Cuda - 从设备全局内存复制到纹理内存

Question

我正在尝试使用Cuda和C++在GPU上执行两个任务(分成2个内核).作为输入,我采用NxM矩阵(作为浮点数组存储在主机的内存中).然后,我将使用对此矩阵执行某些操作的内核使其成为NxMxD矩阵.然后我有一个第二个内核,它对这个3D矩阵执行一些操作(我只是读取值,我不必为其写入值).

在纹理内存中运行似乎对我的任务来说要快得多,所以我的问题是如果可以在内核1之后从设备上的全局内存中复制我的数据并将其直接传输到内核2的纹理内存而不将其带回主机？

UPDATE

我添加了一些代码来更好地说明我的问题.

这是两个内核.第一个只是现在的占位符,并将2D矩阵复制为3D.

__global__ void computeFeatureVector(float* imData3D_dev, int imX, int imY, int imZ) {

//calculate each thread global index  
int xindex=blockIdx.x*blockDim.x+threadIdx.x; 
int yindex=blockIdx.y*blockDim.y+threadIdx.y;     

#pragma unroll
for (int z=0; z<imZ; z++) { 
    imData3D_dev[xindex+yindex*imX + z*imX*imY] = tex2D(texImIp,xindex,yindex);
}
}

第二个将采用这个3D矩阵,现在表示为纹理并对其执行一些操作.现在空白.

__global__ void kernel2(float* resData_dev, int imX) {
//calculate each thread global index  
int xindex=blockIdx.x*blockDim.x+threadIdx.x; 
int yindex=blockIdx.y*blockDim.y+threadIdx.y;     

resData_dev[xindex+yindex*imX] = tex3D(texImIp3D,xindex,yindex, 0);

return; 
} 

然后代码的主体如下:

// declare textures
texture<float,2,cudaReadModeElementType> texImIp; 
texture<float,3,cudaReadModeElementType> texImIp3D; 

void main_fun() {

// constants
int imX = 1024;
int imY = 768;
int imZ = 16;

// input data
float* imData2D  = new float[sizeof(float)*imX*imY];        
for(int x=0; x<imX*imY; x++)
    imData2D[x] = (float) rand()/RAND_MAX;

//create channel to describe data type 
cudaArray* carrayImIp; 
cudaChannelFormatDesc channel; 
channel=cudaCreateChannelDesc<float>();  

//allocate device memory for cuda array 
cudaMallocArray(&carrayImIp,&channel,imX,imY);

//copy matrix from host to device memory  
cudaMemcpyToArray(carrayImIp,0,0,imData2D,sizeof(float)*imX*imY,cudaMemcpyHostToDevice); 

// Set texture properties
texImIp.filterMode=cudaFilterModePoint;
texImIp.addressMode[0]=cudaAddressModeClamp; 
texImIp.addressMode[1]=cudaAddressModeClamp; 

// bind texture reference with cuda array   
cudaBindTextureToArray(texImIp,carrayImIp);

// kernel params
dim3 blocknum; 
dim3 blocksize;
blocksize.x=16; blocksize.y=16; blocksize.z=1; 
blocknum.x=(int)ceil((float)imX/16);
blocknum.y=(int)ceil((float)imY/16);    

// store output here
float* imData3D_dev;        
cudaMalloc((void**)&imData3D_dev,sizeof(float)*imX*imY*imZ); 

// execute kernel
computeFeatureVector<<<blocknum,blocksize>>>(imData3D_dev, imX, imY, imZ); 

//unbind texture reference to free resource 
cudaUnbindTexture(texImIp); 

// check copied ok
float* imData3D  = new float[sizeof(float)*imX*imY*imZ];
cudaMemcpy(imData3D,imData3D_dev,sizeof(float)*imX*imY*imZ,cudaMemcpyDeviceToHost);     
cout << " kernel 1" << endl;
for (int x=0; x<10;x++)
    cout << imData3D[x] << " ";
cout << endl;
delete [] imData3D;


//
// kernel 2
//


// copy data on device to 3d array
cudaArray* carrayImIp3D;
cudaExtent volumesize;
volumesize = make_cudaExtent(imX, imY, imZ);
cudaMalloc3DArray(&carrayImIp3D,&channel,volumesize); 
cudaMemcpyToArray(carrayImIp3D,0,0,imData3D_dev,sizeof(float)*imX*imY*imZ,cudaMemcpyDeviceToDevice); 

// texture params and bind
texImIp3D.filterMode=cudaFilterModePoint;
texImIp3D.addressMode[0]=cudaAddressModeClamp; 
texImIp3D.addressMode[1]=cudaAddressModeClamp; 
texImIp3D.addressMode[2]=cudaAddressModeClamp;
cudaBindTextureToArray(texImIp3D,carrayImIp3D,channel); 

// store output here
float* resData_dev;
cudaMalloc((void**)&resData_dev,sizeof(float)*imX*imY); 

// kernel 2
kernel2<<<blocknum,blocksize>>>(resData_dev, imX); 
cudaUnbindTexture(texImIp3D);

//copy result matrix from device to host memory   
float* resData  = new float[sizeof(float)*imX*imY];
cudaMemcpy(resData,resData_dev,sizeof(float)*imX*imY,cudaMemcpyDeviceToHost); 

// check copied ok
cout << " kernel 2" << endl;
for (int x=0; x<10;x++)
    cout << resData[x] << " ";
cout << endl;


delete [] imData2D;
delete [] resData;
cudaFree(imData3D_dev);  
cudaFree(resData_dev);
cudaFreeArray(carrayImIp); 
cudaFreeArray(carrayImIp3D); 

}

我很高兴第一个内核正常工作,但3D矩阵imData3D_dev似乎没有正确绑定到纹理texImIp3D.

回答

我用cudaMemcpy3D解决了我的问题.这是主函数第二部分的修订代码.imData3D_dev包含来自第一个内核的全局内存中的3D矩阵.

    cudaArray* carrayImIp3D;
cudaExtent volumesize;
volumesize = make_cudaExtent(imX, imY, imZ);
cudaMalloc3DArray(&carrayImIp3D,&channel,volumesize); 
cudaMemcpy3DParms copyparms={0};

copyparms.extent = volumesize;
copyparms.dstArray = carrayImIp3D;
copyparms.kind = cudaMemcpyDeviceToDevice;  
copyparms.srcPtr = make_cudaPitchedPtr((void*)imData3D_dev, sizeof(float)*imX,imX,imY); 
cudaMemcpy3D(&copyparms);

// texture params and bind
texImIp3D.filterMode=cudaFilterModePoint;
texImIp3D.addressMode[0]=cudaAddressModeClamp; 
texImIp3D.addressMode[1]=cudaAddressModeClamp; 
texImIp3D.addressMode[2]=cudaAddressModeClamp;

cudaBindTextureToArray(texImIp3D,carrayImIp3D,channel); 

// store output here
float* resData_dev;
cudaMalloc((void**)&resData_dev,sizeof(float)*imX*imY); 

kernel2<<<blocknum,blocksize>>>(resData_dev, imX); 

    // ... clean up

Answer 1

当这个问题第一次被问到时，各种 cudaMemcpy 例程的命名曾经有些复杂，但自那以后已经被 Nvidia 清理了。

为了在 3D 数组上进行操作，您需要使用cudaMemcpy3D()能够将线性内存中的 3D 数据复制到 3D 数组的工具（在其他数组中）。
cudaMemcpyToArray()曾经是将线性数据复制到二维数组所需的函数，但已被弃用，取而代之的是更一致的命名cudaMemcpy2D()。

但是，如果您使用的设备的计算能力为 2.0 或更高，则您不想使用任何功能cudaMemcpy*()。相反，使用允许您直接写入纹理的表面，而不需要在内核之间进行任何数据复制。（您仍然需要将读取和写入分离到两个不同的内核中，就像现在一样，因为纹理缓存与表面写入不一致，并且仅在内核启动时无效）。