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_DEVICE
  GemmIdentityThreadblockSwizzle() { }

  /// 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_DEVICE
  static 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_DEVICE
  static 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_DEVICE
  static 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 tiles
  CUTLASS_HOST_DEVICE
  static 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 width
  CUTLASS_HOST_DEVICE
  static int get_log_tile(GemmCoord tiled_shape) {
    auto n = tiled_shape.n();
    // Thresholds picked so that it doesn't cause too many no-op CTAs
    if (N >= 8 && n >= 6)
      return 3;
    else if (N >= 4 && n >= 3)
      return 2;
    else if (N >= 2 && n >= 2)
      return 1;
    else
      return 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_DEVICE
  static 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_DEVICE
  static 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。