Implementing CUDA VecAdd from sample code

arrays, c, cuda, gpu, parallel-processing

Solution

First, you have to define the pointers that will hold the data that will be copied to GPU:

In your example, we want to copy the arrays 'a','b' and 'c' from `CPU` to the `GPU's` global memory.

int a[array_size], b[array_size],c[array_size]; // your original arrays
int *a_cuda,*b_cuda,*c_cuda;                    // defining the "cuda" pointers

define the size that each array will occupy.

int size = array_size * sizeof(int); // Is the same for the 3 arrays

Then you will allocate the space to the data that will be used in cuda:

Cuda memory allocation:

msg_erro[0] = cudaMalloc((void **)&a_cuda,size);
msg_erro[1] = cudaMalloc((void **)&b_cuda,size);
msg_erro[2] = cudaMalloc((void **)&c_cuda,size);

Now we need to copy this data from CPU to the GPU:

Copy from CPU to GPU:

msg_erro[3] = cudaMemcpy(a_cuda, a,size,cudaMemcpyHostToDevice);
msg_erro[4] = cudaMemcpy(b_cuda, b,size,cudaMemcpyHostToDevice);
msg_erro[5] = cudaMemcpy(c_cuda, c,size,cudaMemcpyHostToDevice);

Execute the kernel

int blocks = //;
int threads_per_block = //;
VecAdd<<<blocks, threads_per_block>>>(a_cuda, b_cuda, c_cuda);

Copy the results from GPU to CPU (in our example array C):

msg_erro[6] = cudaMemcpy(c,c_cuda,size,cudaMemcpyDeviceToHost);

Free Memory:

cudaFree(a_cuda);
cudaFree(b_cuda);
cudaFree(c_cuda);

For debugging purposes, I normally save the status of the functions on an array, like this:

cudaError_t msg_erro[var];

However, this is not strictly necessary but it will save you time if an error occurs during the allocation or memory transference. You can take out all the 'msg_erro[x] =' from the code above if you wish.

If you mantain the 'msg_erro[x] =', and if a error does occur you can use a function like the one that follows, to print these erros:

void printErros(cudaError_t *erros,int size)
{
 for(int i = 0; i < size; i++)
      printf("{%d} => %s\n",i ,cudaGetErrorString(erros[i]));
}

Problem

I'm trying to test out a sample code from the CUDA site http://docs.nvidia.com/cuda/cuda-c-programming-guide/index.html#kernels. I simply want to add two arrays A and B of size 4, and store it in array C. Here is what I have so far: ``` #include <stdio.h> #include "util.h" void print_array(int* array, int size) { int i; for (i = 0; i < size; i++) { printf("%d ", array[i]); } printf("\n"); } __global__ void VecAdd(int* A, int* B, int* C) { int i = threadIdx.x; C[i] = A[i] + B[i]; } int main(int argc , char **argv) { int N = 4; int i; int *A = (int *) malloc(N * sizeof(int)); int *B = (int *) malloc(N * sizeof(int)); int *C = (int *) malloc(N * sizeof(int)); for (i = 0; i < N; i++) { A[i] = i + 1; B[i] = i + 1; } print_array(A, N); print_array(B, N); VecAdd<<<1, N>>>(A, B, C); print_array(C, N); return 0; } ``` I'm expecting the C array (the last row of the output) to be 2, 4, 6, 8, but it doesn't seem to get added: ``` 1 2 3 4 1 2 3 4 0 0 0 0 ``` What am I missing?

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