CudaMemcpy doesn't copy values from host to device

Question

I've created a simple but complete program basing on this tutorial: http://docs.nvidia.com/cuda/cuda-c-programming-guide/index.html#device-memory

#include <cuda.h>
#include <stdio.h>
#include <stdlib.h>

#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, const 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);
   }
}

//Kernel definition
__global__ void VecAdd(float* A, float* B, float* C,int N)
{
    int i = blockDim.x * blockIdx.x + threadIdx.x;
    if(i < N)
        C[i] = A[i] + B[i];
}

//Host code
int main()
{
    int N = 1000;
    int i;
    FILE *f;
    size_t size = N * sizeof(float);

    //allocate input vectors h_A and h_B in host memory
    float *h_A = (float*)malloc(size);
    float *h_B = (float*)malloc(size);
    float *h_C = (float*)malloc(size);  

    //Initialize input vectors
    f = fopen("A.txt","r");
    for(i=0;i<N;i++)
        fscanf(f,"%f ",&h_A[i]);
    fclose(f);  
    f = fopen("B.txt","r");
    for(i=0;i<N;i++)
        fscanf(f,"%f ",&h_B[i]);
    fclose(f);
    //Allocate vactors in device memory
    float *d_A;
    gpuErrchk(cudaMalloc(&d_A,size));
    float *d_B;
    cudaMalloc(&d_B,size);
    float *d_C;
    cudaMalloc(&d_C,size);

    gpuErrchk(cudaMemcpy(d_A, h_A, size, cudaMemcpyHostToDevice));
    cudaMemcpy(d_B, h_B, size, cudaMemcpyHostToDevice);

    //invoke kernel
    int threadsPerBlock = 256;
    int blocksPerGrid = (N + threadsPerBlock - 1) / threadsPerBlock;

    VecAdd<<<blocksPerGrid, threadsPerBlock>>>(d_A, d_B, d_C, N);

    cudaMemcpy(h_C, d_C, size, cudaMemcpyDeviceToHost);

    f = fopen("C.txt","w");
    printf("%f \n",h_C[i]); 
    for(i=0;i<1000;i++)
        fprintf(f,"%f ",h_C[i]); 
    fclose(f);
    printf("Zakonczono obliczenia\n");
    // Free device memory
    cudaFree(d_A);
    cudaFree(d_B);
    cudaFree(d_C);
    //Free host memory
    free(h_A);
    free(h_B);
    return 0;
}

It should read two vectors from files, add them on device and then print the output into 'C.txt' file. However, it prints one thousand of zeros.

After a little debugging did I find the culprit- the cudaMalloc function.

(cuda-gdb) n
42      cudaMemcpy(d_A, h_A, size, cudaMemcpyHostToDevice);
(cuda-gdb) n
43      cudaMemcpy(d_B, h_B, size, cudaMemcpyHostToDevice);
(cuda-gdb) print d_A[0]
$1 = 0
(cuda-gdb) print h_A[0]
$2 = 3.66192293

I wonder why it doesn't work, this part of code had been raw copied from the tutorial.

@FiddlingBits No, it's a member of 'cudaMemcpyKind' enum. It describes the type of memory transfer. — 0x6B6F77616C74
add proper cuda error checking to your code, and run your code with cuda-memcheck. Your thread index variable i in the kernel will also not let you scale to more than one threadblock's worth of data. You don't seem to have copied that from the "tutorial" correctly. — Robert Crovella
Your code works for me as posted. The only thing that is a little squirrely is this line: printf("%f \n",h_C[i]); I think it should be something like this: printf("%f \n",h_C[0]); But I created A.txt and B.txt files composed of 1000 lines of 0.2 and the resultant C.txt file had a single line of 1000 iterations of 0.400000. And cuda-memcheck reports no errors (in my case). So if your C.txt has garbage then I suspect a machine configuration issue (CUDA not functioning). If you add the proper error checking I linked to, it will probably shed some light on that. — Robert Crovella

0x6B6F77616C74 0x6B6F77616C74 · Accepted Answer · 2015-04-24T13:07:19

It seems that I had to restart my computer after installing CUDA, not only log out and log in. After that my program works, but I can no longer debug it. There is

[Thread debugging using libthread_db enabled] Using host libthread_db library "/lib/x86_64-linux-gnu/libthread_db.so.1". fatal: All CUDA devices are used for display and cannot be used while debugging. (error code = 24).

just after 'run'.

CudaMemcpy doesn't copy values from host to device

1 Answers