Question

下面的代码计算C = A * B，如何将其展开以进行多平铺乘法？ A的一行在每一个循环上乘以一列B。我的问题是如何修改它，使得A的一行与B的多列相乘，这样我们就可以避免再次为同一行重新加载它们。

（代码取自：Dynamic matrix multiplication with CUDA）

// CUDA Kernel
__global__ void matrixMul( float* C, float* A, float* B, int wA, int wB,size_t block_size)
 {
int bx = blockIdx.x;
int by = blockIdx.y;
int tx = threadIdx.x;
int ty = threadIdx.y;

int aBegin = wA * block_size * by;
int aEnd   = aBegin + wA - 1;
int aStep  = block_size;

int bBegin = block_size * bx;

int bStep  = block_size * wB;
float Csub=0;

for (int a = aBegin, b = bBegin; a <= aEnd; a += aStep, b += bStep) 
{
    extern __shared__ float As[];
    extern __shared__ float Bs[];
    extern __shared__ float smem[];

    smem[ty*block_size+tx] = A[a + wA * ty + tx];

    smem[block_size*block_size+ty*block_size+tx]  = B[b + wB * ty + tx];

    __syncthreads();

    for (int k = 0; k < block_size; ++k)
        Csub += smem[ty*block_size+k] * smem[block_size*block_size+k*block_size+tx] ;

    __syncthreads();
}

int c = wB * block_size * by + block_size * bx;
C[c + wB * ty + tx] = Csub;
}

由于

Answer 1

Kirk和Hwu有一本书深入探讨了开发有效的矩阵乘法内核：

http://www.amazon.com/Programming-Massively-Parallel-Processors-Hands/dp/0123814723

此前，Stack Overflow上还有一些关于CUDA上平铺矩阵乘法的问题。请参阅以下内容：

Matrix Multiplication CUDA

Non Square Matrix Multiplication in CUDA

使用多列与CUDA进行矩阵乘法

1 个答案: