本博文主要講解下基於cuda的矩陣相乘,cuda特別擅長的就是矩陣乘法,而且也比較容易實現。通過矩陣乘法的實現,可以比較容易理解cuda的核心思想。網上也有很多基於cuda實現的矩陣乘法,但是感覺都不完成,要不就是有錯,本文給出的代碼都是經過驗證可行的,希望能夠幫助到大家。
矩陣乘法實現方式一:矩陣乘法的逐點實現方式,具體如下圖所示
對應實現代碼:
#include <stdio.h>
#include <stdlib.h>
#include <cuda_runtime.h>
__global__ void MatMul(int *M,int *N,int *P,int width)
{
int x = threadIdx.x;
int y = threadIdx.y;
float Pervalue = 0;
float elem1 = 0.0,elem2 = 0.0,value = 0.0;
for(int i = 0;i < width;i++)
{
elem1 = M[y * width + i];//取M矩陣的一行
elem2 = N[i * width + x];//取N矩陣的一列
value += elem1 * elem2;//求和
}
P[y * width + x] = value;
}
int main()
{
const int ND = 30;
int a[ND][ND],b[ND][ND],c[ND][ND];
int *M,*N,*P;
int width = ND;
int NUM = 900;
dim3 blockSize(ND,ND);
cudaEvent_t start,stop;
float elapsedTime = 0;
cudaEventCreate(&start);
cudaEventCreate(&stop);
//設備端內存分配
cudaMalloc((void**)&M,ND * ND * sizeof(int));
cudaMalloc((void**)&N,ND * ND * sizeof(int));
cudaMalloc((void**)&P,ND * ND * sizeof(int));
//初始化
for(int i = 0;i < ND;i++)
{
for(int j = 0;j < ND;j++)
{
a[i][j] = 2;
b[i][j] = 3;
}
}
int Size = ND * ND;
//數據拷貝,主機到設備
cudaMemcpy(M,a,Size * sizeof(int),cudaMemcpyHostToDevice);
cudaMemcpy(N,b,Size * sizeof(int),cudaMemcpyHostToDevice);
cudaEventRecord(start,0);
MatMul<<<1,blockSize>>>(M,N,P,width);//調用核函數
cudaThreadSynchronize();
cudaEventRecord(stop,0);
cudaEventSynchronize(stop);
cudaEventElapsedTime(&elapsedTime,start,stop);
cudaMemcpy(c,P,Size * sizeof(int),cudaMemcpyDeviceToHost);
printf("c0 = %d \n",c[0][0]);
//釋放設備內存
cudaFree(M);
cudaFree(N);
cudaFree(P);
return 0;
}
運行結果:
矩陣相乘實現方式二:矩陣乘法分塊實現,具體如下圖所示
具體代碼實現:
#include <stdio.h>
#include <stdlib.h>
#include <cuda_runtime.h>
#define TILE_WIDTH 10
//核函數的具體實現
__global__ void matmul(int *M,int *N,int *P,int width)
{
__shared__ float Mds[TILE_WIDTH][TILE_WIDTH];
__shared__ float Nds[TILE_WIDTH][TILE_WIDTH];
int bx = blockIdx.x;
int by = blockIdx.y;
int tx = threadIdx.x;
int ty = threadIdx.y;
int Col = bx * TILE_WIDTH + tx;
int Row = by * TILE_WIDTH + ty;
int Pervalue = 0;
for(int i = 0;i < width / TILE_WIDTH;i++) //有多少個TILE_WIDTH,每個循環計算一個塊的大小
{
Mds[ty][tx] = M[Row * width + (i * TILE_WIDTH + tx)];
Nds[ty][tx] = N[Col + (i * TILE_WIDTH + ty) * width];
__syncthreads();
for(int k = 0;k < TILE_WIDTH;k++) //TILE_WIDTH相乘
Pervalue += Mds[ty][k] * Nds[k][tx];
__syncthreads();
}
P[Row * width + Col] = Pervalue;
}
int main()
{
const int Nd = 30;
int Size = Nd * Nd;
int *M,*N,*P;
int width = Nd / 3;
int a[Nd][Nd];
int b[Nd][Nd];
int c[Nd][Nd];
//線程塊以及線程的劃分
dim3 gridSize(Nd / width,Nd / width);
dim3 blockSize(width,width);
cudaEvent_t start,stop;
float elapsedTime;
cudaEventCreate(&start);
cudaEventCreate(&stop);
//設備內存分配
cudaMalloc((void**)&M,Size * sizeof(int));
cudaMalloc((void**)&N,Size * sizeof(int));
cudaMalloc((void**)&P,Size * sizeof(int));
//初始化
for(int i = 0;i < Nd;i++)
{
for(int j = 0;j < Nd;j++)
{
a[i][j] = 2;
b[i][j] = 3;
}
}
//數據拷貝,主機到設備
cudaMemcpy(M,a,Size * sizeof(int),cudaMemcpyHostToDevice);
cudaMemcpy(N,b,Size * sizeof(int),cudaMemcpyHostToDevice);
cudaEventRecord(start,0);
matmul<<<gridSize,blockSize>>>(M,N,P,Nd); //調用核函數
cudaThreadSynchronize();
cudaEventRecord(stop,0);
cudaEventSynchronize(stop);
cudaEventElapsedTime(&elapsedTime,start,stop);
cudaMemcpy(c,P,Size * sizeof(int),cudaMemcpyDeviceToHost);
printf("c0 = %d\n",c[0][0]);
cudaFree(M);
cudaFree(N);
cudaFree(P);
return 0;
}
運行結果:
本文也參考了網上的一些資料,主要是做了一定的修改以及程序的完備,圖片就直接網上copy的,水平有限,有不當之處,請指教,謝謝!