Accessing dynamically allocated arrays on device (without passing them as kernel arguments) - memory-management

How can an array of structs that has been dynamically allocated on the host be used by a kernel, without passing the array of structs as a kernel argument? This seems like a common procedure with a good amount of documentation online, yet it doesn't work on the following program.
Note: Please note that the following questions have been studied before posting this question:
1) copying host memory to cuda __device__ variable 2) Global variable in CUDA 3) Is there any way to dynamically allocate constant memory? CUDA
So far, unsuccessful attempts have been made to:
Dynamically allocate array of structs with cudaMalloc(), then
Use cudaMemcpyToSymbol() with the pointer returned from cudaMalloc() to copy to a __device__ variable which can be used by the kernel.
Code attempt:
NBody.cu (error checking using cudaStatus has mostly been omitted for better readability, and function to read data from file into dynamic array removed):
#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include <stdio.h>
#include <stdlib.h>
#define BLOCK 256
struct nbody {
float x, y, vx, vy, m;
};
typedef struct nbody nbody;
// Global declarations
nbody* particle;
// Device variables
__device__ unsigned int d_N; // Kernel can successfully access this
__device__ nbody d_particle; // Update: part of problem was here with (*)
// Aim of kernel: to print contents of array of structs without using kernel argument
__global__ void step_cuda_v1() {
int i = threadIdx.x + blockDim.x * blockIdx.x;
if (i < d_N) {
printf("%.f\n", d_particle.x);
}
}
int main() {
unsigned int N = 10;
unsigned int I = 1;
cudaMallocHost((void**)&particle, N * sizeof(nbody)); // Host allocation
cudaError_t cudaStatus;
for (int i = 0; i < N; i++) particle[i].x = i;
nbody* particle_buf; // device buffer
cudaSetDevice(0);
cudaMalloc((void**)&particle_buf, N * sizeof(nbody)); // Allocate device mem
cudaMemcpy(particle_buf, particle, N * sizeof(nbody), cudaMemcpyHostToDevice); // Copy data into device mem
cudaMemcpyToSymbol(d_particle, &particle_buf, sizeof(nbody*)); // Copy pointer to data into __device__ var
cudaMemcpyToSymbol(d_N, &N, sizeof(unsigned int)); // This works fine
int NThreadBlock = (N + BLOCK - 1) / BLOCK;
for (int iteration = 0; iteration <= I; iteration++) {
step_cuda_v1 << <NThreadBlock, BLOCK >> > ();
//step_cuda_v1 << <1, 5 >> > (particle_buf);
cudaDeviceSynchronize();
cudaStatus = cudaGetLastError();
if (cudaStatus != cudaSuccess)
{
fprintf(stderr, "ERROR: %s\n", cudaGetErrorString(cudaStatus));
exit(-1);
}
}
return 0;
}
OUTPUT:
"ERROR: kernel launch failed."
Summary:
How can I print the contents of the array of structs from the kernel, without passing it as a kernel argument?
Coding in C using VS2019 with CUDA 10.2

With the help of #Robert Crovella and #talonmies, here is the solution that outputs a sequence that cycles from 0 to 9 repeatedly.
#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include <stdio.h>
#include <stdlib.h>
#define BLOCK 256
//#include "Nbody.h"
struct nbody {
float x, y, vx, vy, m;
};
typedef struct nbody nbody;
// Global declarations
nbody* particle;
// Device variables
__device__ unsigned int d_N; // Kernel can successfully access this
__device__ nbody* d_particle;
//__device__ nbody d_particle; // Update: part of problem was here with (*)
// Aim of kernel: to print contents of array of structs without using kernel argument
__global__ void step_cuda_v1() {
int i = threadIdx.x + blockDim.x * blockIdx.x;
if (i < d_N) {
printf("%.f\n", d_particle[i].x);
}
}
int main() {
unsigned int N = 10;
unsigned int I = 1;
cudaMallocHost((void**)&particle, N * sizeof(nbody)); // Host allocation
cudaError_t cudaStatus;
for (int i = 0; i < N; i++) particle[i].x = i;
nbody* particle_buf; // device buffer
cudaSetDevice(0);
cudaMalloc((void**)&particle_buf, N * sizeof(nbody)); // Allocate device mem
cudaMemcpy(particle_buf, particle, N * sizeof(nbody), cudaMemcpyHostToDevice); // Copy data into device mem
cudaMemcpyToSymbol(d_particle, &particle_buf, sizeof(nbody*)); // Copy pointer to data into __device__ var
cudaMemcpyToSymbol(d_N, &N, sizeof(unsigned int)); // This works fine
int NThreadBlock = (N + BLOCK - 1) / BLOCK;
for (int iteration = 0; iteration <= I; iteration++) {
step_cuda_v1 << <NThreadBlock, BLOCK >> > ();
//step_cuda_v1 << <1, 5 >> > (particle_buf);
cudaDeviceSynchronize();
cudaStatus = cudaGetLastError();
if (cudaStatus != cudaSuccess)
{
fprintf(stderr, "ERROR: %s\n", cudaGetErrorString(cudaStatus));
exit(-1);
}
}
return 0;
}

Related

CUDA Initialize Array on Device

I am very new to CUDA and I am trying to initialize an array on the device and return the result back to the host to print out to show if it was correctly initialized. I am doing this because the end goal is a dot product solution in which I multiply two arrays together, storing the results in another array and then summing up the entire thing so that I only need to return the host one value.
In the code I am working on all I am only trying to see if I am initializing the array correctly. I am trying to create an array of size N following the patterns of 1,2,3,4,5,6,7,8,1,2,3....
This is the code that I've written and it compiles without issue but when I run it the terminal is hanging and I have no clue why. Could someone help me out here? I'm so incredibly confused :\
#include <stdio.h>
#include <stdlib.h>
#include <chrono>
#define ARRAY_SIZE 100
#define BLOCK_SIZE 32
__global__ void cu_kernel (int *a_d,int *b_d,int *c_d, int size)
{
int x = blockIdx.x * blockDim.x + threadIdx.x;
__shared__ int temp;
if(temp != 8){
a_d[x] = temp;
temp++;
} else {
a_d[x] = temp;
temp = 1;
}
}
int main (int argc, char *argv[])
{
//declare pointers for arrays
int *a_d, *b_d, *c_d, *sum_h, *sum_d,a_h[ARRAY_SIZE];
//set space for device variables
cudaMalloc((void**) &a_d, sizeof(int) * ARRAY_SIZE);
cudaMalloc((void**) &b_d, sizeof(int) * ARRAY_SIZE);
cudaMalloc((void**) &c_d, sizeof(int) * ARRAY_SIZE);
cudaMalloc((void**) &sum_d, sizeof(int));
// set execution configuration
dim3 dimblock (BLOCK_SIZE);
dim3 dimgrid (ARRAY_SIZE/BLOCK_SIZE);
// actual computation: call the kernel
cu_kernel <<<dimgrid, dimblock>>> (a_d,b_d,c_d,ARRAY_SIZE);
cudaError_t result;
// transfer results back to host
result = cudaMemcpy (a_h, a_d, sizeof(int) * ARRAY_SIZE, cudaMemcpyDeviceToHost);
if (result != cudaSuccess) {
fprintf(stderr, "cudaMemcpy failed.");
exit(1);
}
// print reversed array
printf ("Final state of the array:\n");
for (int i =0; i < ARRAY_SIZE; i++) {
printf ("%d ", a_h[i]);
}
printf ("\n");
}
There are at least 3 issues with your kernel code.
you are using shared memory variable temp without initializing it.
you are not resolving the order in which threads access a shared variable as discussed here.
you are imagining (perhaps) a particular order of thread execution, and CUDA provides no guarantees in that area
The first item seems self-evident, however naive methods to initialize it in a multi-threaded environment like CUDA are not going to work. Firstly we have the multi-threaded access pattern, again, Furthermore, in a multi-block scenario, shared memory in one block is logically distinct from shared memory in another block.
Rather than wrestle with mechanisms unsuited to create the pattern you desire, (informed by notions carried over from a serial processing environment), I would simply do something trivial like this to create the pattern you desire:
__global__ void cu_kernel (int *a_d,int *b_d,int *c_d, int size)
{
int x = blockIdx.x * blockDim.x + threadIdx.x;
if (x < size) a_d[x] = (x&7) + 1;
}
Are there other ways to do it? certainly.
__global__ void cu_kernel (int *a_d,int *b_d,int *c_d, int size)
{
int x = blockIdx.x * blockDim.x + threadIdx.x;
__shared__ int temp;
if (!threadIdx.x) temp = blockIdx.x*blockDim.x;
__syncthreads();
if (x < size) a_d[x] = ((temp+threadIdx.x) & 7) + 1;
}
You can get as fancy as you like.
These changes will still leave a few values at zero at the end of the array, which would require changes to your grid sizing. There are many questions about this already, or study a sample code like vectorAdd.

Openacc error ibgomp: while loading libgomp-plugin-host_nonshm.so.1: libgomp-plugin-host_nonshm.so.1: cannot

I want to compile an easy openacc sample (it was attached) , it was correctly compiled but when i run it got an error :
compile with : gcc-5 -fopenacc accVetAdd.c -lm
run with : ./a.out
got error in runtime
error: libgomp: while loading libgomp-plugin-host_nonshm.so.1: libgomp-plugin-host_nonshm.so.1: cannot open shared object file: No such file or directory
I google it and find only one page! then i ask how to fix this problem?
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main(int argc, char* argv[])
{
// Size of vectors
int n = 10000;
// Input vectors
double *restrict a;
double *restrict b;
// Output vector
double *restrict c;
// Size, in bytes, of each vector
size_t bytes = n*sizeof(double);
// Allocate memory for each vector
a = (double*)malloc(bytes);
b = (double*)malloc(bytes);
c = (double*)malloc(bytes);
// Initialize content of input vectors, vector a[i] = sin(i)^2 vector b[i] = cos(i)^2
int i;
for (i = 0; i<n; i++) {
a[i] = sin(i)*sin(i);
b[i] = cos(i)*cos(i);
}
// sum component wise and save result into vector c
#pragma acc kernels copyin(a[0:n],b[0:n]), copyout(c[0:n])
for (i = 0; i<n; i++) {
c[i] = a[i] + b[i];
}
// Sum up vector c and print result divided by n, this should equal 1 within error
double sum = 0.0;
for (i = 0; i<n; i++) {
sum += c[i];
}
sum = sum / n;
printf("final result: %f\n", sum);
// Release memory
free(a);
free(b);
free(c);
return 0;
}
libgomp dynamically loads shared object files for the plugins it supports, such as the one implementing the host_nonshm device. If they're installed in a non-standard directory (that is, not in the system's default search path), you need to tell the dynamic linker where to look for these shared object files: either compile with -Wl,-rpath,[...], or set the LD_LIBRARY_PATH environment variable.

Cuda kernel function only changes matrix's first row

I am trying to sum up two matrices a_h_1 and a_h_2, and writing the result back to a_h_1. But for some reason my kernel function does not change the array members except the first N elements. Even if I write a[8] = 45, for example, it is printed as 8 when it is copied back to host. What is wrong?
#include <stdio.h>
#include <cuda_runtime.h>
#include <device_launch_parameters.h>
#include <cuda.h>
// Kernel that executes on the CUDA device
__global__ void matrix_summation(float *a, float *b, int M, int N)
{
int idx = blockIdx.x * blockDim.x + threadIdx.x;
if (idx<M*N)
{
a[idx] = blockIdx.x;
}
}
// main routine that executes on the host
int main(void)
{
float *a_h_1,*a_h_2, *a_d_1,*a_d_2; // Pointer to host & device arrays
const int N = 5;
const int M = 5;
// Number of elements in arrays
size_t size = (N * M) * sizeof(float);
a_h_1 = (float *)malloc(size); // Allocate array1 on host
a_h_2 = (float *)malloc(size); // Allocate array2 on host
cudaMalloc((void **) &a_d_1, size); // Allocate array1 on device
cudaMalloc((void **) &a_d_2, size); // Allocate array2 on device
// Initialize host array and copy it to CUDA device
for (int i=0; i<N*M; i++){
a_h_1[i] = (float)i;
a_h_2[i] = (float)i;
}
cudaMemcpy(a_d_1, a_h_1, size, cudaMemcpyHostToDevice);
cudaMemcpy(a_d_2, a_h_2, size, cudaMemcpyHostToDevice);
// Do calculation on device:
int block_size = M;
int n_blocks = (M*N)/block_size;
matrix_summation <<< n_blocks, block_size >>> ( a_d_1,a_d_2, M, N));
// Retrieve result from device and store it in host array
cudaMemcpy(a_h_1, a_d_1, sizeof(float)*N, cudaMemcpyDeviceToHost);
// Print results
printf("\n\nROW 1 \n");
for (int i=0; i<(M*N); i++)
{
printf(" %f ", a_h_1[i]);
if((i+1)%N == 0)
{
printf("\nROW %d \n", ((i+1)/N)+1);
}
}
// Cleanup
free(a_h_1);
free(a_h_2);
cudaFree(a_d_1);
cudaFree(a_d_2);
system("pause");
}
Here is the output:
ROW 1
0.0 2.0 4.0 6.0 8.0 < this line is correct but others are not
ROW 2
5.0 6.0 7.0 8.0 9.0
ROW 3
10.0 11.0 12.0 13.0 14.0
ROW 4
15.0 16.0 17.0 18.0 19.0
ROW 5
20.0 21.0 22.0 23.0 24.0
It looks like you're not copying all of the device array to your host array. In this line:
cudaMemcpy(a_h_1, a_d_1, sizeof(float)*N, cudaMemcpyDeviceToHost);
I think you meant to copy sizeof(float)*N*M

Cuda thrust global memory writing very slow

I am currently writing a code, that calculates a integral Histogram on the GPU using the Nvidia thrust library.
Therefore I allocate a continuous Block of device memory which I update with a custom functor all the time.
The problem is, that the write to the device memory is veeery slow, but the reads are actually ok.
The basic setup is the following:
struct HistogramCreation
{
HistogramCreation(
...
// pointer to memory
...
){}
/// The actual summation operator
__device__ void operator()(int index){
.. do the calculations ..
for(int j=0;j<30;j++){
(1) *_memoryPointer = values (also using reads to such locations) ;
}
}
}
void foo(){
cudaMalloc(_pointer,size);
HistogramCreation initialCreation( ... _pointer ...);
thrust::for_each(
thrust::make_counting_iterator(0),
thrust::make_counting_iterator(_imageSize),
initialCreation);
}
if I change the writing in (1) to the following>
unsigned int val = values;
The performance is much better. THis is the only global memory write I have.
Using the memory write I get about 2s for HD Footage.
using the local variable it takes about 50 ms so about a factor of 40 less.
Why is this so slow? how could I improve it?
Just as #OlegTitov said, frequent load/store with global
memory should be avoided as much as possible. When there's a
situation where it's inevitable, then coalesced memory
access can help the execution process not to get too slow;
however in most cases, histogram calculation is pretty tough
to realize the coalesced access.
While most of the above is basically just restating
#OlegTitov's answer, i'd just like to share about an
investigation i did about finding summation with NVIDIA
CUDA. Actually the result is pretty interesting and i hope
it'll be a helpful information for other xcuda developers.
The experiment was basically to run a speed test of finding
summation with various memory access patterns: using global
memory (1 thread), L2 cache (atomic ops - 128 threads), and
L1 cache (shared mem - 128 threads)
This experiment used:
Kepler GTX 680,
1546 cores # 1.06GHz
GDDR5 256-bit # 3GHz
Here are the kernels:
__global__
void glob(float *h) {
float* hist = h;
uint sd = SEEDRND;
uint random;
for (int i = 0; i < NUMLOOP; i++) {
if (i%NTHREADS==0) random = rnd(sd);
int rind = random % NBIN;
float randval = (float)(random % 10)*1.0f ;
hist[rind] += randval;
}
}
__global__
void atom(float *h) {
float* hist = h;
uint sd = SEEDRND;
for (int i = threadIdx.x; i < NUMLOOP; i+=NTHREADS) {
uint random = rnd(sd);
int rind = random % NBIN;
float randval = (float)(random % 10)*1.0f ;
atomicAdd(&hist[rind], randval);
}
}
__global__
void shm(float *h) {
int lid = threadIdx.x;
uint sd = SEEDRND;
__shared__ float shm[NTHREADS][NBIN];
for (int i = 0; i < NBIN; i++) shm[lid][i] = h[i];
for (int i = lid; i < NUMLOOP; i+=NTHREADS) {
uint random = rnd(sd);
int rind = random % NBIN;
float randval = (float)(random % 10)*1.0f ;
shm[lid][rind] += randval;
}
/* reduction here */
for (int i = 0; i < NBIN; i++) {
__syncthreads();
if (threadIdx.x < 64) {
shm[threadIdx.x][i] += shm[threadIdx.x+64][i];
}
__syncthreads();
if (threadIdx.x < 32) {
shm[threadIdx.x][i] += shm[threadIdx.x+32][i];
}
__syncthreads();
if (threadIdx.x < 16) {
shm[threadIdx.x][i] += shm[threadIdx.x+16][i];
}
__syncthreads();
if (threadIdx.x < 8) {
shm[threadIdx.x][i] += shm[threadIdx.x+8][i];
}
__syncthreads();
if (threadIdx.x < 4) {
shm[threadIdx.x][i] += shm[threadIdx.x+4][i];
}
__syncthreads();
if (threadIdx.x < 2) {
shm[threadIdx.x][i] += shm[threadIdx.x+2][i];
}
__syncthreads();
if (threadIdx.x == 0) {
shm[0][i] += shm[1][i];
}
}
for (int i = 0; i < NBIN; i++) h[i] = shm[0][i];
}
OUTPUT
atom: 102656.00 shm: 102656.00 glob: 102656.00
atom: 122240.00 shm: 122240.00 glob: 122240.00
... blah blah blah ...
One Thread: 126.3919 msec
Atomic: 7.5459 msec
Sh_mem: 2.2207 msec
The ratio between these kernels is 57:17:1. Many things can
be analyzed here, and it truly does not mean that using
L1 or L2 memory spaces will always give you more than 10
times speedup of the whole program.
And here's the main and other funcs:
#include <iostream>
#include <cstdlib>
#include <cstdio>
using namespace std;
#define NUMLOOP 1000000
#define NBIN 36
#define SEEDRND 1
#define NTHREADS 128
#define NBLOCKS 1
__device__ uint rnd(uint & seed) {
#if LONG_MAX > (16807*2147483647)
int const a = 16807;
int const m = 2147483647;
seed = (long(seed * a))%m;
return seed;
#else
double const a = 16807;
double const m = 2147483647;
double temp = seed * a;
seed = (int) (temp - m * floor(temp/m));
return seed;
#endif
}
... the above kernels ...
int main()
{
float *h_hist, *h_hist2, *h_hist3, *d_hist, *d_hist2,
*d_hist3;
h_hist = (float*)malloc(NBIN * sizeof(float));
h_hist2 = (float*)malloc(NBIN * sizeof(float));
h_hist3 = (float*)malloc(NBIN * sizeof(float));
cudaMalloc((void**)&d_hist, NBIN * sizeof(float));
cudaMalloc((void**)&d_hist2, NBIN * sizeof(float));
cudaMalloc((void**)&d_hist3, NBIN * sizeof(float));
for (int i = 0; i < NBIN; i++) h_hist[i] = 0.0f;
cudaMemcpy(d_hist, h_hist, NBIN * sizeof(float),
cudaMemcpyHostToDevice);
cudaMemcpy(d_hist2, h_hist, NBIN * sizeof(float),
cudaMemcpyHostToDevice);
cudaMemcpy(d_hist3, h_hist, NBIN * sizeof(float),
cudaMemcpyHostToDevice);
cudaEvent_t start, end;
float elapsed = 0, elapsed2 = 0, elapsed3;
cudaEventCreate(&start);
cudaEventCreate(&end);
cudaEventRecord(start, 0);
atom<<<NBLOCKS, NTHREADS>>>(d_hist);
cudaThreadSynchronize();
cudaEventRecord(end, 0);
cudaEventSynchronize(start);
cudaEventSynchronize(end);
cudaEventElapsedTime(&elapsed, start, end);
cudaEventRecord(start, 0);
shm<<<NBLOCKS, NTHREADS>>>(d_hist2);
cudaThreadSynchronize();
cudaEventRecord(end, 0);
cudaEventSynchronize(start);
cudaEventSynchronize(end);
cudaEventElapsedTime(&elapsed2, start, end);
cudaEventRecord(start, 0);
glob<<<1, 1>>>(d_hist3);
cudaThreadSynchronize();
cudaEventRecord(end, 0);
cudaEventSynchronize(start);
cudaEventSynchronize(end);
cudaEventElapsedTime(&elapsed3, start, end);
cudaMemcpy(h_hist, d_hist, NBIN * sizeof(float),
cudaMemcpyDeviceToHost);
cudaMemcpy(h_hist2, d_hist2, NBIN * sizeof(float),
cudaMemcpyDeviceToHost);
cudaMemcpy(h_hist3, d_hist3, NBIN * sizeof(float),
cudaMemcpyDeviceToHost);
/* print output */
for (int i = 0; i < NBIN; i++) {
printf("atom: %10.2f shm: %10.2f glob:
%10.2f¥n",h_hist[i],h_hist2[i],h_hist3[i]);
}
printf("%12s: %8.4f msec¥n", "One Thread", elapsed3);
printf("%12s: %8.4f msec¥n", "Atomic", elapsed);
printf("%12s: %8.4f msec¥n", "Sh_mem", elapsed2);
return 0;
}
When writing GPU code you should avoid reading and writing to/from global memory. Global memory is very slow on GPU. That's the hardware feature. The only thing you can do is to make neighboring treads read/write in neighboring adresses in global memory. This will cause coalescing and speed up the process. But in general read your data once, process it and write it out once.
Note that NVCC might optimize out a lot of your code after you make the modification - it detects that no write to global memory is made and just removes the "unneeded" code. So this speedup may not be coming out of the global writer per ce.
I would recommend using profiler on your actual code (the one with global write) to see if there's anything like unaligned access or other perf problem.

Breakpoints inside CUDA kernel __global__ not hitting

Using visual studios 2010. Win 7. Nsight 2.1
#include "cuda.h"
#include "cuda_runtime.h"
#include "device_launch_parameters.h"
// incrementArray.cu
#include <stdio.h>
#include <assert.h>
void incrementArrayOnHost(float *a, int N)
{
int i;
for (i=0; i < N; i++) a[i] = a[i]+1.f;
}
__global__ void incrementArrayOnDevice(float *a, int N)
{
int idx = blockIdx.x*blockDim.x + threadIdx.x;
int j = idx;
int i = 2;
i = i+j; //->breakpoint here
if (idx<N) a[idx] = a[idx]+1.f; //->breakpoint here
}
int main(void)
{
float *a_h, *b_h; // pointers to host memory
float *a_d; // pointer to device memory
int i, N = 10;
size_t size = N*sizeof(float);
// allocate arrays on host
a_h = (float *)malloc(size);
b_h = (float *)malloc(size);
// allocate array on device
cudaMalloc((void **) &a_d, size);
// initialization of host data
for (i=0; i<N; i++) a_h[i] = (float)i;
// copy data from host to device
cudaMemcpy(a_d, a_h, sizeof(float)*N, cudaMemcpyHostToDevice);
// do calculation on host
incrementArrayOnHost(a_h, N);
// do calculation on device:
// Part 1 of 2. Compute execution configuration
int blockSize = 4;
int nBlocks = N/blockSize + (N%blockSize == 0?0:1);
// Part 2 of 2. Call incrementArrayOnDevice kernel
incrementArrayOnDevice <<< nBlocks, blockSize >>> (a_d, N);
// Retrieve result from device and store in b_h
cudaMemcpy(b_h, a_d, sizeof(float)*N, cudaMemcpyDeviceToHost);
// check results
for (i=0; i<N; i++) assert(a_h[i] == b_h[i]);
// cleanup
free(a_h); free(b_h); cudaFree(a_d);
return 0;
}
I've tried inserting breakpoints as listed above inside my global void incrementArrayOnDevice(float *a, int N) but they're not hitting.
When I run debug (f5) in visual studios, I tried to step into incrementArrayOnDevice <<< nBlocks, blockSize >>> (a_d, N); but they would skip the entire kernel code section.
tried to add a watch on the variables i and j but there was an error "CXX0017: Error: symbol "i" not found."
Is this issue normal? Can someone please try on their pc and let me know if they can hit the breakpoints? If you can, what possible problem could mine be? Please help! :(
Nsight debugging is different from VS debugging . You need to use Nsight debugging to hit the kernel breakpoints. However, for this you need 2 GPU cards. Do you have 2 cards in the first place? Please check
You can debug on a single GPU but on the following conditions:
You have to be using 5.0 toolkit
You have to be programming on a GPU that suports 303.xx NForceWare or higher

Resources