矩陣乘法CUDA

Question

我一直在閱讀幾個網站，甚至使用NVIDA的代碼作為指南，但我仍然得到了錯誤的答案。 主要詢問用戶的大小，並顯示A和B然后顯示結果矩陣C.但是說我為A和B運行2x2矩陣這是我的樣本輸出：

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

但這是不正確的。 它應該是：

40.000 56.000
16.000 16.000

我將它從小數改為整數，以便更容易檢查，我發現它是不正確的。 我不明白為什么它會不正確，特別是即使我從他們的代碼示例中采取了它。

#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_

主機代碼：

    //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);

謝謝你的幫助，丹

Answer 1

您隱式使用的代碼要求矩陣的大小是塊大小的四倍（在這種情況下為16x16）。 內積計算一次處理一個區塊寬度，而不檢查超出范圍內存訪問。 因此，2x2矩陣將無法正常工作。

如果您嘗試使用16x16輸入運行內核（例如將2x2大小寫填充為16x16的零填充），則應該能夠確認結果。