c - 矩阵乘法 CUDA

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CUDA Matrix Multiplication write to wrong memory location

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4 个月前关闭。




我已经阅读了几个网站，甚至使用 NVIDA's 代码作为指导，但我仍然得到错误的答案。 main 会询问用户大小，然后显示 A 和 B，然后显示结果矩阵 C。但是说我为 A 和 B 运行一个 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。它应该是:

40.000 56.000
16.000 16.000

我把它从小数改为整数，这样更容易检查，我发现它不正确。我不明白为什么它是不正确的，尤其是即使我是从他们的代码示例中获取的。

#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_

主机代码:

    //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);

谢谢你的帮助，
担

最佳答案

您正在使用的代码隐式要求矩阵的大小是块大小的整数倍(在这种情况下为 16x16)。内积计算一次处理一个平铺宽度，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 情况填充为零到 16x16)，您应该能够确认结果。

关于c - 矩阵乘法 CUDA，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/8813750/