[1]) +
(p1.pfDimens[2]-sharedpoints[i-start].pfDimens[2]) * (p1.pfDimens[2]-sharedpoints[i-start].pfDimens[2]);
if (dis <= doubleRadius) {
neighbors[i*N+tid] = true;
} else {
neighbors[i*N+tid] = false;
}
}
tid += blockDim.x;
}
}
int main(){
float *dx, *dy, *dz, *hx, *hy, *hz;
DataPoint *dp, *hp;
bool *dn, *hn1, *hn2, *hn3;
hx =(float *)malloc(N*sizeof(float));
hy =(float *)malloc(N*sizeof(float));
hz =(float *)malloc(N*sizeof(float));
hp =(DataPoint *)malloc(N*sizeof(DataPoint));
hn1=(bool *)malloc(N*N*sizeof(bool));
hn2=(bool *)malloc(N*N*sizeof(bool));
hn3=(bool *)malloc(N*N*sizeof(bool));
cudaMalloc(&dx, N*sizeof(float));
cudaMalloc(&dy, N*sizeof(float));
cudaMalloc(&dz, N*sizeof(float));
cudaMalloc(&dp, N*sizeof(DataPoint));
cudaMalloc(&dn, N*N*sizeof(bool));
for (int i =0; i < N; i++){
hx[i] = rand()/(float)RAND_MAX;
hy[i] = rand()/(float)RAND_MAX;
hz[i] = rand()/(float)RAND_MAX;
hp[i].pfDimens[0] = hx[i];
hp[i].pfDimens[1] = hy[i];
hp[i].pfDimens[2] = hz[i];}
cudaMemcpy(dx, hx, N*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(dy, hy, N*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(dz, hz, N*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(dp, hp, N*sizeof(DataPoint), cudaMemcpyHostToDevice);
#ifdef USE_CONSTANT
cudaMemcpyToSymbol(cpx, hx, N*sizeof(float));
cudaMemcpyToSymbol(cpy, hy, N*sizeof(float));
cudaMemcpyToSymbol(cpz, hz, N*sizeof(float));
#endif
// warm-up
calcNeighbors<<<nBLK, nTPB>>>(dp, THRESH, dn);
cudaDeviceSynchronize();
cudaMemset(dn, 0, N*N*sizeof(bool));
unsigned long long t1 = dtime_usec(0);
calcNeighbors<<<nBLK, nTPB>>>(dp, THRESH, dn);
cudaDeviceSynchronize();
cudaCheckErrors("kernel 1 error");
t1 = dtime_usec(t1);
cudaMemcpy(hn1, dn, N*N*sizeof(bool), cudaMemcpyDeviceToHost);
// warm-up
calcNeighbors2<<<nBLK, nTPB>>>(dx, dy, dz, THRESH, dn);
cudaDeviceSynchronize();
cudaMemset(dn, 0, N*N*sizeof(bool));
unsigned long long t2 = dtime_usec(0);
calcNeighbors2<<<nBLK, nTPB>>>(dx, dy, dz, THRESH, dn);
cudaDeviceSynchronize();
cudaCheckErrors("kernel 2 error");
t2 = dtime_usec(t2);
cudaMemcpy(hn2, dn, N*N*sizeof(bool), cudaMemcpyDeviceToHost);
// warm-up
calcNeighbors3<<<nBLK, nTPB>>>(dp, THRESH, dn);
cudaDeviceSynchronize();
cudaMemset(dn, 0, N*N*sizeof(bool));
unsigned long long t3 = dtime_usec(0);
calcNeighbors3<<<nBLK, nTPB>>>(dp, THRESH, dn);
cudaDeviceSynchronize();
cudaCheckErrors("kernel 3 error");
t3 = dtime_usec(t3);
cudaMemcpy(hn3, dn, N*N*sizeof(bool), cudaMemcpyDeviceToHost);
cudaCheckErrors("some error");
printf("t1: %fs, t2: %fs, t3: %fs\n", t1/(float)USECPSEC, t2/(float)USECPSEC, t3/(float)USECPSEC);
// results validation
for (int i = 0; i < N; i++)
for (int j = i+1; j < N; j++)
if (hn1[i*N+j] != hn2[j*N+i]) {printf("1:2 mismatch at %d, %d, was: %d, should be: %d\n", i, j, hn2[j*N+i], hn1[i*N+j]); return 1;}
for (int i = 0; i < N*N; i++)
if (hn1[i] != hn3[i]) {printf("1:3 mismatch at %d, was: %d, should be: %d\n", i, hn1[i], hn3[i]); return 1;}
return 0;
}
$ nvcc -arch=sm_35 -o t749 t749.cu
$ ./t749
t1: 1.260010s, t2: 0.022661s, t3: 0.029632s
$
For this test, I have changed the data set size to 32768 since that is closer to the range you care about. Your shared memory kernel shows about a 42x speedup over your original kernel, and my kernel shows about a 55x speedup, on my K40c.

Related

Why PyCUDA is faster than C CUDA in this example

I am exploring to move from OpenCL to CUDA, and did a few tests to benchmark the speed of CUDA in various implementations. To my surprise, in the examples below, the PyCUDA implementation is about 20% faster than the C CUDA example.
I read many posts talking about "release build" of C CUDA code. I did try having -Xptxas -O3 in the makefile and that really did not make a difference. I also tried to adjust the block size, with which the kernel was executed. Unfortunately, it did not help improve the speed, either.
My questions here are:
What could be the reasons leading to the speed difference between C CUDA and PYCUDA?
If the "advanced" (lack of a better word) compiling in PYCUDA is one of reasons, how can I optimize the compiling of my C CUDA code?
Are there any other ways to improve the speed of C CUDA in this case?
While I appreciate general comments, I am looking for actionable suggestions that I can validate on my machine. Thanks!
import pycuda.autoinit
import pycuda.driver as drv
import numpy as np
from pycuda.compiler import SourceModule
import time
mod = SourceModule(
"""
__global__ void saxpy(int n, const float a, float *x, float *y)
{
int i = blockIdx.x * blockDim.x + threadIdx.x;
if (i < n){
y[i] = a * x[i] + y[i];
}
}
"""
)
saxpy = mod.get_function("saxpy")
N = 1 << 25
time_elapse = 0.0
for i in range(100):
# print(i)
# print(N)
x = np.ones(N).astype(np.float32)
y = 2 * np.ones(N).astype(np.float32)
start = time.time()
saxpy(
np.int32(N),
np.float32(2.0),
drv.In(x),
drv.InOut(y),
block=(512, 1, 1),
grid=(int(N / 512) + 1, 1),
)
time_elapse += (time.time() - start)
print(time_elapse )
print(y[-100:-1])
print(y.sum())
print(N * 4.0)
#include <stdio.h>
#include <time.h>
#define DIM 512
__global__ void saxpy(int n, float a, float *x, float *y)
{
int i = blockIdx.x * blockDim.x + threadIdx.x;
if (i < n)
y[i] = a * x[i] + y[i];
}
int main(int num_iterations)
{
double start;
double cputime;
int N = 1 << 25;
float *x, *y, *d_x, *d_y;
int i, j;
for (j = 0; j < num_iterations; j++)
{
x = (float *)malloc(N * sizeof(float));
y = (float *)malloc(N * sizeof(float));
cudaMalloc(&d_x, N * sizeof(float));
cudaMalloc(&d_y, N * sizeof(float));
for (i = 0; i < N; i++)
{
x[i] = 1.0f;
y[i] = 2.0f;
}
cudaMemcpy(d_x, x, N * sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(d_y, y, N * sizeof(float), cudaMemcpyHostToDevice);
// Perform SAXPY on 1M elements
start = clock();
saxpy<<<(N + DIM) / DIM, DIM>>>(N, 2.0f, d_x, d_y);
cputime += ((double)(clock() - start) / CLOCKS_PER_SEC);
cudaMemcpy(y, d_y, N * sizeof(float), cudaMemcpyDeviceToHost);
// float maxError = 0.0f;
// for (int i = 0; i < N; i++){
// maxError = max(maxError, abs(y[i] - 4.0f));
// //printf("y[%d]: %f\n", i,y[i]);
// }
// printf("Max error: %f\n", maxError);
cudaFree(d_x);
cudaFree(d_y);
free(x);
free(y);
}
printf("cpu time is %f\n", cputime);
return 0;
}
I saved the above file as cuda_example.cu and compile it with the following commands in a makefile:
nvcc -arch=sm_61 -Xptxas -O3,-v -o main cuda_example.cu
If I execute your CUDA-C code as is, and set num_iterations to 300 like this:
int num_iterations =300;
then the execution of your program takes about 60s on a Geforce GTX 1650. Your code is extremely inefficient, as you copy data back and forth between GPU and device at every iteration.
So, lets restrict the loop to just the kernel execution:
#include <stdio.h>
#include <time.h>
#define DIM 512
__global__ void saxpy(int n, float a, float *x, float *y)
{
int i = blockIdx.x * blockDim.x + threadIdx.x;
if (i < n)
y[i] = a * x[i] + y[i];
}
int main()
{
double start = clock();
int N = 1 << 25;
float *x, *y, *d_x, *d_y;
int i, j;
int num_iterations = 300;
x = (float *)malloc(N * sizeof(float));
y = (float *)malloc(N * sizeof(float));
cudaMalloc(&d_x, N * sizeof(float));
cudaMalloc(&d_y, N * sizeof(float));
for (i = 0; i < N; i++)
{
x[i] = 1.0f;
y[i] = 2.0f;
}
cudaMemcpy(d_x, x, N * sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(d_y, y, N * sizeof(float), cudaMemcpyHostToDevice);
for (j = 0; j < num_iterations; j++){
saxpy<<<(N + DIM) / DIM, DIM>>>(N, 2.0f, d_x, d_y);
cudaDeviceSynchronize();
}
cudaMemcpy(y, d_y, N * sizeof(float), cudaMemcpyDeviceToHost);
cudaFree(d_x);
cudaFree(d_y);
free(x);
free(y);
double cputime = ((double)(clock() - start) / CLOCKS_PER_SEC);
printf("cpu time is %f\n", cputime);
return 0;
}
If I do that, then the execution time becomes 1.36 seconds. Doing sth similar to the PyCUDA code I got about 19s of execution time.

Incorrect addition of Prime numbers in CUDA [duplicate]

This question already has an answer here:
How to find the sum of array in CUDA by reduction
(1 answer)
Closed 3 years ago.
I use reduction logic in code by referring How to find the sum of array in CUDA by reduction.
But It is giving some errors. I am not getting my mistake, could you please help me out??
required specification:
1.Cuda toolkit v6.5
2. graphics: GTX 210 (compute capability 1.2)
3. visual studio 2013
#include<stdio.h>
#include<cuda.h>
#include<malloc.h>
#include<conio.h>
#include<time.h>
#include<windows.h>
#define SIZE 10
#define N 100
__global__ void vectoreAdd(int *d_a, int *d_b, int *d_c)
{
__shared__ int sdata[256];
int i = threadIdx.x + (blockIdx.x*blockDim.x);
sdata[threadIdx.x] = d_a[i];
__syncthreads();
if (i<SIZE)
for (i = 2; i<SIZE; i++)
{
int counter = 0;
for (int j = 2; j<d_a[i]; j++)
{
if (d_a[i] % j == 0)
{
counter = 1; break;
}
}
if (counter == 0)
{
d_b[i] = d_a[i];
}
}
// do reduction in shared mem
for (int s = 1; s < blockDim.x; s *= 2)
{
int index = 2 * s * threadIdx.x;;
if (index < blockDim.x)
{
sdata[index] += sdata[index + s];
}
__syncthreads();
}
// write result for this block to global mem
if (threadIdx.x == 0)
atomicAdd(d_c, sdata[0]);
}
}
int main()
{
clock_t tic = clock();
int *a, *b, *summation=0, sum = 0,count=-1; //declare summation as double/long if needed
int *d_a, *d_b, *d_c;
//int blocks, block_size = 512;
int size = N * sizeof(int);
a = (int *)malloc(SIZE*sizeof(int));
b = (int *)malloc(SIZE*sizeof(int));
summation = (int *)malloc(SIZE*sizeof(int));
cudaMalloc((void**)&d_a, SIZE * sizeof(int));
cudaMalloc((void**)&d_b, SIZE * sizeof(int));
cudaMalloc((void**)&d_c, SIZE * sizeof(int));
for (int i = 1; i<SIZE; i++)
{
a[i] = i;
b[i] = 0;
}
cudaMemcpy(d_a, a, SIZE*sizeof(int), cudaMemcpyHostToDevice);
cudaMemcpy(d_b, b, SIZE*sizeof(int), cudaMemcpyHostToDevice);
cudaMemcpy(d_c, c, SIZE*sizeof(int), cudaMemcpyHostToDevice);
/*blocks = SIZE / block_size;
if (SIZE% block_size != 0)
blocks++; */
dim3 blocksize(256); // create 1D threadblock
dim3 gridsize(N / blocksize.x); //create 1D grid
vectoreAdd << < gridsize, blocksize >> >(d_a, d_b, d_c);
//cudaThreadSynchronize();
cudaMemcpy(b, d_b, SIZE*sizeof(int), cudaMemcpyDeviceToHost);
cudaMemcpy(summation, d_c, SIZE*sizeof(int), cudaMemcpyDeviceToHost);
for (int m = 0; m < SIZE; m++)
{
if (b[m] != 0)
{
printf("\n prime no is:%d", b[m]);
count = count + 1;
}
}
printf("\n\n Total prime no. are: %d", count);
/* for (int j = 1; j<SIZE; j++)
{
sum = sum + b[j];
}*/
printf("\n \nsum of all prime no upto %d is:%d", SIZE, summation);
clock_t toc = clock();
printf("\n\nElapsed: %f seconds\n", (double)(toc - tic) / CLOCKS_PER_SEC);
free(a); free(b); free(summation);
cudaFree(d_a); cudaFree(d_b); cudaFree(d_c);
getchar(); return 0;
}
There are lots of mistakes in your code :
cudaMalloc((void**)&d_a, SIZE * sizeof(int));
should be :
cudaMalloc((void**)&d_a, N * sizeof(int)); //OR
cudaMalloc((void**)&d_a, size);
as you already calculated but didnt passed it. same in case of malloc() //Host code

The efficiency and performance of ILP for the NVIDIA Kepler architecture

Quoting the "Kepler Tuning Guide" provided by NVIDIA:
Also note that Kepler GPUs can utilize ILP in place of
thread/warp-level parallelism (TLP) more readily than Fermi GPUs can.
In my opinion, the following code snippet
a = .....;
a2 = f(a);
a3 = g(a2);
can be improved as follows
a = ...;
b = ....;
a2 = f(a);
b2 = f(b);
a3 = g(a2);
b3 = g(b2);
So in my projects, I have a section of code as follows (example 1)
if(x < src.cols && y < src.rows)
{
if(!mask(y,x))
{
src.ptr(y)[x] = make_short4(0,0,0,0);
}
}
and I rewrite it as follows (example2)
if(x < src.cols && y < src.rows)
{
if(!mask(y,x))
{
short4 t;
t.x = 0;
t.y = 0;
t.z = 0;
t.w = 0;
src.ptr(y)[x].x = t.x;
src.ptr(y)[x].y = t.y;
src.ptr(y)[x].z = t.z;
src.ptr(y)[x].w = t.w;
}
}
In the Kepler architecture, the example2 will be more efficient and exhibit better performance than example1, is that right?
A good explanation on Instruction Level Parallelism (ILP) can be found at CUDA Performance: Maximizing Instruction-Level Parallelism.
It has been pointed out by Robert Crovella and talonmies, and it has been recognized by yourself, that your example above does not reach ILP.
Concerning how implementing ILP, I'm showing below the classical example, translated from the PyCUDA code at numbapro-examples, which I have tested for a Fermi and for a Kepler GPU. Please, notice that for the latter case I have not observed relevant speedups.
THE CODE
#include <stdio.h>
#include <time.h>
#define BLOCKSIZE 64
/*******************/
/* iDivUp FUNCTION */
/*******************/
int iDivUp(int a, int b){
return ((a % b) != 0) ? (a / b + 1) : (a / b);
}
/********************/
/* 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);
}
}
/************************************/
/* NO INSTRUCTION LEVEL PARALLELISM */
/************************************/
__global__ void ILP0(float* d_a, float* d_b, float* d_c) {
int i = threadIdx.x + blockIdx.x * blockDim.x;
d_c[i] = d_a[i] + d_b[i];
}
/************************************/
/* INSTRUCTION LEVEL PARALLELISM X2 */
/************************************/
__global__ void ILP2(float* d_a, float* d_b, float* d_c) {
// --- Loading the data
int i = threadIdx.x + blockIdx.x * blockDim.x;
float ai = d_a[i];
float bi = d_b[i];
int stride = gridDim.x * blockDim.x;
int j = i + stride;
float aj = d_a[j];
float bj = d_b[j];
// --- Computing
float ci = ai + bi;
float cj = aj + bj;
// --- Writing the data
d_c[i] = ci;
d_c[j] = cj;
}
/************************************/
/* INSTRUCTION LEVEL PARALLELISM X4 */
/************************************/
__global__ void ILP4(float* d_a, float* d_b, float* d_c) {
// --- Loading the data
int i = threadIdx.x + blockIdx.x * blockDim.x;
float ai = d_a[i];
float bi = d_b[i];
int stride = gridDim.x * blockDim.x;
int j = i + stride;
float aj = d_a[j];
float bj = d_b[j];
int k = j + stride;
float ak = d_a[k];
float bk = d_b[k];
int l = k + stride;
float al = d_a[l];
float bl = d_b[l];
// --- Computing
float ci = ai + bi;
float cj = aj + bj;
float ck = ak + bk;
float cl = al + bl;
// --- Writing the data
d_c[i] = ci;
d_c[j] = cj;
d_c[k] = ck;
d_c[l] = cl;
}
/************************************/
/* INSTRUCTION LEVEL PARALLELISM X8 */
/************************************/
__global__ void ILP8(float* d_a, float* d_b, float* d_c) {
// --- Loading the data
int i = threadIdx.x + blockIdx.x * blockDim.x;
float ai = d_a[i];
float bi = d_b[i];
int stride = gridDim.x * blockDim.x;
int j = i + stride;
float aj = d_a[j];
float bj = d_b[j];
int k = j + stride;
float ak = d_a[k];
float bk = d_b[k];
int l = k + stride;
float al = d_a[l];
float bl = d_b[l];
int m = l + stride;
float am = d_a[m];
float bm = d_b[m];
int n = m + stride;
float an = d_a[n];
float bn = d_b[n];
int p = n + stride;
float ap = d_a[p];
float bp = d_b[p];
int q = p + stride;
float aq = d_a[q];
float bq = d_b[q];
// --- Computing
float ci = ai + bi;
float cj = aj + bj;
float ck = ak + bk;
float cl = al + bl;
float cm = am + bm;
float cn = an + bn;
float cp = ap + bp;
float cq = aq + bq;
// --- Writing the data
d_c[i] = ci;
d_c[j] = cj;
d_c[k] = ck;
d_c[l] = cl;
d_c[m] = cm;
d_c[n] = cn;
d_c[p] = cp;
d_c[q] = cq;
}
/********/
/* MAIN */
/********/
void main() {
float timing;
cudaEvent_t start, stop;
const int N = 65536*4; // --- ASSUMPTION: N can be divided by BLOCKSIZE
float* a = (float*)malloc(N*sizeof(float));
float* b = (float*)malloc(N*sizeof(float));
float* c = (float*)malloc(N*sizeof(float));
float* c_ref = (float*)malloc(N*sizeof(float));
srand(time(NULL));
for (int i=0; i<N; i++) {
a[i] = rand() / RAND_MAX;
b[i] = rand() / RAND_MAX;
c_ref[i] = a[i] + b[i];
}
float* d_a; gpuErrchk(cudaMalloc((void**)&d_a,N*sizeof(float)));
float* d_b; gpuErrchk(cudaMalloc((void**)&d_b,N*sizeof(float)));
float* d_c0; gpuErrchk(cudaMalloc((void**)&d_c0,N*sizeof(float)));
float* d_c2; gpuErrchk(cudaMalloc((void**)&d_c2,N*sizeof(float)));
float* d_c4; gpuErrchk(cudaMalloc((void**)&d_c4,N*sizeof(float)));
float* d_c8; gpuErrchk(cudaMalloc((void**)&d_c8,N*sizeof(float)));
gpuErrchk(cudaMemcpy(d_a, a, N*sizeof(float), cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_b, b, N*sizeof(float), cudaMemcpyHostToDevice));
/******************/
/* ILP0 TEST CASE */
/******************/
cudaEventCreate(&start);
cudaEventCreate(&stop);
cudaEventRecord(start, 0);
ILP0<<<iDivUp(N,BLOCKSIZE),BLOCKSIZE>>>(d_a, d_b, d_c0);
gpuErrchk(cudaPeekAtLastError());
gpuErrchk(cudaDeviceSynchronize());
cudaEventRecord(stop, 0);
cudaEventSynchronize(stop);
cudaEventElapsedTime(&timing, start, stop);
printf("Elapsed time - ILP0: %3.3f ms \n", timing);
gpuErrchk(cudaMemcpy(c, d_c0, N*sizeof(float), cudaMemcpyDeviceToHost));
// --- Checking the results
for (int i=0; i<N; i++)
if (c[i] != c_ref[i]) {
printf("Error!\n");
return;
}
printf("Test passed!\n");
/******************/
/* ILP2 TEST CASE */
/******************/
cudaEventRecord(start, 0);
ILP2<<<(N/2)/BLOCKSIZE,BLOCKSIZE>>>(d_a, d_b, d_c2);
gpuErrchk(cudaPeekAtLastError());
gpuErrchk(cudaDeviceSynchronize());
cudaEventRecord(stop, 0);
cudaEventSynchronize(stop);
cudaEventElapsedTime(&timing, start, stop);
printf("Elapsed time - ILP2: %3.3f ms \n", timing);
gpuErrchk(cudaMemcpy(c, d_c2, N*sizeof(float), cudaMemcpyDeviceToHost));
// --- Checking the results
for (int i=0; i<N; i++)
if (c[i] != c_ref[i]) {
printf("Error!\n");
return;
}
printf("Test passed!\n");
/******************/
/* ILP4 TEST CASE */
/******************/
cudaEventRecord(start, 0);
ILP4<<<(N/4)/BLOCKSIZE,BLOCKSIZE>>>(d_a, d_b, d_c4);
gpuErrchk(cudaPeekAtLastError());
gpuErrchk(cudaDeviceSynchronize());
cudaEventRecord(stop, 0);
cudaEventSynchronize(stop);
cudaEventElapsedTime(&timing, start, stop);
printf("Elapsed time - ILP4: %3.3f ms \n", timing);
gpuErrchk(cudaMemcpy(c, d_c4, N*sizeof(float), cudaMemcpyDeviceToHost));
// --- Checking the results
for (int i=0; i<N; i++)
if (c[i] != c_ref[i]) {
printf("Error!\n");
return;
}
printf("Test passed!\n");
/******************/
/* ILP8 TEST CASE */
/******************/
cudaEventRecord(start, 0);
ILP8<<<(N/8)/BLOCKSIZE,BLOCKSIZE>>>(d_a, d_b, d_c8);
gpuErrchk(cudaPeekAtLastError());
gpuErrchk(cudaDeviceSynchronize());
cudaEventRecord(stop, 0);
cudaEventSynchronize(stop);
cudaEventElapsedTime(&timing, start, stop);
printf("Elapsed time - ILP8: %3.3f ms \n", timing);
gpuErrchk(cudaMemcpy(c, d_c8, N*sizeof(float), cudaMemcpyDeviceToHost));
// --- Checking the results
for (int i=0; i<N; i++)
if (c[i] != c_ref[i]) {
printf("%f %f\n",c[i],c_ref[i]);
printf("Error!\n");
return;
}
printf("Test passed!\n");
}
PERFORMANCE
Card Kernel Time [ms] Speedup
GeForce GT540M ILP0 4.609 1
" ILP2 2.666 1.72
" ILP4 1.675 2.76
" ILP8 1.477 3.12
Kepler K20c ILP0 0.045
" ILP2 0.043
" ILP4 0.043
" ILP8 0.042

Matrix Multiplication giving wrong output [duplicate]

This question already has an answer here:
Unable to execute device kernel in CUDA
(1 answer)
Closed 7 years ago.
What I am attempting to do is Multiply Matrix A & Matrix B and then from the product matrix I get the index of the maximum value per column. But unfortunately, only the first 128*128 values of the matrix multiplication are correct while others are just garbage. I do not quite understand how this works. I request you to kindly guide me with this ..
#include<stdio.h>
#include "cuda.h"
#include<stdlib.h>
#define blockD 32
const int wA = 128;
const int hA = 4096;
const int wB = 4096;
const int hB = wA;
main(void){
void MatrixMultiplication(float *, float *, float *, float *);
int size_A = wA * hA * sizeof(float);
int size_B = wB * hB * sizeof(float);
int size_C = wB * hA * sizeof(float);
int size_max = 2 * wB * sizeof(float);
float *M, *N, *P, *C;
// allocate memory on the CPU
M = (float*)malloc(size_A);
N = (float*)malloc(size_B);
P = (float*)malloc(size_max);
C = (float*)malloc(size_C);
// initialize the matrices
for (int y=0; y < hA; y++) {
for (int x=0; x < wA; x++){
M[y*wA + x] = 32; //x + y*wA;
}
}
for (int y=0; y<hB; y++) {
for (int x=0; x<wB; x++){
N[y*wB + x] = 21; //x + y*wB;
}
}
MatrixMultiplication(M, N, P, C);
//Write
FILE *f1;
int i,j;
f1 = fopen("C.txt","w");
for(i = hA - 2 ; i < hA; i ++){
for(j = 0; j < wB; j++){
fprintf(f1,"%d\t",int(C[i*wB + j]));
}
fprintf(f1,"\n");
}
fclose(f1);
// free the memory allocated on the CPU
free( M );
free( N );
free( P );
free( C );
cudaDeviceReset();
return 0;
}
__device__ void MaxFunction(float* Pd, float* max)
{
int x = (threadIdx.x + blockIdx.x * blockDim.x);
int y = (threadIdx.y + blockIdx.y * blockDim.y);
int k = 0;
int temp = 0; int temp_idx = 0;
for (k = 0; k < wB; ++k) {
if(Pd[x*wB + k] > temp){
temp = Pd[x*wB + k];
temp_idx = x*wB + k;
}
}
max[y*2 + 0] = temp;
max[y*2 + 1] = temp_idx;
}
__global__ void MatrixMulKernel(float* Md, float* Nd, float* Pd, float* max)
{
// declare cache in the shared memory
__shared__ float Mds[blockD][blockD];
__shared__ float Nds[blockD][blockD];
float Pvalue = 0;
// Loop over the Md and Nd block dimension required to compute the Pd element
for (int m = (wA * blockD * blockIdx.y), n = (blockD * blockIdx.x);
m < ((wA * blockD * blockIdx.y)+wA-1);
m += blockD, n += (blockD*hB)){
// collaboratively loading of Md and Nd blocks into shared memory
Mds[threadIdx.y][threadIdx.x] = Md[m + wA * threadIdx.y + threadIdx.x];
Nds[threadIdx.y][threadIdx.x] = Nd[n + wA * threadIdx.y + threadIdx.x];
__syncthreads();
// keep track of the running sum
for (int k = 0; k < blockD; k++)
Pvalue += Mds[threadIdx.y][k] * Nds[k][threadIdx.x];
__syncthreads();
}
// write back to the global memory
int p = hB * blockD * blockIdx.y + blockD * blockIdx.x;
Pd[p + hB * threadIdx.y + threadIdx.x] = Pvalue;
__syncthreads();
MaxFunction(Pd, max);
}
void MatrixMultiplication(float *M, float *N, float *P, float *C) {
int size_A = wA * hA * sizeof(float);
int size_B = wB * hB * sizeof(float);
int size_C = wB * hA * sizeof(float);
int size_max = 2 * wB * sizeof(float);
float *Md, *Nd, *Pd, *max;
// allocate memory on the GPU
cudaMalloc((void**)&Md, size_A);
cudaMalloc((void**)&Nd, size_B);
cudaMalloc((void**)&Pd, size_C);
cudaMalloc((void**)&max, size_max);
// transfer M and N to device memory
cudaMemcpy(Md, M, size_A, cudaMemcpyHostToDevice);
cudaMemcpy(Nd, N, size_B, cudaMemcpyHostToDevice);
// kernel invocation code
dim3 dimBlock(blockD, blockD);
dim3 dimGrid(wA/blockD, hB/blockD);
//Execute Kernel
MatrixMulKernel<<<dimGrid, dimBlock>>>( Md, Nd, Pd, max);
// transfer P from device
cudaMemcpy(P, max, size_max, cudaMemcpyDeviceToHost);
cudaMemcpy(C, Pd, size_C, cudaMemcpyDeviceToHost);
// free the memory allocated on the GPU
cudaFree(Md);
cudaFree(Nd);
cudaFree(Pd);
cudaFree(max);
}
In your code you seem to have more than one problem. One of the problems is, in place of this:
dim3 dimGrid(wA/blockD, hB/blockD);
You should have this:
dim3 dimGrid(wB/blockD, hA/blockD);
Ultimately you need one thread in your grid for each output point. Your formulation was giving you a grid of 4 blocks by 4 blocks, whereas you need a grid of 128 blocks by 128 blocks.
The other problem I found with your code was in these lines in the kernel:
int p = hB * blockD * blockIdx.y + blockD * blockIdx.x;
Pd[p + hB * threadIdx.y + threadIdx.x] = Pvalue;
They are not indexing properly through the output array. Rather than try to sort it out using your scheme, I used this instead:
Pd[(threadIdx.x + (blockIdx.x * blockDim.x)) + ((threadIdx.y + (blockIdx.y * blockDim.y))*(gridDim.x*blockDim.x))] = Pvalue;
When I made the above two changes to your code, I got what I believe are correct results throughout the array. And it took about 32 seconds on my machine to run it. (Note that I haven't tried fixing your original max-finding code -- see below for a better approach.)
Based on your previous question, you seemed to be concerned about speed. If you want to do fast matrix multiply, you should use cublas. The following code shows how to use cublas to multiply two ordinary C-style matrices (they don't have to be square). I've also included a column-max finding kernel that will be fast when the number of columns is large (say, over 500 or so. You have 4096 columns in your example). For small numbers of columns, there may be quicker ways to perform this function, but small numbers of columns also suggests that the overall problem size may be small and so speed (of this piece of code) will not really be an issue.
Here's the code:
#include <stdio.h>
#include <cublas_v2.h>
#define VERBOSE 1
#define nTPB 64
#define ROW_A 4
#define COL_A 4
#define ROW_B COL_A
#define COL_B 4
#define ROW_C ROW_A
#define COL_C COL_B
#define SIZ_A (ROW_A*COL_A)
#define SIZ_B (ROW_B*COL_B)
#define SIZ_C (ROW_C*COL_C)
// error check macros
#define cudaCheckErrors(msg) \
do { \
cudaError_t __err = cudaGetLastError(); \
if (__err != cudaSuccess) { \
fprintf(stderr, "Fatal error: %s (%s at %s:%d)\n", \
msg, cudaGetErrorString(__err), \
__FILE__, __LINE__); \
fprintf(stderr, "*** FAILED - ABORTING\n"); \
exit(1); \
} \
} while (0)
// for CUBLAS V2 API
#define cublasCheckErrors(fn) \
do { \
cublasStatus_t __err = fn; \
if (__err != CUBLAS_STATUS_SUCCESS) { \
fprintf(stderr, "Fatal cublas error: %d (at %s:%d)\n", \
(int)(__err), \
__FILE__, __LINE__); \
fprintf(stderr, "*** FAILED - ABORTING\n"); \
exit(1); \
} \
} while (0)
__global__ void col_max(float *mat, float *max, unsigned int *midx, unsigned int rows, unsigned int cols){
int idx = threadIdx.x + blockDim.x*blockIdx.x;
if (idx < cols){
float tempmax = mat[idx];
unsigned int tempmidx = 0;
for (int i = 1; i< rows; i++)
if (mat[idx + (i*cols)] > tempmax){
tempmax = mat[idx + (i*cols)];
tempmidx = i;}
max[idx] = tempmax;
midx[idx] = tempmidx;
}
}
int main(){
float *h_A, *h_B, *h_C, *d_A, *d_B, *d_C, *h_max, *d_max;
unsigned int *h_idx, *d_idx;
h_A = (float *)malloc(SIZ_A*sizeof(float));
if (h_A==0) {printf("malloc fail\n"); return -1;}
h_B = (float *)malloc(SIZ_B*sizeof(float));
if (h_B==0) {printf("malloc fail\n"); return -1;}
h_C = (float *)malloc(SIZ_C*sizeof(float));
if (h_C==0) {printf("malloc fail\n"); return -1;}
h_max = (float *)malloc(COL_C*sizeof(float));
if (h_max==0) {printf("malloc fail\n"); return -1;}
h_idx = (unsigned int*)malloc(COL_C*sizeof(unsigned int));
if (h_idx==0) {printf("malloc fail\n"); return -1;}
cudaMalloc((void **)&d_A, SIZ_A*sizeof(float));
cudaMalloc((void **)&d_B, SIZ_B*sizeof(float));
cudaMalloc((void **)&d_C, SIZ_C*sizeof(float));
cudaMalloc((void **)&d_max, COL_C*sizeof(float));
cudaMalloc((void **)&d_idx, COL_C*sizeof(unsigned int));
cudaCheckErrors("cuda malloc fail");
// initialize data
for (int i=0; i< SIZ_A; i++) h_A[i] = (float)(i+1);
for (int i=0; i< SIZ_B; i++) h_B[i] = (float)(i+2);
cudaMemcpy(d_A, h_A, SIZ_A*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(d_B, h_B, SIZ_B*sizeof(float), cudaMemcpyHostToDevice);
cudaCheckErrors("cuda memcpy 1 fail");
const float alpha = 1.0f;
const float beta = 0.0f;
cublasHandle_t handle;
cublasCheckErrors(cublasCreate(&handle));
// C = A*B
// due to cublas expecting column-major storage, parameters
// are scrambled
cublasCheckErrors(cublasSgemm(handle, CUBLAS_OP_N, CUBLAS_OP_N, COL_B, ROW_A, COL_A, &alpha, d_B, COL_B, d_A, COL_A, &beta, d_C, COL_C));
cudaMemcpy(h_C, d_C, SIZ_C*sizeof(float), cudaMemcpyDeviceToHost);
cudaCheckErrors("cuda memcpy 2 fail");
col_max<<<(COL_C + nTPB - 1)/nTPB, nTPB>>>(d_C, d_max, d_idx, ROW_C, COL_C);
cudaCheckErrors("kernel launch fail");
cudaMemcpy(h_max, d_max, COL_C*sizeof(float), cudaMemcpyDeviceToHost);
cudaMemcpy(h_idx, d_idx, COL_C*sizeof(unsigned int), cudaMemcpyDeviceToHost);
cudaCheckErrors("cuda memcpy 3 fail/kernel fail");
if (VERBOSE){
printf("A: \n");
for (int i=0; i< ROW_A; i++){
for (int j=0; j< COL_A; j++)
printf("%7.5G", h_A[j+(i*COL_A)]);
printf("\n");}
printf("B: \n");
for (int i=0; i< ROW_B; i++){
for (int j=0; j< COL_B; j++)
printf("%7.5G", h_B[j+(i*COL_B)]);
printf("\n");}
printf("C = A*B: \n");
for (int i=0; i< ROW_C; i++){
for (int j=0; j< COL_C; j++)
printf("%7.5G", h_C[j+(i*COL_C)]);
printf("\n");}
printf("COLUMN MAX:\n");
for (int i=0; i< COL_C; i++)
printf("%7.5G", h_max[i]);
printf("\nCOLUMN MAX IDX:\n");
for (int i=0; i< COL_C; i++)
printf("%7d", h_idx[i]);
}
printf("\n finished!\n");
return 0;
}
Here's what I used to compile:
$ nvcc -arch=sm_20 -O3 -o t221 t221.cu -lcublas
And here's the sample output:
$ cuda-memcheck ./t221
========= CUDA-MEMCHECK
A:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
B:
2 3 4 5
6 7 8 9
10 11 12 13
14 15 16 17
C = A*B:
100 110 120 130
228 254 280 306
356 398 440 482
484 542 600 658
COLUMN MAX:
484 542 600 658
COLUMN MAX IDX:
3 3 3 3
finished!
========= ERROR SUMMARY: 0 errors
$
When I extended my code to handle the same sizes you indicated, (A = 4096x128, B=128x4096) it took about 1 second on my machine. So it's much faster than your code. However, when I take your code and comment out your call to MaxFunction in the kernel, it also only takes about 1 second to compute the matrix multiply result. So if you wanted to keep your matrix multiply code (i.e. not use cublas) you could break the code into 2 kernels, and use your multiply routine in the first kernel with my max-finding routine (col_max) in the second kernel, and also probably get a pretty fast result.
As #talonmies indicated, if you are running on a windows machine, be sure you are aware of the ramifications of windows TDR. (search that in the upper right corner search box if needed)

Cannot read out Values from Texture Memory

Hi I'm writing a simple Program for practicing to work with texture memory. I Just want to write my data into Texture Memory and write it back into Global Memory. But i cannont read out the Values. Here is the code.
#include <stdio.h>
#include <iostream>
#include "cuda.h"
#include <stdlib.h>
#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include "HelloWorld.h"
#include "linearInterpolation_kernel4.cu"
using namespace std;
using std::cout;
const int blocksize = 16;
__global__
void hello(char *a, int *b) {
a[threadIdx.x] += b[threadIdx.x];
}
////////////////////////////////////////////////////////////////////////////////
// These are CUDA Helper functions
// This will output the proper CUDA error strings in the event that a CUDA host call returns an error
#define checkCudaErrors(err) __checkCudaErrors (err, __FILE__, __LINE__)
inline void __checkCudaErrors( cudaError err, const char *file, const int line )
{
if( cudaSuccess != err) {
printf("%s(%i) : CUDA Runtime API error %d: %s.\n",file, line, (int)err, cudaGetErrorString( err ) );
}
}
// This will output the proper error string when calling cudaGetLastError
#define getLastCudaError(msg) __getLastCudaError (msg, __FILE__, __LINE__)
inline void __getLastCudaError( const char *errorMessage, const char *file, const int line )
{
cudaError_t err = cudaGetLastError();
if( cudaSuccess != err) {
printf("%s(%i) : getLastCudaError() CUDA error : %s : (%d) %s.\n", file, line, errorMessage, (int)err, cudaGetErrorString( err ) );
}
}
int main()
{
int N = 40;
float *A;
A = (float *) malloc(N*sizeof(float));
float *B;
B = (float *) malloc(N*sizeof(float));
float *result;
result = (float *) malloc(N*sizeof(float));
float angle = 0.8f;
for(int i = 0; i < N; i++){
A[i] = i; //(float)rand();
B[i] = i+1; //(float)rand();
}
ipLinearTexture2(A,B,result,angle,N);
float result2;
result2 = (angle)*A[4] + (1-angle)*B[4];
printf(" A %f B %f Result %f\n", A[4], B[4], result[4]);
cout << result2 << endl;
return 1;
}
void ipLinearTexture2(float *A, float* B, float* result, float angle, int N)
{
float cuTime;
int N2 = N * 2;
float *dev_result;
float **AB;
AB = (float **) malloc( N * sizeof(float *));
if(AB)
{
for(int i = 0; i < N; i++)
{
AB[i] = (float *) malloc( 2 * sizeof(float *));
}
}
for (int i = 0; i < N; i = i++)
{
AB[i][0] = A[i];
AB[i][1] = B[i];
}
cudaMalloc(&dev_result, N * sizeof(float));
unsigned int size = N2 * sizeof(float);
//cudaChannelFormatDesc channelDesc = cudaCreateChannelDesc(32, 0, 0, 0, cudaChannelFormatKindFloat);
cudaChannelFormatDesc channelDesc = cudaCreateChannelDesc<float>();
cudaArray* cu_array;
checkCudaErrors(cudaMallocArray( &cu_array, &channelDesc,N,2));
cudaMemcpy2DToArray(cu_array,0,0,AB,N * sizeof(float), N * sizeof(float), 2, cudaMemcpyHostToDevice);
// set texture parameters
tex2.normalized = true;
tex2.filterMode = cudaFilterModeLinear;
tex2.addressMode[0] = cudaAddressModeWrap; //cudaAddressModeWrap;
tex2.addressMode[1] = cudaAddressModeWrap; //cudaAddressModeClamp;
checkCudaErrors(cudaBindTextureToArray( tex2, cu_array, channelDesc));
dim3 dimBlock(10, 1, 1);
dim3 dimGrid((int)ceil((double)N*2/dimBlock.x), 1, 1);
transformKernel4<<< 256, 256, 0 >>>( dev_result, N, 2, angle);
checkCudaErrors(cudaMemcpy(result, dev_result, N * sizeof(float), cudaMemcpyDeviceToHost));
cout << "==================================================" << endl;
for (int i = 0 ; i < N ;i++)
{
cout << result[i] << " on " << i << endl;
}
cout << "==================================================" << endl;
checkCudaErrors(cudaUnbindTexture(tex));
checkCudaErrors(cudaFree(dev_result));
checkCudaErrors(cudaFreeArray(cu_array));
}
and here is the kernel code
#ifndef _SIMPLETEXTURE_KERNEL5_H_
#define _SIMPLETEXTURE_KERNEL5_H_
// Texture references
texture<float, 2, cudaReadModeElementType> tex2;
__global__ void
transformKernel4(float* g_odata, int width, int height, float theta)
{
unsigned int xid = blockIdx.x * blockDim.x + threadIdx.x;
unsigned int yid = blockIdx.y * blockDim.y + threadIdx.y;
if (xid >= width || yid >= height) return;
float dx = 1.0f / (float)width;
float dy = 1.0f / (float)height;
float x = ((float)xid + 0.5f) * dx;
float y = ((float)yid + 0.5f) * dy;
float value = tex2D(tex2, x , y);
printf("wert %f xid %i yid %i \n",value, xid, yid);
g_odata[yid * width + xid] = value;
}
#endif // #ifndef _SIMPLETEXTURE_KERNEL_H_
Can somebody tell what i am doing wrong?
I have edited it to remove the first 2 logical mistake. Put why am I need able to print out my data?
It was the wrong binding of the Arrays. You can not use multidimensional Arrays in C that can be copied. You have to use a onedimensional array that respresents a multidimensional.
I can see 2 logical errors here.
The first one is the one pointed out by #asm.
The output should be stored by calculating linear index from 2D x and y indices.
outputIndex = yid * width + xid;
The second one is that the memory allocation for the cudaArray structure is internally aligned.
You should consider using cudaMemcpy2DToArray function to avoid erroneous data copying.
cudaMemcpy2DToArray(cu_array,0,0,AB,N * sizeof(float), N * sizeof(float), 2, cudaMemcpyHostToDevice);