CUDA:我如何使这个代码并行？

Question

我正在努力使代码并行运行(CUDA).简化的代码是:

float sum = ...         //sum = some number
for (i = 0; i < N; i++){
    f = ...             // f = a function that returns a float and puts it into f
    sum += f;
}

我遇到的问题是sum+=f因为它需要sum在线程之间共享.我__shared__在声明sum(__shared__ float sum)时尝试使用该参数,但这不起作用(它没有给我正确的结果).我也听说过减少(并知道如何在OpenMP上使用它),但不知道如何在这里应用它.

任何帮助将不胜感激.谢谢!

Answer 1

这是代码:

#include <stdio.h>
__global__ void para(float* f, int len) {
    int i = threadIdx.x + blockDim.x * blockIdx.x;
    if (i < len ){
        // calculate f[i], f in iteration ith
        f[i] = i;
    }
}

int main(int argc, char ** argv) {
    int inputLength=1024;
    float * h_f;
    float * d_f;
    int size = inputLength*sizeof(float);

    h_f = (float *) malloc(size);
    cudaMalloc((void**)&d_f , size);
    cudaMemcpy(d_f, h_f, size, cudaMemcpyHostToDevice);

    dim3 DimGrid((inputLength)/256 +1 , 1 , 1);
    dim3 DimBlock(256 , 1, 1);

    para<<<DimGrid , DimBlock>>>(d_f , inputLength);
    cudaThreadSynchronize();

    cudaMemcpy(h_f, d_f, size , cudaMemcpyDeviceToHost);

    cudaFree(d_f);

    // do parallel reduction
    int i;
    float sum=0;
    for(i=0; i<inputLength; i++)
        sum+=h_f[i];

    printf("%6.4f\n",sum);

    free(h_f);

    return 0;
}

平行减少部分可以用工作CUDA并行减少(例如这个)来代替.很快我会花时间改变它.

编辑:

以下是使用CUDA执行并行缩减的代码:

#include <stdio.h>
__global__ void para(float* f, int len) {
    int i = threadIdx.x + blockDim.x * blockIdx.x;
    if (i < len ){
        // calculate f[i], f in iteration ith
        f[i] = i;
    }
}
__global__ void strideSum(float *f, int len, int strid){
    int i = threadIdx.x + blockDim.x * blockIdx.x;
    if(i+strid<len){
        f[i]=f[i]+f[i+strid];
    }
}

#define BLOCKSIZE 256
int main(int argc, char ** argv) {
    int inputLength=4096;
    float * h_f;
    float * d_f;
    int size = inputLength*sizeof(float);

    h_f = (float *) malloc(size);
    cudaMalloc((void**)&d_f , size);
    cudaMemcpy(d_f, h_f, size, cudaMemcpyHostToDevice);

    dim3 DimGrid((inputLength)/BLOCKSIZE +1 , 1 , 1);
    dim3 DimBlock(BLOCKSIZE , 1, 1);

    para<<<DimGrid , DimBlock>>>(d_f , inputLength);
    cudaThreadSynchronize();

    int i;
    float sum=0, d_sum=0;
    // serial sum on host. YOU CAN SAFELY COMMENT FOLLOWING COPY AND LOOP. intended for sum validity check.
    cudaMemcpy(h_f, d_f, size , cudaMemcpyDeviceToHost);
    for(i=0; i<inputLength; i++)
        sum+=h_f[i];

    // parallel reduction on gpu
    for(i=inputLength; i>1; i=i/2){
        strideSum<<<((i/BLOCKSIZE)+1),BLOCKSIZE>>>(d_f,i,i/2);
        cudaThreadSynchronize();
    }
    cudaMemcpy(&d_sum, d_f, 1*sizeof(float) , cudaMemcpyDeviceToHost);

    printf("Host -> %6.4f, Device -> %6.4f\n",sum,d_sum);

    cudaFree(d_f);
    free(h_f);

    return 0;
}