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CUDA, Supercomputing for the Masses: Part 3

Shared Memory Version

The following source listing is for, which I discuss in the next installment. I provide it now for your convenience so you can see how to use shared memory on this problem right now.

// includes, system
#include <stdio.h>
#include <assert.h>

// Simple utility function to check for CUDA runtime errors
void checkCUDAError(const char* msg);

// Part 2 of 2: implement the fast kernel using shared memory
__global__ void reverseArrayBlock(int *d_out, int *d_in)
    extern __shared__ int s_data[];

    int inOffset  = blockDim.x * blockIdx.x;
    int in  = inOffset + threadIdx.x;

    // Load one element per thread from device memory and store it 
    // *in reversed order* into temporary shared memory
    s_data[blockDim.x - 1 - threadIdx.x] = d_in[in];

    // Block until all threads in the block have 
    // written their data to shared mem

    // write the data from shared memory in forward order, 
    // but to the reversed block offset as before

    int outOffset = blockDim.x * (gridDim.x - 1 - blockIdx.x);

    int out = outOffset + threadIdx.x;
    d_out[out] = s_data[threadIdx.x];
// Program main
int main( int argc, char** argv) 
    // pointer for host memory and size
    int *h_a;
    int dimA = 256 * 1024; // 256K elements (1MB total)

    // pointer for device memory
    int *d_b, *d_a;

    // define grid and block size
    int numThreadsPerBlock = 256;

    // Compute number of blocks needed based on array size 
    // and desired block size
    int numBlocks = dimA / numThreadsPerBlock;  

    // Part 1 of 2: Compute number of bytes of shared memory needed
    // This is used in the kernel invocation below
    int sharedMemSize = numThreadsPerBlock * sizeof(int);

    // allocate host and device memory
    size_t memSize = numBlocks * numThreadsPerBlock * sizeof(int);
    h_a = (int *) malloc(memSize);
    cudaMalloc( (void **) &d_a, memSize );
    cudaMalloc( (void **) &d_b, memSize );

    // Initialize input array on host
    for (int i = 0; i < dimA; ++i)
        h_a[i] = i;

    // Copy host array to device array
    cudaMemcpy( d_a, h_a, memSize, cudaMemcpyHostToDevice );

    // launch kernel
    dim3 dimGrid(numBlocks);
    dim3 dimBlock(numThreadsPerBlock);
    reverseArrayBlock<<< dimGrid, dimBlock, 
             sharedMemSize >>>( d_b, d_a );

    // block until the device has completed

    // check if kernel execution generated an error
    // Check for any CUDA errors
    checkCUDAError("kernel invocation");

    // device to host copy
    cudaMemcpy( h_a, d_b, memSize, cudaMemcpyDeviceToHost );

    // Check for any CUDA errors

    // verify the data returned to the host is correct
    for (int i = 0; i < dimA; i++)
        assert(h_a[i] == dimA - 1 - i );

    // free device memory

    // free host memory

    // If the program makes it this far, then results are correct and
    // there are no run-time errors.  Good work!

    return 0;
void checkCUDAError(const char *msg)
    cudaError_t err = cudaGetLastError();
    if( cudaSuccess != err) 
        fprintf(stderr, "Cuda error: %s: %s.\n", msg, 
                             cudaGetErrorString( err) );

In the next column, I begin looking at the use of shared memory to increase performance. Until then, I suggest looking into the CUDA memory types -- specifically __shared__, __constant__, and register memory.

For More Information

Click here for more information on CUDA and here for more information on NVIDIA.

Rob Farber is a senior scientist at Pacific Northwest National Laboratory. He has worked in massively parallel computing at several national laboratories and as co-founder of several startups. He can be reached at [email protected].

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