這里解決一個(gè)問(wèn)題。通過(guò)兩個(gè)程序，討論共享內(nèi)存的優(yōu)勢(shì)。

共享內(nèi)存預(yù)計(jì)比全局內(nèi)存快得多。它可以用作暫存器內(nèi)存（或軟件托管的高速緩存），以最大程度地減少來(lái)自CUDA塊的全局內(nèi)存訪問(wèn).

一 全局內(nèi)存

// Matrices are stored in row-major order:

// M(row, col) = *(M.elements + row * M.width + col)

typedef struct {

int width;

int height;

float* elements;

} Matrix;

// 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 = 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 = 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 = 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)

{

// Each thread computes one element of C

// by accumulating results into Cvalue

float Cvalue = 0;

int row = blockIdx.y * blockDim.y + threadIdx.y;

int col = blockIdx.x * blockDim.x + threadIdx.x;

for (int e = 0; e < A.width; ++e)

Cvalue += A.elements[row * A.width + e]

* B.elements[e * B.width + col];

C.elements[row * C.width + col] = Cvalue;