矩阵乘法CUDA [英] Matrix Multiplication CUDA
问题描述
我一直在读通过几个网站，甚至使用 NVIDA的 code作为一个指南，但我仍然得到错误的答案。主会要求大小的用户，并显示A和B则显示结果矩阵C.但是说我运行一个2×2矩阵A和B这是我的输出示例：
I have been reading through several websites and even used NVIDA's code as a guide but I am still getting the wrong answer. The main will ask the user for size, and will display A and B then display the resulting matrix C. However say I run a 2x2 matrix for both A and B this is my sample output:
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
但是，这是不正确。它应该是：
But that's incorrect. It should be:
40.000 56.000
16.000 16.000
我从小数到整数改变它，这样它会更容易来检查，我发现这是不正确。我不明白为什么它会是不正确的，特别是，即使我把它直接从他们的code样品。
I changed it from decimals to whole numbers so that it would be easier to check, and I found that it's incorrect. I do not understand why it would be incorrect, especially even though I took it right from their code sample.
#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code：
//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，
丹
Thanks for your help, Dan
推荐答案
您使用的是隐式的code要求矩阵的大小是块大小的圆倍数（16×16在这种情况下）。内积的计算处理同时瓷砖的宽度，不检查越界内存访问。出于这个原因，2×2矩阵将无法工作。
The code you are using implicitly requires that the size of the matrices are round multiples of the block size (16x16 in this case). The inner product calculation processes a tile width at a time without checking for out of bounds memory access. For this reason, 2x2 matrices will not work.
如果你尝试用一个16x16的投入运行的内核（例如零填充您的2x2的情况下16×16），你应该能够确认的结果。
If you try running kernel with a 16x16 input (for example zero padding your 2x2 case to 16x16), you should be able to confirm the result.
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