Question

我是cuda和C ++的新手，似乎无法解决这个问题。

我想要做的是将2d数组A复制到设备，然后将其复制回相同的数组B.

我希望B数组的值与A相同，但有些东西我做错了。

CUDA - 4.2，编译为win32,64位机器，NVIDIA Quadro K5000

这是代码。

void main(){

cout<<"Host main" << endl;

// Host code
const int width = 3;
const int height = 3;
float* devPtr;
float a[width][height]; 

//load and display input array
cout << "a array: "<< endl;
for (int i = 0 ; i < width; i ++)
{
    for (int j = 0 ; j < height; j ++)
    {
        a[i][j] = i + j;
        cout << a[i][j] << " ";

    }
    cout << endl;
}
cout<< endl;


//Allocating Device memory for 2D array using pitch
size_t host_orig_pitch = width * sizeof(float); //host original array pitch in bytes
size_t pitch;// pitch for the device array 

cudaMallocPitch(&devPtr, &pitch, width * sizeof(float), height);

cout << "host_orig_pitch: " << host_orig_pitch << endl;
cout << "sizeof(float): " << sizeof(float)<< endl;
cout << "width: " << width << endl;
cout << "height: " << height << endl;
cout << "pitch:  " << pitch << endl;
cout << endl;

cudaMemcpy2D(devPtr, pitch, a, host_orig_pitch, width, height, cudaMemcpyHostToDevice);

float b[width][height];
//load b and display array
cout << "b array: "<< endl;
for (int i = 0 ; i < width; i ++)
{
    for (int j = 0 ; j < height; j ++)
    {
        b[i][j] = 0;
        cout << b[i][j] << " ";
    }
    cout << endl;
}
cout<< endl;


//MyKernel<<<100, 512>>>(devPtr, pitch, width, height);
//cudaThreadSynchronize();


//cudaMemcpy2d(dst, dPitch,src ,sPitch, width, height, typeOfCopy )
cudaMemcpy2D(b, host_orig_pitch, devPtr, pitch, width, height, cudaMemcpyDeviceToHost);


// should be filled in with the values of array a.
cout << "returned array" << endl;
for(int i = 0 ; i < width ; i++){
    for (int j = 0 ; j < height ; j++){
        cout<< b[i][j] << " " ;
    }
    cout<<endl;
}

cout<<endl;
system("pause");

}

这是输出。

主机主阵列0 1 2 1 2 3 2 3 4

host_orig_pitch：12 sizeof（float）：4宽度：3高度：3间距：512

b数组：0 0 0 0 0 0 0 0 0

返回数组0 0 0   1.17549e-038 0 0 0 0 0

按任意键继续。。

如果需要更多信息，请告诉我，我会发布。任何帮助将不胜感激。

Answer 1

正如评论中所指出的那样，原始海报正在为cudaMemcpy2D电话提供不正确的参数。传输的width参数始终以字节为单位，因此在上面的代码中：

cudaMemcpy2D(b, host_orig_pitch, devPtr, pitch, width, height, cudaMemcpyDeviceToHost);

应该是

cudaMemcpy2D(b, host_orig_pitch, devPtr, pitch, width * sizeof(float), height, cudaMemcpyDeviceToHost);

请注意，此答案已添加为社区维基，以便将此问题从未答复的列表中删除

我如何使用cudaMemcpy2D（）DeviceToHost

1 个答案: