发表博客之：gemm/threadblock/threadblock_swizzle.h 文件夹讲解，cutlass深入讲解

发表博客之：gemm/threadblock/threadblock_swizzle.h 文件夹讲解，cutlass深入讲解

• 在CSDN著名文章发表博客之：cutlass demo讲解，在sm75机器上用cuda core计算fp32矩阵乘！深入理解cutlass::gemm::device::Gemm类 ,感兴趣的老乡别走开！！里面我们介绍了cutlass::gemm::device::Gemm的使用方式，以及这个模版类的一些参数，里面有一个模版参数叫ThreadblockSwizzle，并且当时他还有一个默认值typename threadblock::GemmIdentityThreadblockSwizzle<>,，不知道各位看官是否还记得，现在我要告诉你这个模版参数的准确作用！开心吗？

 

• 首先这个文件的github地址是cutlass/gemm/threadblock/threadblock_swizzle.h

 

• 我们知道，cuda 处理问题都是将一个很大规模的问题分成很多个小问题，每个小问题由一个ThreadBlock来处理，而ThreadblockSwizzle就是负责将逻辑上的小问题映射到cuda上的ThreadBlock上。
• 或者直接引用这个文件上的注释吧！
• Implements several possible threadblock-swizzling functions mapping blockIdx to GEMM problems.

 

先来看一下最简单的struct GemmIdentityThreadblockSwizzle结构体

• 这个结构体有一个默认参数是1。

template <int N = 1>
struct GemmIdentityThreadblockSwizzle {CUTLASS_HOST_DEVICEGemmIdentityThreadblockSwizzle() { }/// Returns the shape of the problem in units of logical tiles/// *Gemm* problem size: gemm(M, N, K)/// 这个函数的作用是简单的。/// 就是以tile_size为逻辑单元，整个问题的逻辑shape！CUTLASS_HOST_DEVICEstatic GemmCoord get_tiled_shape(GemmCoord problem_size,GemmCoord tile_size,int split_k_slices) {return GemmCoord((problem_size.m() + tile_size.m() - 1) / tile_size.m(),(problem_size.n() + tile_size.n() - 1) / tile_size.n(),split_k_slices);}/// Returns the shape of the problem in units of logical tiles/// *ImplicitGemm* Conv2d problem size: conv_operator(NPQK, NHWC, KRSC)CUTLASS_HOST_DEVICEstatic GemmCoord get_tiled_shape(cutlass::conv::Operator conv_operator,cutlass::conv::Conv2dProblemSize const &problem_size,GemmCoord tile_size,int split_k_slices) {gemm::GemmCoord implicit_gemm_problem_size = cutlass::conv::implicit_gemm_problem_size(conv_operator, problem_size);return get_tiled_shape(implicit_gemm_problem_size, tile_size, split_k_slices);}/// Returns the shape of the problem in units of logical tiles/// *ImplicitGemm* Conv3d problem size: conv_operator(NZPQK, NDHWC, KTRSC)CUTLASS_HOST_DEVICEstatic GemmCoord get_tiled_shape(cutlass::conv::Operator conv_operator,cutlass::conv::Conv3dProblemSize const &problem_size,GemmCoord tile_size,int split_k_slices) {gemm::GemmCoord implicit_gemm_problem_size = cutlass::conv::implicit_gemm_problem_size(conv_operator, problem_size);return get_tiled_shape(implicit_gemm_problem_size, tile_size, split_k_slices);}/// 这个函数是获得物理shape！也就是三对三对<<<>>>下的grid_shape!/// Computes CUDA grid dimensions given a size in units of logical tilesCUTLASS_HOST_DEVICEstatic dim3 get_grid_shape(GemmCoord tiled_shape) {int tile = 1 << get_log_tile(tiled_shape);return dim3(tiled_shape.m() * tile, (tiled_shape.n() + tile - 1) / tile, tiled_shape.k());}

• 下面的这个函数来获得最好的get_log_tile!

  /// 这个是防止函数是防止逻辑shape上的n过大，导致grid的第2维过大！/// Calculates optimal swizzle widthCUTLASS_HOST_DEVICEstatic int get_log_tile(GemmCoord tiled_shape) {auto n = tiled_shape.n();// Thresholds picked so that it doesn't cause too many no-op CTAsif (N >= 8 && n >= 6)return 3;else if (N >= 4 && n >= 3)return 2;else if (N >= 2 && n >= 2)return 1;elsereturn 0;}

• 下面两个函数是同一个名字，get_tile_offset，但是参数不同。
  • 他们的共同作用根据物理id是获取 逻辑上Tile的偏移量！
• 但是第二个函数好像很少用到的样子！

  /// Obtains the threadblock offset (in units of threadblock-scoped tiles)CUTLASS_DEVICEstatic GemmCoord get_tile_offset(int log_tile) {int block_idx_x = RematerializeBlockIdxX();int block_idx_y = RematerializeBlockIdxY();int block_idx_z = RematerializeBlockIdxZ();return GemmCoord{(block_idx_x >> log_tile),  //(block_idx_y << log_tile) + ((block_idx_x) & ((1 << (log_tile)) - 1)),block_idx_z};}/// Obtains the threadblock offset (in units of threadblock-scoped tiles)CUTLASS_DEVICEstatic GemmCoord get_tile_offset(GemmCoord tiled_shape) {int const kTile = N;int block_idx_x = RematerializeBlockIdxX();int block_idx_y = RematerializeBlockIdxY();if ((tiled_shape.m() < kTile) || (tiled_shape.n() < kTile))return GemmCoord{block_idx_x, block_idx_y, RematerializeBlockIdxZ()};return GemmCoord{(block_idx_x / kTile),(block_idx_y * kTile) + (block_idx_x % kTile),RematerializeBlockIdxZ()};}
};

• 举个例子，假设N=1，并且$C$输出矩阵被分成下面这样的逻辑shape，
• 那么三对<<<>>>发射的grid就是(4,4,1)!
• 那么每个Tile被映射到的ThreadBlock id如下图所示。

• 如果$N=2$，

• 那么三对<<<>>>发射的grid就是(8,2,1)!

• 那么每个Tile被映射到的ThreadBlock id如下图所示。