cuda内核中的动态共享内存

问题描述

我试图更好地了解cuda内核中的动态共享内存。我复制了此example并对其进行了修改，以允许我更改工作项的数量。它适用于

代码在讨论过程中已更新

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <assert.h>

#define gpuErrChk(ans) { gpuAssert((ans),__FILE__,__LINE__); }

inline void gpuAssert(cudaError_t code,const char *file,int line,bool abort=true) {
        if (code != cudaSuccess) {
                fprintf(stderr,"GPUassert: %s %s %d\n",cudaGetErrorString(code),file,line);
                if (abort) exit(code);
        }
}

__global__ void dynamicAddNine(int *d,int n) {
        extern __shared__ int s[];
        int t = blockIdx.x * blockDim.x * blockDim.y + threadIdx.y * blockDim.x + threadIdx.x;
        s[t] = d[t] + 9;
        __syncthreads();
        d[t] = s[t];
}

static void parseArgs(int argc,char **argv) {
        if (argc != 2) {
                printf("Usage : %s <N>,where N (+ve int) is the required array size\n",argv[0]);
                exit(1);
        }
        if (atoi(argv[1]) < 0 ) {
                printf("Usage : %s <N>,argv[0]);
                exit(1);
        }
}

int main(int argc,char *argv[]) {
        parseArgs(argc,argv);
        const int n = atoi(argv[1]);
        int T = 1024; // maximum number of threads
        if (n < 1024) { T = n;} // if no elements required < 1024

        int a[n],r[n],d[n];

        for (int i = 0; i < n; i++) {
                a[i] = i;
                r[i] = i+9;
                d[i] = 0;
        }

        int *d_d;
        gpuErrChk(cudaMalloc(&d_d,n*sizeof(int)));
        int B = (int)ceil((float)n/(float)T);  //calculate no of blocks required

        // run version with dynamic shared memory
        std::cout << "B: " << B << ",T: " << T << std::endl;

        gpuErrChk(cudaMemcpy(d_d,a,n*sizeof(int),cudaMemcpyHostToDevice));
        dynamicAddNine<<<B,T,n*sizeof(int)>>>(d_d,n);
        gpuErrChk(cudaPeekAtLastError());
        gpuErrChk(cudaDeviceSynchronize());

        gpuErrChk(cudaMemcpy(d,d_d,cudaMemcpyDeviceToHost));

        for (int i = 0; i < n; i++) {
                if (d[i] != r[i]) printf("Error: d[%d] != r[%d] (%d,%d)\n",i,d[i],r[i]);
        }
}

解决方法

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