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做一个已经有10个同名的论坛帖子我感觉有点糟糕,但是在检查了所有这些之后,连同周围的大多数指南,我仍然无法解决问题。

我有一个 char 数组 [40090][11],我想对其两个元素的每个可能的组合进行自定义操作(我将整个 11 字节串视为一个元素)。我知道这是一种矩阵乘法,矩阵是单列单行的。

按照 SDK 手册,我正在考虑每个输出元素有 1 个线程。由于 40090=19*2110,我正在使用:

dim3 threadsperblock(19,19);
dim3 blocksingrid(2110,2110);
xkernel<<<blocksingrid, threadsperblock>>>(dev_b2);

问题1:这样好吗?

好吧,那么,我认为我正在忠实地遵循 SDK 的 maunal 示例(不是使用共享内存的示例)但是,每当我敢于对数据进行部分我想要的操作时,我都会收到一个非常无用的错误 30 返回:未知错误。那么,问题2:我做错了什么?注意:忽略内核没有在任何地方保存任何东西。

#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include <cstdlib>
#include <iostream>
#include <fstream>
#include <iomanip>
#include <ctime>
#include <stdio.h>
using namespace std;

cudaError_t cudafunct(void);
__global__ void xkernel(char * dev_b2);
__device__ unsigned char typecheck(unsigned char type1,unsigned char type2);


#define b2c 40090
unsigned char block2[b2c][11];//
//unsigned int i,b1,b2,counter=0;//Block(2),Piece,Rotation,Type(of block2),InterconnectinTriangle
//unsigned char *block4,type=0;
ofstream ofile;




int main()
{
     ifstream block2file("2.blk",ios::binary);
     block2file.read((char*)(&block2),b2c*11);
     block2file.close();
     //block4=new unsigned char[200000000];//200MB will do, better than doing constant reallocs

    cudaError_t cudaStatus = cudafunct();
    if (cudaStatus != cudaSuccess) {
        fprintf(stderr, "cudafunct failed!");
        system("PAUSE");
        return 1;
    }
    /*

    // cudaDeviceReset must be called before exiting in order for profiling and
    // tracing tools such as Nsight and Visual Profiler to show complete traces.
    cudaStatus = cudaDeviceReset();
    if (cudaStatus != cudaSuccess) {
        fprintf(stderr, "cudaDeviceReset failed!");
        return 1;
    }*/
     cout<<"Sequence end. Saving to file...\n";     
     //ofile.open("blk4.et2",ios::binary);
     //ofile.write((char*)block4,17*counter);   
     //ofile.close(); 
     int t=clock();
     //cout<<"\nFound a total of "<<counter<<" block4s.\nTime elapsed: "<<t<<" clocks / "<<(double)t/(double)CLOCKS_PER_SEC<<" seconds\n";
     system("PAUSE");
}

// Helper function for using CUDA to add vectors in parallel.
cudaError_t cudafunct(void)
{
    char *dev_b2 = 0;
    cudaError_t cudaStatus;

    cudaStatus = cudaMalloc((void**)&dev_b2, sizeof(block2));
    if (cudaStatus != cudaSuccess) {
        fprintf(stderr, "cudaMalloc failed!");
        goto Error;
    }

    cudaStatus = cudaMemcpy(dev_b2, block2, sizeof(block2), cudaMemcpyHostToDevice);
    if (cudaStatus != cudaSuccess) {
        fprintf(stderr, "cudaMemcpy failed!");
        goto Error;
    }

    dim3 threadsperblock(19,19);
    dim3 blocksingrid(2110,2110);
    xkernel<<<blocksingrid, threadsperblock>>>(dev_b2);

    // cudaDeviceSynchronize waits for the kernel to finish, and returns
    // any errors encountered during the launch.
    cudaStatus = cudaDeviceSynchronize();
    if (cudaStatus != cudaSuccess) {
        fprintf(stderr, "cudaDeviceSynchronize returned error code %d after launching xkernel!\n", cudaStatus);
        goto Error;
    }
    /*
    // Copy output vector from GPU buffer to host memory.
    cudaStatus = cudaMemcpy(c, dev_c, size * sizeof(int), cudaMemcpyDeviceToHost);
    if (cudaStatus != cudaSuccess) {
        fprintf(stderr, "cudaMemcpy failed!");
        goto Error;
    }*/

Error:
    cudaFree(dev_b2);
    return cudaStatus;
}


__global__ void xkernel(char *dev_b2)
{
        int i = blockIdx.x * blockDim.x + threadIdx.x; 
        int j = blockIdx.y * blockDim.y + threadIdx.y;
        /*for(int k=0;k<11;k++)
        {
            lb2[0][k]=dev_b2[i*b2c+k];
            lb2[1][k]=dev_b2[j*b2c+k];
        }*/
        int b00;
        b00=dev_b2[i*b2c];

        //int type=typecheck(dev_b2[i*b2c+4],dev_b2[j*b2c+4]);
        //if(!j && !(i % 100))cout<<setw(6)<<i<<" / "<<jc<<" ("<<setw(10)<<(float)100*i/jc<<" % )"<<endl;     
        /*if(
            (dev_b2[i*b2c+7]!=dev_b2[j*b2c+9])||//SW~NW     
            (dev_b2[i*b2c+6]!=dev_b2[j*b2c+10])//SE~NE                                                                                         
        ) return;
        if( (type=typecheck(dev_b2[i*b2c+4],dev_b2[j*b2c+4]) ) ==255) return;*/
        /*if(
            (dev_b2[i*b2c+0]==dev_b2[j*b2c+0])||//1st=3rd
            (dev_b2[i*b2c+0]==dev_b2[j*b2c+2])||//1st=4th
            (dev_b2[i*b2c+2]==dev_b2[j*b2c+0])||//2nd=3rd
            (dev_b2[i*b2c+2]==dev_b2[j*b2c+2])//2nd=4th
        ) return;*/
        /*
        *(block4+counter*17+0)=b2[i][0];//1st piece
        *(block4+counter*17+1)=b2[i][1];//1st rotation
        *(block4+counter*17+2)=b2[i][2];//2nd piece
        *(block4+counter*17+3)=b2[i][3];//2nd rotation
        *(block4+counter*17+4)=b2[j][0];//3rd piece
        *(block4+counter*17+5)=b2[j][1];//3rd rotation
        *(block4+counter*17+6)=b2[j][2];//4th piece
        *(block4+counter*17+7)=b2[j][3];//4th rotation
        *(block4+counter*17+8)=type;
        *(block4+counter*17+9)=b2[i][5];//Right frame colours, down->up
        *(block4+counter*17+10)=b2[j][5];
        *(block4+counter*17+11)=b2[j][6];//Up frame colours, right->left
        *(block4+counter*17+12)=b2[j][7];
        *(block4+counter*17+13)=b2[j][8];//Left frame colours, up->down
        *(block4+counter*17+14)=b2[i][8];
        *(block4+counter*17+15)=b2[i][9];//Down frame colours, left->right
        *(block4+counter++*17+16)=b2[i][10];*/
}  



__device__ unsigned char typecheck(unsigned char type1,unsigned char type2)
{//Warning! Previous error! First partenthesis is t*2* = upper piece!
       if( (type1==4) && (type2==0) ) return  0;  
       if( (type1==6) && (type2==1) ) return  1;  
       if( (type1==2) && (type2==6) ) return  2;  
       if( (type1==3) && (type2==4) ) return  3;  
       if( (type1==4) && (type2==4) ) return  4;  
       if( (type1==8) && (type2==5) ) return  5;  
       if( (type1==6) && (type2==6) ) return  6;  
       if( (type1==7) && (type2==8) ) return  7;  
       if( (type1==8) && (type2==8) ) return  8;  
       if( (type1==9) && (type2==8) ) return  9;  
       if( (type1==10) && (type2==8) ) return  10;  
       if( (type1==8) && (type2==11) ) return  11;  
       if( (type1==8) && (type2==12) ) return  12;  
       if( (type1==8) && (type2==13) ) return  13;  
       return 255;
}
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1 回答 1

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dev_b2我有一种感觉,您从数组 中读取越界。blockIdx.x在 的范围内[0..2110],因此变量i在 的范围内[0..23210]。但是然后你将它与b2c. 因此,您读取的最高地址将是b2c*23210 = 930488900.

dev_b2只有b2c*11 = 440990.

于 2012-10-11T08:57:22.543 回答