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

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

  • 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。
cpp 复制代码
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!
cpp 复制代码
  /// 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!
cpp 复制代码
  /// 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 C 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 N=2 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。

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