这里解决一个问题。通过两个程序，讨论共享内存的优势。
共享内存预计比全局内存快得多。它可以用作暂存器内存（或软件托管的高速缓存），以最大程度地减少来自CUDA块的全局内存访问.
一 全局内存
// 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<<
// 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;