下面你可以找到我砍死在一起,以測試多線程並行線程的性能的代碼。我沒有清理它,也沒有進行優化。所以代碼是raw。
代碼保存計算Mandelbrot集爲位圖是不是我的,你可以使用該程序具有下列參數發現here
#include <cstdlib> //for atoi
#include <iostream>
#include <iomanip> //for setw and setfill
#include <vector>
#include "bitmap_Image.h" //for saving the mandelbrot as a bmp
#include <pthread.h>
pthread_mutex_t mutexCounter;
int sharedCounter(0);
int percent(0);
int horizPixels(0);
int vertPixels(0);
int maxiter(0);
//doesn't need to be locked
std::vector<std::vector<int> > result; //create 2 dimensional vector
void *DoThread(void *null) {
double curX,curY,xSquare,ySquare,x,y;
int i, intx, inty, counter;
counter = 0;
do {
counter++;
pthread_mutex_lock (&mutexCounter); //lock
intx = int((sharedCounter/vertPixels) + 0.5);
inty = sharedCounter % vertPixels;
sharedCounter++;
pthread_mutex_unlock (&mutexCounter); //unlock
//exit thread when finished
if (intx >= horizPixels) {
std::cout << "exited thread - I did " << counter << " calculations" << std::endl;
pthread_exit((void*) 0);
}
//set x and y to the correct value now -> in the range like singlethread
x = (3.0/horizPixels) * (intx - (horizPixels/1.5));
y = (3.0/vertPixels) * (inty - (vertPixels/2));
curX = x + x*x - y*y;
curY = y + x*y + x*y;
ySquare = curY*curY;
xSquare = curX*curX;
for (i=0; i<maxiter && ySquare + xSquare < 4;i++){
ySquare = curY*curY;
xSquare = curX*curX;
curY = y + curX*curY + curX*curY;
curX = x - ySquare + xSquare;
}
result[intx][inty] = i;
} while (true);
}
int DoSingleThread(const double x, const double y) {
double curX,curY,xSquare,ySquare;
int i;
curX = x + x*x - y*y;
curY = y + x*y + x*y;
ySquare = curY*curY;
xSquare = curX*curX;
for (i=0; i<maxiter && ySquare + xSquare < 4;i++){
ySquare = curY*curY;
xSquare = curX*curX;
curY = y + curX*curY + curX*curY;
curX = x - ySquare + xSquare;
}
return i;
}
void SingleThreaded(std::vector<std::vector<int> >& result) {
for(int x = horizPixels - 1; x != -1; x--) {
for(int y = vertPixels - 1; y != -1; y--) {
//3.0 -> so we always have -1.5 -> 1.5 as the window; (x - (horizPixels/2) will go from -horizPixels/2 to +horizPixels/2
result[x][y] = DoSingleThread((3.0/horizPixels) * (x - (horizPixels/1.5)),(3.0/vertPixels) * (y - (vertPixels/2)));
}
}
}
void MultiThreaded(int threadCount, std::vector<std::vector<int> >& result) {
/* Initialize and set thread detached attribute */
pthread_t thread[threadCount];
pthread_attr_t attr;
pthread_attr_init(&attr);
pthread_attr_setdetachstate(&attr, PTHREAD_CREATE_JOINABLE);
for (int i = 0; i < threadCount - 1; i++) {
pthread_create(&thread[i], &attr, DoThread, NULL);
}
std::cout << "all threads created" << std::endl;
for(int i = 0; i < threadCount - 1; i++) {
pthread_join(thread[i], NULL);
}
std::cout << "all threads joined" << std::endl;
}
int main(int argc, char* argv[]) {
//first arg = length along horizontal axis
horizPixels = atoi(argv[1]);
//second arg = length along vertical axis
vertPixels = atoi(argv[2]);
//third arg = iterations
maxiter = atoi(argv[3]);
//fourth arg = threads
int threadCount = atoi(argv[4]);
result = std::vector<std::vector<int> >(horizPixels, std::vector<int>(vertPixels,21)); // init 2-dimensional vector
if (threadCount <= 1) {
SingleThreaded(result);
} else {
MultiThreaded(threadCount, result);
}
//TODO: remove these lines
bitmapImage image(horizPixels, vertPixels);
for(int y = 0; y < vertPixels; y++) {
for(int x = 0; x < horizPixels; x++) {
image.setPixelRGB(x,y,16777216*result[x][y]/maxiter % 256, 65536*result[x][y]/maxiter % 256, 256*result[x][y]/maxiter % 256);
//std::cout << std::setw(2) << std::setfill('0') << std::hex << result[x][y] << " ";
}
std::cout << std::endl;
}
image.saveToBitmapFile("~/Desktop/test.bmp",32);
}
良好的效果可以得到:
mandelbrot 5120 3840 256 3
那樣你會得到5 * 1024寬的圖像; 5 * 768高256色(唉你只會得到1或2)和3個線程(1主線程除了創建工作線程和2個工作線程以外沒有任何工作)
你已經勾起了我的興趣!你說你有作爲並行算法研究員的經驗。當然,你可以給我一些有趣的問題或算法嗎?是的,你是正確的,基準測試對自己的意義很小,這就是爲什麼我不會花太多時間在他們身上 – sven 2008-11-29 10:05:41
已經有一段時間了!我基本上採用了一種有效的順序算法,首先依次測量它的性能,然後對照不斷增加的no。的CPU。 – 2008-11-29 10:16:19
(如在並行版本...) – 2008-11-29 10:18:15