使用 2 个视频卡进行 CUDA C 编程

Question

我是 CUDA 编程的新手，正在阅读 nvidia 提供的“CUDA C 编程指南”。 ( http://developer.download.nvidia.com/compute/DevZone/docs/html/C/doc/CUDA_C_Programming_Guide.pdf )

在第 25 页中，它具有以下执行矩阵乘法的 C 代码。您能告诉我如何使该代码在两个设备上运行吗？ (如果我的电脑上安装了两张支持 nvida CUDA 的卡)。你能给我举个例子吗？

// Matrices are stored in row-major order: 
// M(row, col) = *(M.elements + row * M.stride + col) 
typedef struct { 
    int width; 
    int height; 
    int stride; 
    float* elements; 
} Matrix; 

// Get a matrix element 
__device__ float GetElement(const Matrix A, int row, int col) 
{ 
    return A.elements[row * A.stride + col]; 
} 

// Set a matrix element 
__device__ void SetElement(Matrix A, int row, int col, float value) 
{ 
    A.elements[row * A.stride + col] = value; 
} 

// Get the BLOCK_SIZExBLOCK_SIZE sub-matrix Asub of A that is 
// located col sub-matrices to the right and row sub-matrices down 
// from the upper-left corner of A 
__device__ Matrix GetSubMatrix(Matrix A, int row, int col) 
{ 
    Matrix Asub; 
    Asub.width = BLOCK_SIZE; 
    Asub.height = BLOCK_SIZE; 
    Asub.stride = A.stride; 
    Asub.elements = &A.elements[A.stride * BLOCK_SIZE * row + BLOCK_SIZE * col]; 
    return Asub;
    } 

// Thread block size 
#define BLOCK_SIZE 16 

// Forward declaration of the matrix multiplication kernel 
__global__ void MatMulKernel(const Matrix, const Matrix, Matrix); 

// Matrix multiplication - Host code
// Matrix dimensions are assumed to be multiples of BLOCK_SIZE 
void MatMul(const Matrix A, const Matrix B, Matrix C) 
{ 
    // Load A and B to device memory 
    Matrix d_A; 
    d_A.width = d_A.stride = A.width; d_A.height = A.height; 
    size_t size = A.width * A.height * sizeof(float); 
    cudaMalloc(&d_A.elements, size); 
    cudaMemcpy(d_A.elements, A.elements, size, cudaMemcpyHostToDevice); 
    Matrix d_B; 
    d_B.width = d_B.stride = B.width; d_B.height = B.height; 
    size = B.width * B.height * sizeof(float); 
    cudaMalloc(&d_B.elements, size); 
    cudaMemcpy(d_B.elements, B.elements, size, cudaMemcpyHostToDevice); 

    // Allocate C in device memory 
    Matrix d_C; 
    d_C.width = d_C.stride = C.width; d_C.height = C.height; 
    size = C.width * C.height * sizeof(float); 
    cudaMalloc(&d_C.elements, size); 

    // Invoke kernel 
    dim3 dimBlock(BLOCK_SIZE, BLOCK_SIZE); 
    dim3 dimGrid(B.width / dimBlock.x, A.height / dimBlock.y); 
    MatMulKernel<<<dimGrid, dimBlock>>>(d_A, d_B, d_C); 

    // Read C from device memory 
    cudaMemcpy(C.elements, d_C.elements, size, cudaMemcpyDeviceToHost); 

    // Free device memory 
    cudaFree(d_A.elements); 
    cudaFree(d_B.elements); 
    cudaFree(d_C.elements); 
} 

// Matrix multiplication kernel called by MatMul() 
__global__ void MatMulKernel(Matrix A, Matrix B, Matrix C) 
{ 
    // Block row and column 
    int blockRow = blockIdx.y; 
    int blockCol = blockIdx.x; 

    // Each thread block computes one sub-matrix Csub of C 
    Matrix Csub = GetSubMatrix(C, blockRow, blockCol);

    // Each thread computes one element of Csub 
    // by accumulating results into Cvalue 
    float Cvalue = 0; 

    // Thread row and column within Csub 
    int row = threadIdx.y; 
    int col = threadIdx.x; 

    // Loop over all the sub-matrices of A and B that are 
    // required to compute Csub 
    // Multiply each pair of sub-matrices together 
    // and accumulate the results 
    for (int m = 0; m < (A.width / BLOCK_SIZE); ++m) 
    { 
        // Get sub-matrix Asub of A 
        Matrix Asub = GetSubMatrix(A, blockRow, m); 
        // Get sub-matrix Bsub of B 
        Matrix Bsub = GetSubMatrix(B, m, blockCol); 

        // Shared memory used to store Asub and Bsub respectively 
        __shared__ float As[BLOCK_SIZE][BLOCK_SIZE]; 
        __shared__ float Bs[BLOCK_SIZE][BLOCK_SIZE]; 

        // Load Asub and Bsub from device memory to shared memory 
        // Each thread loads one element of each sub-matrix 
        As[row][col] = GetElement(Asub, row, col); 
        Bs[row][col] = GetElement(Bsub, row, col); 

        // Synchronize to make sure the sub-matrices are loaded 
        // before starting the computation 
        __syncthreads(); 

        // Multiply Asub and Bsub together 
        for (int e = 0; e < BLOCK_SIZE; ++e) 
            Cvalue += As[row][e] * Bs[e][col]; 

        // 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 Csub to device memory 
    // Each thread writes one element 
    SetElement(Csub, row, col, Cvalue); 
}

Answer 1

没有在多个 GPU 上运行 CUDA 内核的“自动”方式。

您需要设计一种方法将矩阵乘法问题分解为可以并行运行的独立操作(因此每个 GPU 上并行一个)。举个简单的例子:

C = A.B 等同于 C = [A].[B1|B2] = [A.B1|A.B2] 其中 B1 和 B2 是大小合适的矩阵，包含矩阵 B 的列，| 表示按列串联。您可以将 A.B1 和 A.B2 作为单独的矩阵乘法运算进行计算，然后在将生成的子矩阵复制回主机内存时执行串联。

一旦您有了合适的分解方案，您就可以使用 CUDA 4.x API 中的标准多 GPU 设施来实现它。要全面了解使用 CUDA API 进行多 GPU 编程，我建议观看 Paulius Micikevicius 在 GTC 2012 上的精彩演讲，该演讲以流媒体视频和 PDF 的形式提供 here .