Programming With CUDA

1. Chris Kerkhoff Matthew Sullivan 12/2/2009 Programming With CUDA

2. Basic Flow • The host computer initializes an array with data. • The array is copied from the main memory to the memory on the GPU. • The GPU performs operations on the array. • The array is copied back to the main memory on the computer.

3. #include<stdio.h> #include<cuda.h> // Kernel that executes on the CUDA device: __global__ voidcube_array(float *a, int N) {intidx = blockIdx.x * blockDim.x + threadIdx.x; if (idx<N) a[idx] = a[idx] * a[idx] * a[idx]; } int main(void) {// main routine that executes on the host float *a_h, *a_d; // Pointer to host & device arrays constint N = 10; // Number of elements in arrays size_t size = N * sizeof(float); a_h = (float *)malloc(size); // Allocate array on host cudaMalloc((void **) &a_d, size); // Allocate array on device // Initialize host array and copy it to CUDA device: for (inti=0; i<N; i++) a_h[i] = (float)i; cudaMemcpy(a_d, a_h, size, cudaMemcpyHostToDevice); // Do calculation on device: intblock_size = 4; intn_blocks = N/block_size + (N%block_size == 0 ? 0:1); cube_array <<< n_blocks, block_size >>> (a_d, N); // Retrieve result from device and store it in host array: cudaMemcpy(a_h, a_d, sizeof(float)*N, cudaMemcpyDeviceToHost); for (inti=0; i<N; i++) printf("%d %f\n", i, a_h[i]); //Print results free(a_h); cudaFree(a_d);}//Cleanup

4. 0 0.000000 1 1.000000 2 8.000000 3 27.000000 4 64.000000 5 125.000000 6 216.000000 7 343.000000 8 512.000000 9 729.000000 Press any key to continue . . .

5. Only if you have an Nvidia card