我正在努力熟悉CUDA编程，并且有一段非常有趣的时间。我目前正在研究处理矩阵乘法的this pdf，有和没有共享内存。可以找到两个版本的完整代码here。该代码几乎与CUDA矩阵乘法样本中的代码完全相同。虽然非共享内存版本具有以任何矩阵大小运行的能力，但无论块大小如何，共享内存版本必须与块大小的倍数（我设置为4，默认最初为16）的矩阵一起使用。
pdf结尾处提出的一个问题是更改它，以便共享内存版本也可以使用块大小的非倍数。我认为这将是一个简单的索引检查，就像在非共享版本中一样：
int row = blockIdx.y * blockDim.y + threadIdx.y;
int col = blockIdx.x * blockDim.x + threadIdx.x;
if(row > A.height || col > B.width) return;
但这不起作用。这是完整的代码，减去主要的方法（有点乱，对不起），我已经有所修改了一些：
void MatMul(const Matrix A, const Matrix B, Matrix C) {
// Load A and B to device memory
Matrix d_A;
d_A.width = d_A.stride = A.width;
d_A.height = A.height;
size_t size = A.width * A.height * sizeof(float);
cudaError_t err = cudaMalloc(&d_A.elements, size);
printf("CUDA malloc A: %s\n",cudaGetErrorString(err));
err = cudaMemcpy(d_A.elements, A.elements, size, cudaMemcpyHostToDevice);
printf("Copy A to device: %s\n",cudaGetErrorString(err));
Matrix d_B;
d_B.width = d_B.stride = B.width;
d_B.height = B.height;
size = B.width * B.height * sizeof(float);
err = cudaMalloc(&d_B.elements, size);
printf("CUDA malloc B: %s\n",cudaGetErrorString(err));
err = cudaMemcpy(d_B.elements, B.elements, size, cudaMemcpyHostToDevice);
printf("Copy B to device: %s\n",cudaGetErrorString(err));
Matrix d_C;
d_C.width = d_C.stride = C.width;
d_C.height = C.height;
size = C.width * C.height * sizeof(float);
err = cudaMalloc(&d_C.elements, size);
printf("CUDA malloc C: %s\n",cudaGetErrorString(err));
dim3 dimBlock(BLOCK_SIZE, BLOCK_SIZE);
dim3 dimGrid((B.width + dimBlock.x - 1) / dimBlock.x, (A.height + dimBlock.y-1) / dimBlock.y);
MatMulKernel<<<dimGrid, dimBlock>>>(d_A, d_B, d_C);
err = cudaThreadSynchronize();
printf("Run kernel: %s\n", cudaGetErrorString(err));
// Read C from device memory
err = cudaMemcpy(C.elements, d_C.elements, size, cudaMemcpyDeviceToHost);
printf("Copy C off of device: %s\n",cudaGetErrorString(err));
// Free device memory
cudaFree(d_A.elements);
cudaFree(d_B.elements);
cudaFree(d_C.elements);
}
// Get a matrix element
__device__ float GetElement(const Matrix A, int row, int col) {
return A.elements[row * A.stride + col];
}
// Set a matrix element
__device__ void SetElement(Matrix A, int row, int col, float value) {
A.elements[row * A.stride + col] = value;
}
// Get the BLOCK_SIZExBLOCK_SIZE sub-matrix Asub of A that is
// located col sub-matrices to the right and row sub-matrices down
// from the upper-left corner of A
__device__ Matrix GetSubMatrix(Matrix A, int row, int col) {
Matrix Asub;
Asub.width = BLOCK_SIZE;
Asub.height = BLOCK_SIZE;
Asub.stride = A.stride;
Asub.elements = &A.elements[A.stride * BLOCK_SIZE * row + BLOCK_SIZE * col];
return Asub;
}
// Matrix multiplication kernel called by MatMul()
__global__ void MatMulKernel(Matrix A, Matrix B, Matrix C) {
// Block row and column
int blockRow = blockIdx.y;
int blockCol = blockIdx.x;
int rowTest = blockIdx.y * blockDim.y + threadIdx.y;
int colTest = blockIdx.x * blockDim.x + threadIdx.x;
if (rowTest>A.height || colTest>B.width)
return;
// Each thread block computes one sub-matrix Csub of C
Matrix Csub = GetSubMatrix(C, blockRow, blockCol);
// Each thread computes one element of Csub
// by accumulating results into Cvalue
float Cvalue = 0.0;
// Thread row and column within Csub
int row = threadIdx.y;
int col = threadIdx.x;
// Loop over all the sub-matrices of A and B that are
// required to compute Csub
// Multiply each pair of sub-matrices together
// and accumulate the results
for (int m = 0; m < (BLOCK_SIZE + A.width - 1)/BLOCK_SIZE; ++m) {
// Get sub-matrix Asub of A
Matrix Asub = GetSubMatrix(A, blockRow, m);
// Get sub-matrix Bsub of B
Matrix Bsub = GetSubMatrix(B, m, blockCol);
// Shared memory used to store Asub and Bsub respectively
__shared__ float As[BLOCK_SIZE][BLOCK_SIZE];
__shared__ float Bs[BLOCK_SIZE][BLOCK_SIZE];
// Load Asub and Bsub from device memory to shared memory
// Each thread loads one element of each sub-matrix
As[row][col] = GetElement(Asub, row, col);
Bs[row][col] = GetElement(Bsub, row, col);
// Synchronize to make sure the sub-matrices are loaded
// before starting the computation
__syncthreads();
// Multiply Asub and Bsub together
for (int e = 0; e < BLOCK_SIZE; ++e)
{
Cvalue += As[row][e] * Bs[e][col];
}
// Synchronize to make sure that the preceding
// computation is done before loading two new
// sub-matrices of A and B in the next iteration
__syncthreads();
}
// Write Csub to device memory
// Each thread writes one element
SetElement(Csub, row, col, Cvalue);
}
我改变的值得注意的事情：我在MatMulKernel中添加了一个检查，检查我们当前的线程是否正在尝试在C中不存在的某个位置上工作。这似乎不起作用。虽然它确实改变了结果，但是这些更改似乎没有任何模式，除了稍后（更高的x或y值）条目似乎更受影响（并且我得到更多的非整数结果）。我还改变了给定的dimGrid计算方法和MatMulKernel中m的循环条件（在它只是宽度或高度除以块大小之前，这似乎是错误的）。
即使我为本指南找到的解决方案指南似乎也建议它应该只是一个简单的索引检查，所以我想我错过了一些非常基本的东西。
答案 0 :(得分：21)
当矩阵尺寸不是瓷砖尺寸的倍数时，可能会发生一些瓷砖仅部分覆盖矩阵。落在不完全重叠的瓷砖之外的瓷砖元素应该适当地归零。因此，将代码扩展到任意大小的矩阵很容易，但不需要进行简单的索引检查。下面，我正在复制并粘贴我的版本的平铺矩阵 - 矩阵乘法内核和任意大小的矩阵
__global__ void MatMul(float* A, float* B, float* C, int ARows, int ACols, int BRows,
int BCols, int CRows, int CCols)
{
float CValue = 0;
int Row = blockIdx.y*TILE_DIM + threadIdx.y;
int Col = blockIdx.x*TILE_DIM + threadIdx.x;
__shared__ float As[TILE_DIM][TILE_DIM];
__shared__ float Bs[TILE_DIM][TILE_DIM];
for (int k = 0; k < (TILE_DIM + ACols - 1)/TILE_DIM; k++) {
if (k*TILE_DIM + threadIdx.x < ACols && Row < ARows)
As[threadIdx.y][threadIdx.x] = A[Row*ACols + k*TILE_DIM + threadIdx.x];
else
As[threadIdx.y][threadIdx.x] = 0.0;
if (k*TILE_DIM + threadIdx.y < BRows && Col < BCols)
Bs[threadIdx.y][threadIdx.x] = B[(k*TILE_DIM + threadIdx.y)*BCols + Col];
else
Bs[threadIdx.y][threadIdx.x] = 0.0;
__syncthreads();
for (int n = 0; n < TILE_DIM; ++n)
CValue += As[threadIdx.y][n] * Bs[n][threadIdx.x];
__syncthreads();
}
if (Row < CRows && Col < CCols)
C[((blockIdx.y * blockDim.y + threadIdx.y)*CCols) +
(blockIdx.x * blockDim.x)+ threadIdx.x] = CValue;
}