CUDA“未指定的启动失败”访问内存

Question

我想做的事情很简单。 每个线程从存储在全局内存中的全局数组中读取子数组。 然后它进行一些计算并将结果存储在静态数组中。 最后输出被存储回全局内存中的另一个数组中 当我注释将静态数组写入全局数组的行时，内核将运行。如代码所示。 有什么想法吗？

GPU内核：

#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功能：

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;

}

现在清楚了吗？（编辑后）

Answer 1

尝试通过 cuda-memcheck 运行您的程序（或者启用内存检查，如果您正在使用 Parallel Nsight）。