使用2个视频卡进行CUDA C编程

时间:2012-07-16 09:27:33

标签: c cuda nvidia

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,它具有以下用于矩阵乘法的C代码。你能否告诉我如何在两台设备上运行该代码? (如果我的计算机上安装了两个支持nvida CUDA的卡)。能告诉我一个例子吗?

// 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); 
}

2 个答案:

答案 0 :(得分:6)

没有“自动”方式在多个GPU上运行CUDA内核。

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运算(因此每个GPU上并行运行一个)。举个简单的例子:

C = A.B相当于C = [A].[B1|B2] = [A.B1|A.B2],其中B1B2是大小合适的矩阵,包含矩阵B|的列表示列式连接。您可以将A.B1A.B2计算为单独的矩阵乘法运算,然后在将生成的子矩阵复制回主机内存时执行连接。

一旦有了合适的分解方案,就可以使用CUDA 4.x API中的标准multi-gpu工具实现它。有关使用CUDA API进行多GPU编程的精彩概述,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格式提供here

答案 1 :(得分:2)

CUDA C Programming Guide under section 3.2.6.

中介绍了基础知识

基本上,您可以通过调用cudaSetDevice()来设置当前主机线程在哪个GPU上运行。仍然需要编写自己的代码,将您的例程分解为多个GPU。