矩阵乘法CUDA
我一直在阅读几个网站，甚至使用NVIDA的代码作为指南，但我仍然得到了错误的答案。 main将询问用户的大小，并显示A和B然后显示结果矩阵C.但是，我说A和B都运行2×2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
我将它从小数改为整数，以便更容易检查，我发现它是不正确的。 我不明白为什么它会不正确，特别是即使我从他们的代码示例中采取了它。
#ifndef _MATRIXMUL_KERNEL_H_ #define _MATRIXMUL_KERNEL_H_ #include // 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<<>>(deviceMatrixC, deviceMatrixA, deviceMatrixB, a.colSize, b.colSize);
谢谢你的帮助，丹
您隐式使用的代码要求矩阵的大小是块大小的四倍（在这种情况下为16×16）。 内积计算一次处理一个tile宽度，而不检查越界内存访问。 因此，2×2矩阵将无法正常工作。
如果您尝试使用16×16输入运行内核（例如将2×2大小写填充到16×16的零填充），您应该能够确认结果。