Matrix not copied back from device to host successfully in CUDA

Question

I am new to cuda. I wrote a kernel to create an identity matrix(GPUsetIdentity) of dimension sizeXsize. Further inside a function GPUfunctioncall, I called my kernel. The identity matrix should be stored in dDataInv. But when I copy it back to dataOut sizexsize , all the values are zero. I know, I am doing something very stupid somewhere, but couldnt get it, I am new to cuda, if anyone can point my mistake. Thanks.

#include <stdio.h>
#include <malloc.h>
#include <memory.h>
#include <math.h>
#include <stdlib.h>
#include <iostream>
#include <stdlib.h>
#include <string>
#include <fstream>
#include <iterator>
#include <sstream>
#include <vector>
#include <cstring>
#include <cstdlib>
#include <ctime>  
#include <stdlib.h>
#include <cuda_runtime.h>
#include "cuda.h"

#define BLOCKSIZE 16


using namespace std;

__global__ void GPUsetIdentity (float* matrix, int width)

{
        int tx = threadIdx.x;
        int bx = blockIdx.x;
        int offset = bx * BLOCKSIZE + tx;
        matrix[offset + width * offset] = 1;

}


void print_matrix_host(float* A , int nr_rows_A, int nr_cols_A) {

        for(int i = 0; i < nr_rows_A; ++i){
                for(int j = 0; j < nr_cols_A; ++j){
                        std::cout << A[i * nr_rows_A + j ]  << " ";
                }
                std::cout << std::endl;

        }
        std::cout << std::endl;
}

int GPUfunctioncall (float* hDataOut, int size){

        float *dDataInv;


        cudaMalloc ((void **) &dDataInv, size);
        cudaMemset ((void *) dDataInv, 0, size);



        dim3 idyThreads (BLOCKSIZE);
        dim3 idyBlocks (size / BLOCKSIZE);


        GPUsetIdentity <<< idyBlocks, idyThreads >>> (dDataInv, size);
        cudaThreadSynchronize ();

        cudaMemcpy ((void *) hDataOut, (void *) dDataInv, size, cudaMemcpyDeviceToHost);
        cudaFree (dDataInv);

        return 0;

}

int main()

{
        int size = 4;
        float* dataOut;

        dataOut = new float[size*size];

        GPUfunctioncall(dataOut, size);
        print_matrix_host(dataOut, size, size);


}

Ask yourself what the value of (size/BLOCKSIZE) is. – talonmies Oct 04 '14 at 14:54 — talonmies, Oct 04 '14 at 14:54

score 1 · Answer 1 · edited May 23 '17 at 12:20

Any time you are having trouble with a CUDA code, it's good practice to use proper cuda error checking. You can also run your code with cuda-memcheck to get a quick read on whether there are any errors.

Using either of these methods, you would have discovered an "invalid configuration error" on your kernel launch. This usually means that the parameters in the <<< >>> syntax are incorrect. When you run into this type of error, simply printing out those values may indicate the problem.

In your case, this line of code:

    dim3 idyBlocks (size / BLOCKSIZE);

results in a value of 0 for idyBlocks when size is 4 and BLOCKSIZE is 16. So you are requesting a kernel launch of 0 blocks which is illegal. Therefore your kernel is not running and your results are not what you expect.

There are a variety of ways to solve this, many of them involving detecting this condition and adding an "extra block" when size is not evenly divisible by BLOCKSIZE. Using this approach, we may be launching "extra threads", so we must include a "thread check" in the kernel to prevent those extra threads from doing anything (such as accessing arrays out of bounds). For this, we often need to know the intended size in the kernel, and we can pass this value as an extra kernel parameter.

You've also made some errors in your handling of device variables. The following code:

    dataOut = new float[size*size];

allocates enough space for a square matrix of dimension size. But the following code:

    cudaMalloc ((void **) &dDataInv, size);

only allocates enough space for size bytes. You want size*size*sizeof(float) instead of size here, and likewise you want it in the following cudaMemset and cudaMemcpy operations. cudaMalloc, cudaMemset and cudaMemcpy require a size parameter in bytes, just like malloc, memset, and memcpy. This error is found in your usage of cudaMemset and cudaMemcpy as well.

The following code has those modifications, and seems to work correctly for me:

$ cat t580.cu
#include <stdio.h>
#include <stdlib.h>
#include <iostream>

#define BLOCKSIZE 16


using namespace std;

__global__ void GPUsetIdentity (float* matrix, int width, int size)

{
        int tx = threadIdx.x;
        int bx = blockIdx.x;
        int offset = bx * BLOCKSIZE + tx;
        if (tx < size)
          matrix[offset + width * offset] = 1;

}


void print_matrix_host(float* A , int nr_rows_A, int nr_cols_A) {

        for(int i = 0; i < nr_rows_A; ++i){
                for(int j = 0; j < nr_cols_A; ++j){
                        std::cout << A[i * nr_rows_A + j ]  << " ";
                }
                std::cout << std::endl;

        }
        std::cout << std::endl;
}

int GPUfunctioncall (float* hDataOut, int size){

        float *dDataInv;


        cudaMalloc ((void **) &dDataInv, size*size*sizeof(float));
        cudaMemset ((void *) dDataInv, 0, size*size*sizeof(float));



        dim3 idyThreads (BLOCKSIZE);
        int num_blocks = size/BLOCKSIZE + (size%BLOCKSIZE)?1:0;
        dim3 idyBlocks (num_blocks);


        GPUsetIdentity <<< idyBlocks, idyThreads >>> (dDataInv, size, size);
        cudaThreadSynchronize ();

        cudaMemcpy ((void *) hDataOut, (void *) dDataInv, size*size*sizeof(float), cudaMemcpyDeviceToHost);
        cudaFree (dDataInv);

        return 0;

}

int main()

{
        int size = 4;
        float* dataOut;

        dataOut = new float[size*size];

        GPUfunctioncall(dataOut, size);
        print_matrix_host(dataOut, size, size);


}
$ nvcc -arch=sm_20 -o t580 t580.cu
$ cuda-memcheck ./t580
========= CUDA-MEMCHECK
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1

========= ERROR SUMMARY: 0 errors
$

Note that it may be redundant to pass size twice to the kernel. For this particular example, we could have easily used the width parameter to do our kernel "thread check". But for educational purposes, I chose to call it out as a separate parameter, because in the general case you will often pass it as a separate parameter to other kernels that you write.

Finally, note that cudaThreadSynchronize() is deprecated and should be replaced with cudaDeviceSynchronize() instead. In this particular example, niether are actually necessary, as the next cudaMemcpy operation will force the same kind of synchronization, but you may use it if you decide to add cuda error checking to your code (recommended).

Oh, Thanks, I really did some stupid mistakes, and Thanks, next time I will refer to cuda error checking ........ — Nishant Saurabh, Oct 04 '14 at 15:04

Matrix not copied back from device to host successfully in CUDA

1 Answers1