矩阵乘法 CUDA

Question

我已经阅读了几个网站，甚至使用 NVIDA 的代码作为指导，但我仍然得到错误的答案。 main 将询问用户大小，并显示 A 和 B，然后显示结果矩阵 C。但是，假设我为 A 和 B 运行 2x2 矩阵，这是我的示例输出：

Matrix A
0.000000 8.000000
2.000000 2.000000


Matrix B
3.000000 1.000000
5.000000 7.000000


Matrix C (Results)
0.000000 9.000000
7.000000 4.000000

但这是不正确的。应该是：

40.000 56.000
16.000 16.000

我把小数改成了整数，这样方便检查，结果发现不对。我不明白为什么它是不正确的，特别是即使我直接从他们的代码示例中获取它。

#ifndef _MATRIXMUL_KERNEL_H_
#define _MATRIXMUL_KERNEL_H_

#include <stdio.h>

// Thread block size
#define BLOCK_SIZE 16
#define TILE_SIZE  16



// CUDA Kernel
__global__ void matrixMul( float* C, float* A, float* B, int wA, int wB)
{
    // Block index
    int bx = blockIdx.x;
    int by = blockIdx.y;

// Thread index
int tx = threadIdx.x;
int ty = threadIdx.y;

// Index of the first sub-matrix of A processed 
// by the block
int aBegin = wA * BLOCK_SIZE * by;

// Index of the last sub-matrix of A processed 
// by the block
int aEnd   = aBegin + wA - 1;

// Step size used to iterate through the 
// sub-matrices of A
int aStep  = BLOCK_SIZE;

// Index of the first sub-matrix of B processed 
// by the block
int bBegin = BLOCK_SIZE * bx;

// Step size used to iterate through the 
// sub-matrices of B
int bStep  = BLOCK_SIZE * wB;
float Csub=0;
// Loop over all the sub-matrices of A and B
// required to compute the block sub-matrix
for (int a = aBegin, b = bBegin; a <= aEnd; a += aStep, b += bStep) 
{
    // Declaration of the shared memory array As 
    // used to store the sub-matrix of A
    __shared__ float As[BLOCK_SIZE][BLOCK_SIZE];

    // Declaration of the shared memory array Bs 
    // used to store the sub-matrix of B
    __shared__ float Bs[BLOCK_SIZE][BLOCK_SIZE];

    // Load the matrices from global memory
    // to shared memory; each thread loads
    // one element of each matrix
    As[ty][tx] = A[a + wA * ty + tx];
    Bs[ty][tx] = B[b + wB * ty + tx];

    // Synchronize to make sure the matrices 
    // are loaded
    __syncthreads();

    // Multiply the two matrices together;
    // each thread computes one element
    // of the block sub-matrix
    for (int k = 0; k < BLOCK_SIZE; ++k)
        Csub += As[ty][k] * Bs[k][tx];

    // 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 the block sub-matrix to device memory;
// each thread writes one element
int c = wB * BLOCK_SIZE * by + BLOCK_SIZE * bx;
C[c + wB * ty + tx] = Csub;
}

#endif // #ifndef _MATRIXMUL_KERNEL_H_

楼主代码：

    //perform the calculation
    //setup execution parameters
    dim3 threads(BLOCK_SIZE, BLOCK_SIZE);
    dim3 grid(c.colSize / threads.x, c.rowSize / threads.y);

    //   execute the kernel
    matrixMul<<< grid, threads >>>(deviceMatrixC, deviceMatrixA, deviceMatrixB, a.colSize, b.colSize);

感谢您的帮助， 担

Answer 1

您使用的代码隐式要求矩阵的大小是块大小的整数倍（在本例中为 16x16）。内积计算一次处理一个图块宽度，而不检查越界内存访问。因此，2x2 矩阵将不起作用。

如果您尝试使用 16x16 输入运行内核（例如将 2x2 大小写零填充为 16x16），您应该能够确认结果。