CUDA编程(六): 利用好shared memory
CUDA编程(五): 并行规约优化
CUDA编程(四): CPU与GPU的矩阵乘法对比
CUDA编程(三): GPU架构了解一下!
CUDA编程(二): Ubuntu下的CUDA10.x环境搭建
CUDA编程(一): 老黄和他的核弹们
目录
- 前言
- 计时函数
- CPU代码
- GPU代码
- 最后
前言
在上一篇的最后, 我提到了一个矩阵乘法, 这次与CPU进行对比, 从中可以很明显GPU在并行计算上的优势.
计时函数
在贴出代码之前, 来看下我常用的计时函数, 可以精确到微秒级. 首先头文件是
#include. 结构体为:
struct timeval{
long tv_sec; /*秒*/
long tv_usec; /*微秒*/
};
来看下使用的小栗子:
struct timeval start, end;
double timeuse;
int sum = 0;
gettimeofday (&start, NULL);
for (int i = 0; i < 10000; i++){
sum += I;
}
gettimeofday (&end, NULL);
timeuse = end.tv_sec - start.tv_sec + (end.tv_usec - start.tv_usec)/1000000.0;
printf("Use Time:%f\n",timeuse);
CPU代码
CPU也是可以多线程操作的, 所以我第一反应也是多线程进行计算, 但是结果并不理想, 因为线程操作也是有不少开销的, 所以在计算量有限的情况下, 没有线程操作的单线程反而更快. 单线程代码较为简单, 如下:
#include
#include
#include
#include
#define w 8000
struct Matrix
{
int width;
int height;
float *elements;
};
void matMul(float * M, float * N, float * P, int width){
for (int i = 0; i < width; i++){
for (int j = 0; j < width; j++){
float sum = 0;
for (int k = 0; k < width; k++){
float a = M[i * width + k];
float b = N[k * width + j];
sum += a * b;
}
P[i * width + j] = sum;
}
}
}
int main(){
int width = w;
int height = w;
float * m = (float *)malloc (width * height * sizeof (float));
float * n = (float *)malloc (width * height * sizeof (float));
float * p = (float *)malloc (width * height * sizeof (float));
for (int i = 0; i < width * height; i++){
m[i] = 1.0;
n[i] = 2.0;
}
struct timeval t1,t2;
gettimeofday(&t1,NULL);
double timeuse;
matMul(m, n, p, w);
gettimeofday(&t2,NULL);
timeuse = t2.tv_sec - t1.tv_sec + (t2.tv_usec - t1.tv_usec)/1000000.0;
printf("Use Time:%f\n",timeuse);
return 0;
}
多线程部分代码较长, 思路其实并不复杂, 每个线程负责部分计算, 比方说双线程就一个计算偶数行, 一个计算奇数行:
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LOG_
#define SIZE 8000
int * A, * B; // 计算矩阵
int * result, * result2, * result3, * result4; // 结果矩阵
/*
int A[SIZE][SIZE];
int B[SIZE][SIZE];
int result[SIZE][SIZE];
int result2[SIZE][SIZE];
int result3[SIZE][SIZE];
int result4[SIZE][SIZE];
*/
int size; // 矩阵阶数
pthread_t tid2[2]; // 双线程id
pthread_t tid3[3]; // 三线程id
pthread_t tid4[4]; // 四线程id
/* 双线程函数 */
void twoThread1(){
int i, j, k;
for (i = 0; i < size; I++)
for (j = 0; j < size; j++)
for (k = 0; k < size; k++){
if (i % 2 == 0)
result2[i * size + j] += A[i * size + k] * B[k * size + j];
// result2[i][j] += A[i][k] * B[k][j];
}
}
void twoThread2(){
int i, j, k;
for (i = 0; i < size; I++)
for (j = 0; j < size; j++)
for (k = 0; k < size; k++){
if (i % 2 != 0)
result2[i * size + j] += A[i * size + k] * B[k * size + j];
// result2[i][j] += A[i][k] * B[k][j];
}
}
/* 双线程函数 end */
/* 三线程函数 */
void threeThread1(){
int i, j, k;
for (i = 0; i < size; I++)
for (j = 0; j < size; j++)
for (k = 0; k < size; k++){
if (i % 3 == 0)
result3[i * size + j] += A[i * size + k] * B[k * size + j];
// result3[i][j] += A[i][k] * B[k][j];
}
}
void threeThread2(){
int i, j, k;
for (i = 0; i < size; I++)
for (j = 0; j < size; j++)
for (k = 0; k < size; k++){
if (i % 3 != 0 && i % 2 != 0)
result3[i * size + j] += A[i * size + k] * B[k * size + j];
// result3[i][j] += A[i][k] * B[k][j];
}
}
void threeThread3(){
int i, j, k;
for (i = 0; i < size; I++)
for (j = 0; j < size; j++)
for (k = 0; k < size; k++){
if (i % 3 != 0 && i % 2 == 0)
result3[i * size + j] += A[i * size + k] * B[k * size + j];
// result3[i][j] += A[i][k] * B[k][j];
}
}
/* 三线程函数 end */
/* 四线程函数 */
void fourThread1(){
int i, j, k;
for (i = 0; i < size; I++)
for (j = 0; j < size; j++)
for (k = 0; k < size; k++){
if (i % 2 == 0 && i % 4 != 0)
result4[i * size + j] += A[i * size + k] * B[k * size + j];
// result4[i][j] += A[i][k] * B[k][j];
}
}
void fourThread2(){
int i, j, k;
for (i = 0; i < size; I++)
for (j = 0; j < size; j++)
for (k = 0; k < size; k++){
if (i % 4 == 0)
result4[i * size + j] += A[i * size + k] * B[k * size + j];
// result4[i][j] += A[i][k] * B[k][j];
}
}
void fourThread3(){
int i, j, k;
for (i = 0; i < size; I++)
for (j = 0; j < size; j++)
for (k = 0; k < size; k++){
if (i % 2 != 0 && i % 3 == 0)
result4[i * size + j] += A[i * size + k] * B[k * size + j];
// result4[i][j] += A[i][k] * B[k][j];
}
}
void fourThread4(){
int i, j, k;
for (i = 0; i < size; I++)
for (j = 0; j < size; j++)
for (k = 0; k < size; k++){
if (i % 2 != 0 && i % 3 != 0)
result4[i * size + j] += A[i * size + k] * B[k * size + j];
// result4[i][j] += A[i][k] * B[k][j];
}
}
/* 四线程函数 end */
int main(){
int i, j, k, m, n; // 循环变量
struct timeval t1, t2;
double timeuse; // 计时
char sizeChars[8]; // 阶数写入字符串
char timeChars[16]; // 耗时写入字符串
// 申请空间, 计算矩阵和结果矩阵
A = (int *)malloc (sizeof (int) * SIZE * SIZE);
B = (int *)malloc (sizeof (int) * SIZE * SIZE);
result = (int *)malloc (sizeof (int) * SIZE * SIZE);
result2 = (int *)malloc (sizeof (int) * SIZE * SIZE);
result3 = (int *)malloc (sizeof (int) * SIZE * SIZE);
result4 = (int *)malloc (sizeof (int) * SIZE * SIZE);
for (i = 0; i < SIZE; I++)
for (j = 0; j < SIZE; j++){
/*
A[i][j] = 1;
B[i][j] = 2;
result[i][j] = 0;
result2[i][j] = 0;
result3[i][j] = 0;
result4[i][j] = 0;
*/
A[i * SIZE + j] = 1;
B[i * SIZE + j] = 2;
result[i * SIZE + j] = 0;
result2[i * SIZE + j] = 0;
result3[i * SIZE + j] = 0;
result4[i * SIZE + j] = 0;
}
int fd;
fd = open ("./pthreadTime.txt", O_WRONLY | O_CREAT);
lseek(fd, 0, SEEK_SET);
for (size = 200; size <= SIZE; size += 200){
printf ("当前阶数: %d\n", size);
sprintf (sizeChars, "%d, ", size);
write (fd, sizeChars, strlen (sizeChars));
#ifdef LOG
printf ("A矩阵: \n");
for (i = 0; i < size; i++){
for (j = 0; j < size; j++){
printf ("%d ", A[i * size + j]);
// printf ("%d ", A[i][j]);
}
printf ("\n");
}
printf ("B矩阵: \n");
for (i = 0; i < size; i++){
for (j = 0; j < size; j++){
printf ("%d ", B[i * size + j]);
// printf ("%d ", B[i][j]);
}
printf ("\n");
}
#endif
/* 单线程 */
gettimeofday (&t1, NULL);
for (i = 0; i < size; I++)
for (j = 0; j < size; j++)
for (k = 0; k < size; k++){
result[i * size + j] += A[i * size + k] * B[k * size +j];
// result[i][j] += A[i][k] * B[k][j];
}
gettimeofday(&t2, NULL);
timeuse = t2.tv_sec - t1.tv_sec + (t2.tv_usec - t1.tv_usec)/1000000.0;
#ifdef LOG
printf ("单线程结果矩阵: \n");
for (i = 0; i < size; i++){
for (j = 0; j < size; j++){
printf ("%d ", result[i * size + j]);
// printf ("%d ", result[i][j]);
}
printf ("\n");
}
#endif
printf("单线程耗时: %fs\n", timeuse);
sprintf (timeChars, "%lf, ", timeuse);
write (fd, timeChars, strlen (timeChars));
for (i = 0; i < size; I++)
for (j = 0; j < size; j++){
result[i * size + j] = 0;
result2[i * size + j] = 0;
result3[i * size + j] = 0;
result4[i * size + j] = 0;
/*
result[i][j] = 0;
result2[i][j] = 0;
result3[i][j] = 0;
result4[i][j] = 0;
*/
}
/* 单线程 end */
/* 双线程 */
gettimeofday (&t1, NULL);
pthread_create (&tid2[0], NULL, (void *)twoThread1, NULL);
pthread_join (tid2[0], NULL);
pthread_create (&tid2[1], NULL, (void *)twoThread2, NULL);
pthread_join (tid2[1], NULL);
gettimeofday (&t2, NULL);
timeuse = t2.tv_sec - t1.tv_sec + (t2.tv_usec - t1.tv_usec)/1000000.0;
#ifdef LOG
printf ("双线程结果矩阵: \n");
for (i = 0; i < size; i++){
for (j = 0; j < size; j++){
printf ("%d ", result2[i * size + j]);
// printf ("%d ", result2[i][j]);
}
printf ("\n");
}
#endif
printf("双线程耗时: %fs\n", timeuse);
sprintf (timeChars, "%lf, ", timeuse);
write (fd, timeChars, strlen (timeChars));
for (i = 0; i < size; I++)
for (j = 0; j < size; j++){
result[i * size + j] = 0;
result2[i * size + j] = 0;
result3[i * size + j] = 0;
result4[i * size + j] = 0;
/*
result[i][j] = 0;
result2[i][j] = 0;
result3[i][j] = 0;
result4[i][j] = 0;
*/
}
/* 双线程 end */
/* 三线程 */
gettimeofday (&t1, NULL);
pthread_create (&tid3[0], NULL, (void *)threeThread1, NULL);
pthread_join (tid3[0], NULL);
pthread_create (&tid3[1], NULL, (void *)threeThread2, NULL);
pthread_join (tid3[1], NULL);
pthread_create (&tid3[2], NULL, (void *)threeThread3, NULL);
pthread_join (tid3[2], NULL);
gettimeofday (&t2, NULL);
timeuse = t2.tv_sec - t1.tv_sec + (t2.tv_usec - t1.tv_usec)/1000000.0;
#ifdef LOG
printf ("三线程结果矩阵: \n");
for (i = 0; i < size; i++){
for (j = 0; j < size; j++){
printf ("%d ", result3[i * size + j]);
}
printf ("\n");
}
#endif
printf("三线程耗时: %fs\n", timeuse);
sprintf (timeChars, "%lf, ", timeuse);
write (fd, timeChars, strlen (timeChars));
for (i = 0; i < size; I++)
for (j = 0; j < size; j++){
result[i * size + j] = 0;
result2[i * size + j] = 0;
result3[i * size + j] = 0;
result4[i * size + j] = 0;
/*
result[i][j] = 0;
result2[i][j] = 0;
result3[i][j] = 0;
result4[i][j] = 0;
*/
}
/* 三线程 end */
/* 四线程 */
gettimeofday (&t1, NULL);
pthread_create (&tid4[0], NULL, (void *)fourThread1, NULL);
pthread_join (tid4[0], NULL);
pthread_create (&tid4[1], NULL, (void *)fourThread2, NULL);
pthread_join (tid4[1], NULL);
pthread_create (&tid4[2], NULL, (void *)fourThread3, NULL);
pthread_join (tid4[2], NULL);
pthread_create (&tid4[3], NULL, (void *)fourThread4, NULL);
pthread_join (tid4[3], NULL);
gettimeofday (&t2, NULL);
timeuse = t2.tv_sec - t1.tv_sec + (t2.tv_usec - t1.tv_usec)/1000000.0;
#ifdef LOG
printf ("四线程结果矩阵: \n");
for (i = 0; i < size; i++){
for (j = 0; j < size; j++){
printf ("%d ", result4[i * size + j]);
}
printf ("\n");
}
#endif
printf("四线程耗时: %fs\n", timeuse);
sprintf (timeChars, "%lf\n", timeuse);
write (fd, timeChars, strlen (timeChars));
for (i = 0; i < size; I++)
for (j = 0; j < size; j++){
result[i * size + j] = 0;
result2[i * size + j] = 0;
result3[i * size + j] = 0;
result4[i * size + j] = 0;
/*
result[i][j] = 0;
result2[i][j] = 0;
result3[i][j] = 0;
result4[i][j] = 0;
*/
}
/* 四线程 end */
}
// 释放空间
free (A);
free (B);
free (result);
free (result2);
free (result3);
free (result4);
A = NULL;
B = NULL;
result = NULL;
result2 = NULL;
result3 = NULL;
result4 = NULL;
// 关闭文件
close (fd);
return 0;
}
cuda部分的代码直接贴出来, 解析可以看之前的文章. 其实就是为每一次的乘法开一个线程, 这在CPU中是无法想象的, 如此多的线程, 开销太大. 但是在GPU中, 这里运行速度可以用快得无法想象来形容.
#include
#include
#include
#include
#define w 8000
struct Matrix
{
int width;
int height;
float *elements;
};
__device__ float getElement(Matrix *A, int row, int col)
{
return A->elements[row * A->width + col];
}
__device__ void setElement(Matrix *A, int row, int col, float value)
{
A->elements[row * A->width + col] = value;
}
__global__ void matMulKernel(Matrix *A, Matrix *B, Matrix *C)
{
float Cvalue = 0.0;
int row = threadIdx.y + blockIdx.y * blockDim.y;
int col = threadIdx.x + blockIdx.x * blockDim.x;
for (int i = 0; i < A->width; ++i)
{
Cvalue += getElement(A, row, i) * getElement(B, i, col);
}
setElement(C, row, col, Cvalue);
}
int main()
{
int width = w;
int height = w;
Matrix *A, *B, *C;
cudaMallocManaged((void**)&A, sizeof(Matrix));
cudaMallocManaged((void**)&B, sizeof(Matrix));
cudaMallocManaged((void**)&C, sizeof(Matrix));
int nBytes = width * height * sizeof(float);
cudaMallocManaged((void**)&A->elements, nBytes);
cudaMallocManaged((void**)&B->elements, nBytes);
cudaMallocManaged((void**)&C->elements, nBytes);
A->height = height;
A->width = width;
B->height = height;
B->width = width;
C->height = height;
C->width = width;
for (int i = 0; i < width * height; ++i)
{
A->elements[i] = 1.0;
B->elements[i] = 2.0;
}
dim3 blockSize(32, 32);
dim3 gridSize((width + blockSize.x - 1) / blockSize.x,
(height + blockSize.y - 1) / blockSize.y);
struct timeval t1,t2;
gettimeofday(&t1,NULL);
double timeuse;
matMulKernel << < gridSize, blockSize >> >(A, B, C);
cudaDeviceSynchronize();
gettimeofday(&t2,NULL);
timeuse = t2.tv_sec - t1.tv_sec + (t2.tv_usec - t1.tv_usec)/1000000.0;
printf("Use Time:%fs\n", timeuse);
return 0;
}
来看下结果图, 首先是CPU的. 8000阶数的算了一两个小时, 这里可以看到, 多线程在阶数增长起来之后, 双线程有时比单线程耗时要少, 而四线程也有时耗时要少于三线程.
CPU多线程
然后是GPU的, 一般情况下45s左右, 快的时候可以10s就算完, 真可谓是如斯恐怖:
GPU多线程
gpu是gt750m, cpu是i7-4700mq. 其实cpu是比gpu好很多的, 但是并行计算上gpu的优势依旧明显.
最后
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