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  • CUDA纹理绑定

     

    纹理绑定的一般步骤:
    size_t fea_pitch;
    texture<unsigned char, 2> features2D;
    cudaMallocPitch((void**)(&dev_features), &fea_pitch, sizeof(unsigned char) * sfeaturesw, sfeaturesh);
    cudaChannelFormatDesc feaDesc = cudaCreateChannelDesc<unsigned char>();
    cudaMemcpy2D(dev_features, fea_pitch, sfeatures, sizeof(unsigned char) * sfeaturesw, sizeof(unsigned char) * sfeaturesw, sfeaturesh, cudaMemcpyHostToDevice);
    cudaBindTexture2D(NULL, features2D, dev_features, feaDesc, sfeaturesw, sfeaturesh, fea_pitch);//绑定
    
    纹理拾取(读取)的步骤
    point1=tex2D(imageData2D,box_x+x1,box_y+y1);//第y1行,第x1列,cpu版即这个意思,此处尤须注意,和一般数据结构不同

    另外纹理features2D不能直接修改,修改的话必须修改features2D所指的内存,即修改dev_features

    一:数组int a [ 3][3]={1,2,3,
                                       4,5,6,
                                       7,8,9};
    给多维数组定义,可以不指定行,但必须指定列
    a[x][y]存储的是a的第x行,第y列。不是按数学坐标系中的坐标来的

    二:
    tex2D( rT, x, y );取的是rT所绑定的二维数组T的T[y][x],即T的第y行第x列

    三:
    关于opencv里的Mat,如果一个图像宽320,高240,那么存到Mat里后,Mat.rows是图像的行数,也即高240;Mat.cols是图像的列数,也即宽320。
    Mat.at(x,y),是Mat的第x行,第y列,也即图像的第x行,第y列

    绑定纹理的例子
    cudaMallocPitch((void**)(&dev_features), &fea_pitch, sizeof(unsigned char) * sfeaturesw, sfeaturesh);
    cudaChannelFormatDesc feaDesc = cudaCreateChannelDesc<unsigned char>();
    cudaMemcpy2D(dev_features, fea_pitch, sfeatures, sizeof(unsigned char) * sfeaturesw, sizeof(unsigned char) * sfeaturesw, sfeaturesh, cudaMemcpyHostToDevice);
    cudaBindTexture2D(NULL, features2D, dev_features, feaDesc, sfeaturesw, sfeaturesh, fea_pitch);
    
    --------------------------------------------------------------------------------
    int sfeatures_size = sizeof(unsigned char) * sfeaturesw * sfeaturesh;
    cudaChannelFormatDesc chDesc2 = cudaCreateChannelDesc<unsigned char>();
    cudaMallocArray(&featuresArray, &chDesc2, sfeaturesw, sfeaturesh);
    cudaMemcpyToArray( featuresArray, 0, 0, sfeatures, sfeatures_size, cudaMemcpyHostToDevice );
    cudaBindTextureToArray( features2D, featuresArray);   
    -------------------------------------------------------------------------------------
    int grid_data_size = sizeof(float) * gridl;
    cudaMalloc((void**)&dev_grid,grid_data_size);
    cudaMemcpy(dev_grid,sgrid,grid_data_size,cudaMemcpyHostToDevice);
    cudaBindTexture(0,gridData1D,dev_grid);

    对于一维纹理,不管是Linear Memory还是使用cudaMallocPitch的,都可以使用tex1Dfetch和tex1D
    而对于二维纹理,不管是cudaArray还是cudaMallocPitch都是使用tex2D


    下面是关于cudaMemcpy2D和cudaMallocPitch两个函数的参数和用法


    最近学习了下CUDA矩阵内存对齐分配的方法,主要是cudaMemcpy2D和cudaMallocPitch两个函数的用法,先看看cudalibrary中如何定义的这两个函数:

    cudaError_t cudaMallocPitch ( void **  devPtr,
        size_t *  pitch,
        size_t  width,
        size_t  height  
      )      

    Allocates at least widthInBytes * height bytes of linear memory on the device and returns in *devPtr a pointer to the allocated memory. The function may pad the allocation to ensure that corresponding pointers in any given row will continue to meet the alignment requirements for coalescing as the address is updated from row to row. The pitch returned in *pitch by cudaMallocPitch() is the width in bytes of the allocation. The intended usage of pitch is as a separate parameter of the allocation, used to compute addresses within the 2D array. Given the row and column of an array element of type T, the address is computed as:

    T* pElement = (T*)((char*)BaseAddress + Row * pitch) + Column;

    For allocations of 2D arrays, it is recommended that programmers consider performing pitch allocations using cudaMallocPitch(). Due to pitch alignment restrictions in the hardware, this is especially true if the application will be performing 2D memory copies between different regions of device memory (whether linear memory or CUDA arrays).

     

    Parameters:
      devPtr  - Pointer to allocated pitched device memory
      pitch  - Pitch for allocation
      width  - Requested pitched allocation width
      height 

    - Requested pitched allocation height

     

     

    cudaError_t cudaMemcpy2D ( void *  dst,
        size_t  dpitch,
        const void *  src,
        size_t  spitch,
        size_t  width,
        size_t  height,
        enum cudaMemcpyKind  kind  
      )      

     

    Copies a matrix (height rows of width bytes each) from the memory area pointed to by src to the memory area pointed to by dst, where kind is one ofcudaMemcpyHostToHost, cudaMemcpyHostToDevice, cudaMemcpyDeviceToHost, or cudaMemcpyDeviceToDevice, and specifies the direction of the copy. dpitch and spitch are the widths in memory in bytes of the 2D arrays pointed to by dst and src, including any padding added to the end of each row. The memory areas may not overlap. Calling cudaMemcpy2D() with dst and src pointers that do not match the direction of the copy results in an undefined behavior. cudaMemcpy2D() returns an error if dpitch or spitch is greater than the maximum allowed.

     

     

    Parameters:
      dst  - Destination memory address
      dpitch  - Pitch of destination memory
      src  - Source memory address
      spitch  - Pitch of source memory
      width  - Width of matrix transfer (columns in bytes)
      height  - Height of matrix transfer (rows)
      kind  - Type of transfer

     

    由此,可以对这两个函数有个充分的认识。此外,cudaMallocPitch和cudaMemcpy2D,一般用于二维数组各维度size不是2的幂次方的问题。使用cudaMallocPitch()那么该数组的对齐、大小、起始址等就自动做好了,其返回的pitch就是真正分配给数组的size(往往大于其真正申请的大小)。

    PS:

    patch的理解:

    C语言申请2维内存时,一般是连续存放的。a[y][x]存放在第y*widthofx*sizeof(元素)+x*sizeof(元素)个字节。但在cuda的global memory访问中,从256字节对齐的地址(addr=0, 256, 512, ...)开始的连续访问是最有效率的。 这样,为了提高内存访问的效率,有了cudaMallocPitch函数。 cudaMallocPitch函数分配的内存中,数组的每一行的第一个元素的开始地址都保证是对齐的。因为每行有多少个数据是不确定的widthofx*sizeof(元素)不一定是256的倍数。故此,为保证数组的每一行的第一个元素的开始地址对齐,cudaMallocPitch在分配内存时,每行会多分配一些字节,以保证widthofx*sizeof(元素)+多分配的字节是256的倍数(对齐)。这样,y*widthofx*sizeof(元素)+x*sizeof(元素)来计算a[y][x]的地址就不正确了。而应该是y*[widthofx*sizeof(元素)+多分配的字节]+x*sizeof(元素)。而函数中返回的pitch的值就是widthofx*sizeof(元素)+多分配的字节。


    一、内存对齐的原因
    大部分的参考资料都是如是说的:
    1、平台原因(移植原因):不是所有的硬件平台都能访问任意地址上的任意数据 的;某些硬件平台只能在某些地址处取某些特定类型的数据,否则抛出硬件异常。
    2、性能原因:数据结构(尤其是栈)应该尽可能地在自然边界上对齐。 原因在于,为了访问未对齐的内存,处理器需要作两次内存访问;而对齐的内存访问仅需要一次访问。



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  • 原文地址:https://www.cnblogs.com/riasky/p/3430748.html
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