CUDA「不明な起動失敗」メモリへのアクセス

Question

私がやろうとしていることはとても簡単です。 各スレッドは、グローバルメモリに格納されたグローバル配列からサブ配列を読み取ります。 それはいくつかの計算を行い、静的配列に結果を格納します。 最後に出力はグローバルメモリ の別の配列に格納されています。コードに示されているカーネルのrun.asというグローバル配列に静的配列を書き込む行をコメントします。 アイデア？

GPUカーネル：

#ifndef _TEMPLATE_KERNEL_H_ 
#define _TEMPLATE_KERNEL_H_ 

#include <stdio.h> 

__device__ void 
DecompressBlockGPU(unsigned char *compressed_block,unsigned char *compressed_size, 
        int array_length,unsigned char *decompressed_block) 
{  
    int j = 0; 

    for(int i = 0 ; i < array_length ;i++) 
    { 
     for(int idx = 0 ; idx < compressed_size[i]; idx++) 
     { 
      decompressed_block[j] = compressed_block[i]; 
      j++; 
     } 
    } 
} 
__global__ void 

gpu_test(unsigned char *compressed_data,int *OffsetsArray,int xBlocks,int yBlocks, 
     unsigned char *output, int BlockSize,int BlockWidth,int BlockHeight, 
     int cols,int xTB,int yTB,int xTH,int yTH,unsigned char *aux_array) 
{ 
    int x_max = xBlocks ; 
    int y_max = yBlocks ; 

    int x_block = blockIdx.x ; 
    int y_block = blockIdx.y ; 

    x_max = gridDim.x*blockDim.x ; 
    y_max = gridDim.y*blockDim.y ; 

    x_block = (blockIdx.x*xTH); 
    y_block = (blockIdx.y*yTH); 
    int x_block1 = x_block + threadIdx.x; 
    int y_block1 = y_block + threadIdx.y; 

    int block_idx = y_block1*xBlocks + x_block1; 
    unsigned char *temp_ptr = compressed_data + OffsetsArray[block_idx];   
    int *array_length = (int *)temp_ptr; 
    unsigned char *compressed_size = compressed_data + OffsetsArray[block_idx] + 
           array_length[0] +sizeof(int)/sizeof(unsigned char); 
    unsigned char *compressed_block = compressed_data + OffsetsArray[block_idx] + 
           sizeof(int)/sizeof(unsigned char); 

    aux_array = aux_array + (BlockWidth+2)*(BlockHeight+2)*block_idx; 
    aux_array[block_idx]=array_length[0]; 

    unsigned char decompressed_block[72]; 
    unsigned char extracted_block[32]; 

    DecompressBlockGPU(compressed_block,compressed_size,array_length[0], 
          &decompressed_block[0]); 

    if(block_idx == 0) 
    { 
     for(int i=0;i<16;i++) aux_array[i]= decompressed_block[i]; //fails 
     for(int i=16;i<16*36;i++) aux_array[i]=1;//works 
    } 
} 
#endif 

CPU機能：？

unsigned char *runGPU(unsigned char *d_compressed_data,int *OffsetsArray,int xBlocks,int yBlocks,unsigned char *h_output) 
{ 


    printf("xBlocks =%d yBlocks =%d \n",xBlocks,yBlocks); 



    int xTB = 4; 
    int yTB = 4; 
    int xTH = 1; 
    int yTH = 1; 



    unsigned char *d_output; 
    unsigned char *d_aux_array; 
    unsigned char *h_aux_array; 

    int mem_size = image_len*sizeof(unsigned char); 
    int big_mem_size = sizeof(unsigned char)*xBlocks*yBlocks*(BlockWidth+2)*(BlockHeight+2); 

    cutilSafeCall(cudaMalloc((void**) &d_output, mem_size)); 
    cutilSafeCall(cudaMalloc((void**) &d_aux_array,big_mem_size)); 
    h_aux_array = (unsigned char *)malloc(big_mem_size); 


    float time = 0; 
    float totalTime = 0; 
    cudaEvent_t start_event4, stop_event4; 
    cutilSafeCall(cudaEventCreate(&start_event4)); 
    cutilSafeCall(cudaEventCreate(&stop_event4)); 
    cutilSafeCall(cudaEventRecord(start_event4, 0)); 

    dim3 grid(xTB,yTB, 1); 
    dim3 threads(xTH, yTH, 1); 

    gpu_test<<<grid,threads>>>(d_compressed_data,OffsetsArray,xBlocks,yBlocks,d_output,BlockSize,BlockWidth,BlockHeight,cols,xTB,yTB,xTH,yTH,d_aux_array); 
    cudaThreadSynchronize(); 

    cutilSafeCall(cudaEventRecord(stop_event4, 0)); 
    cutilSafeCall(cudaEventSynchronize(stop_event4)); 
    time = 0; 
    cutilSafeCall(cudaEventElapsedTime(&time, start_event4, stop_event4)); 
    totalTime += time; 
    totalTime /= (1.0e3 * 1); 
    shrLogEx(LOGBOTH | MASTER, 0, "GPU decompression Time = %.5f \n",totalTime); 

    cutilSafeCall(cudaMemcpy(h_output,d_output, mem_size, cudaMemcpyDeviceToHost)); 
    cutilSafeCall(cudaMemcpy(h_aux_array,d_aux_array, big_mem_size, cudaMemcpyDeviceToHost)); 


    cudaFree(d_output); 
    cudaFree(d_aux_array); 

    return h_aux_array; 

} 

それが今明らかである

Answer 1

がcuda-memcheckを通して、あなたのプログラムを実行してみてください（またはメモリがあなたの場合はチェックを有効（編集後） Parallel Nsightを使用しています）。