Question

I have a small piece of code which runs perfectly on Nvidia old architecture (Tesla T10 processor) but not on Fermi (Tesla M2090)

I learned that Fermi behaves slightly differently. Due to which unsafe code might work correctly on old architectures, while on Fermi it catches the bug.

But I don’t know how to resolve it.

Here is my code:

__global__ void exec (int *arr_ptr, int size, int *result) {

    int tx = threadIdx.x;
    int ty = threadIdx.y;

    *result = arr_ptr[-2];

}

void run(int *arr_dev, int size, int *result) {

    cudaStream_t stream = 0;
    int *arr_ptr = arr_dev + 5;

    dim3 threads(1,1,1);
    dim3 grid (1,1);

    exec<<<grid, threads, 0, stream>>>(arr_ptr, size, result);

}

since I am accessing arr_ptr[-2], the fermi throws CUDA_EXCEPTION_10, Device Illegal Address. But it is not. The address is legal.

Can anyone help me on this.

---

My driver code is

int main(){
    int *arr;
    int *arr_dev = NULL;
    int result = 1;

    arr = (int*)malloc(10*sizeof(int));

    for(int i = 0; i < 10; i++)
            arr[i] = i;

    if(arr_dev == NULL)
    {
            cudaMalloc((void**)&arr_dev, 10);
            cudaMemcpy(arr_dev, arr, 10*sizeof(int), cudaMemcpyHostToDevice);
    }

    run(arr_dev, 10, &result);
    printf("%d \n", result);
    return 0;

}

Answer 1

Fermi cards have much better memory protection on the device and will detect out of bounds conditions which will appear to “work” on older cards. Use cuda-memchk (or the cuda-memchk mode in cuda-gdb) to get a better handle on what is going wrong.

---

EDIT:

This is the culprit:

cudaMalloc((void**)&arr_dev, 10);

which should be

cudaMalloc((void**)&arr_dev, 10*sizeof(int));

This will result in this code

int *arr_ptr = arr_dev + 5;

passing a pointer to the device which is out of bounds.