P89 bank.cu的疑问

文中提到了改写bank.cu第15-20行的代码后矩阵转置就变成了矩阵复制，但我的理解下改写后的代码仍然可以实现矩阵转置，同时我测试了两个转置函数的输出结果，发现两者都可以实现转置，函数2并不是对矩阵进行复制 这是我的代码： 转置矩阵1，按照bank.cu中的实现 ``` __global__ void transpose_1(const float* A, float* B, int N) { __shared__ float S[16][16]; int bx = blockIdx.x * 16; int by = blockIdx.y * 16; int x = bx + threadIdx.x; int y = by + threadIdx.y; if (x < N && y < N) S[threadIdx.y][threadIdx.x] = A[y * N + x]; __syncthreads(); int nx = bx + threadIdx.y; int ny = by + threadIdx.x; if (nx < N && ny < N) B[nx * N + ny] = S[threadIdx.x][threadIdx.y]; } ``` 转置矩阵2，按照文中所说的进行改写 ```__global__ void transpose_2(const float* A, float* B, int N) { __shared__ float S[16][16]; int bx = blockIdx.x * 16; int by = blockIdx.y * 16; int x = bx + threadIdx.x; int y = by + threadIdx.y; if (x < N && y < N) S[threadIdx.y][threadIdx.x] = A[y * N + x]; __syncthreads(); int nx = bx + threadIdx.x; int ny = by + threadIdx.y; if (nx < N && ny < N) B[nx * N + ny] = S[threadIdx.y][threadIdx.x]; } ``` 主函数 ``` int main() { const int N = 200; // 生成随机矩阵 float h_A[N * N] = { 0 }; for (int i = 0; i < N; ++i) for (int j = 0; j < N; ++j) h_A[i * N + j] = (float)rand() / RAND_MAX; float h_B_1[N * N] = { 0 }; float h_B_2[N * N] = { 0 }; float* d_A, * d_B; cudaMalloc(&d_A, sizeof(h_A)); cudaMalloc(&d_B, sizeof(h_B_1)); // Copy input matrix to device cudaMemcpy(d_A, h_A, sizeof(h_A), cudaMemcpyHostToDevice); dim3 block(16, 16); dim3 grid((N + 15) / 16, (N + 15) / 16); // Run 1 version transpose_1 << <grid, block >> > (d_A, d_B, N); cudaMemcpy(h_B_1, d_B, sizeof(h_B_1), cudaMemcpyDeviceToHost); // Run 2 version transpose_2 << <grid, block >> > (d_A, d_B, N); cudaMemcpy(h_B_2, d_B, sizeof(h_B_2), cudaMemcpyDeviceToHost); //// Output //print_matrix(h_A, N, "Input A"); //print_matrix(h_B_1, N, "B from transpose_1"); //print_matrix(h_B_2, N, "B from transpose_2"); // Check results bool passed = true; for (int i = 0; i < N; ++i) for (int j = 0; j < N; ++j) if (h_B_1[j * N + i] != h_B_2[j * N + i]) { passed = false; printf("Error: B_1[%d][%d] = %f, B_2[%d][%d] = %f\n", i, j, h_B_1[j * N + i], j, i, h_B_2[j * N + i]); break; } if (passed) printf("Test passed!\n"); cudaFree(d_A); cudaFree(d_B); return 0; } ``` 两个函数的输出结果是完全一样的，都实现了转置