我正在通过将模板梯度信息移动到整个目标的梯度图像上(每次旋转(-60至60))来实现模式匹配算法。我已经保存了每次旋转的模板信息,即已经对121个模板进行了预处理和保存。
但是问题是,这消耗大量时间(大约110毫秒),因此决定将一组旋转(-60至-30,-30至0、0至30和30至60)的匹配项拆分为4个线程,但是线程处理比单个进程花费更多的时间(大约115ms至120ms)。
代码段是...
#define MAXTARGETNUM 64
MatchResultA totalResultsTemp[MAXTARGETNUM];
void CShapeMatch::match(ShapeInfo *ShapeInfoVec, search_region SearchRegion, float MinScore, float Greediness, int width,int height, int16_t *pBufGradX ,int16_t *pBufGradY,float *pBufMag, bool corr)
{
MatchResultA resultsPerDeg[MAXTARGETNUM];
....
....
int startX = SearchRegion.StartX;
int startY = SearchRegion.StartY;
int endX = SearchRegion.EndX;
int endY = SearchRegion.EndY;
float AngleStep = SearchRegion.AngleStep;
float AngleStart = SearchRegion.AngleStart;
float AngleStop = SearchRegion.AngleStop;
int startIndex = (int)(ShapeInfoVec[0].AngleNum/2) + ShapeInfoVec[0].AngleNum%2+(int)AngleStart/AngleStep;
int stopIndex = (int)(ShapeInfoVec[0].AngleNum/2) + ShapeInfoVec[0].AngleNum%2+(int)AngleStop/AngleStep;
for (int k = startIndex; k < stopIndex ; k++){
....
for(int j = startY; j < endY; j++){
for(int i = startX; i < endX; i++){
for(int m = 0; m < ShapeInfoVec[k].NoOfCordinates; m++)
{
curX = i + (ShapeInfoVec[k].Coordinates + m)->x; // template X coordinate
curY = j + (ShapeInfoVec[k].Coordinates + m)->y ; // template Y coordinate
iTx = *(ShapeInfoVec[k].EdgeDerivativeX + m); // template X derivative
iTy = *(ShapeInfoVec[k].EdgeDerivativeY + m); // template Y derivative
iTm = *(ShapeInfoVec[k].EdgeMagnitude + m); // template gradients magnitude
if(curX < 0 ||curY < 0||curX > width-1 ||curY > height-1)
continue;
offSet = curY*width + curX;
iSx = *(pBufGradX + offSet); // get corresponding X derivative from source image
iSy = *(pBufGradY + offSet); // get corresponding Y derivative from source image
iSm = *(pBufMag + offSet);
if (PartialScore > MinScore)
{
float Angle = ShapeInfoVec[k].Angel;
bool hasFlag = false;
for(int n = 0; n < resultsNumPerDegree; n++)
{
if(abs(resultsPerDeg[n].CenterLocX - i) < 5 && abs(resultsPerDeg[n].CenterLocY - j) < 5)
{
hasFlag = true;
if(resultsPerDeg[n].ResultScore < PartialScore)
{
resultsPerDeg[n].Angel = Angle;
resultsPerDeg[n].CenterLocX = i;
resultsPerDeg[n].CenterLocY = j;
resultsPerDeg[n].ResultScore = PartialScore;
break;
}
}
}
if(!hasFlag)
{
resultsPerDeg[resultsNumPerDegree].Angel = Angle;
resultsPerDeg[resultsNumPerDegree].CenterLocX = i;
resultsPerDeg[resultsNumPerDegree].CenterLocY = j;
resultsPerDeg[resultsNumPerDegree].ResultScore = PartialScore;
resultsNumPerDegree ++;
}
minScoreTemp = minScoreTemp < PartialScore ? PartialScore : minScoreTemp;
}
}
}
for(int i = 0; i < resultsNumPerDegree; i++)
{
mtx.lock();
totalResultsTemp[totalResultsNum] = resultsPerDeg[i];
totalResultsNum++;
mtx.unlock();
}
n++;
}
void CallerFunction(){
int16_t *pBufGradX = (int16_t *) malloc(bufferSize * sizeof(int16_t));
int16_t *pBufGradY = (int16_t *) malloc(bufferSize * sizeof(int16_t));
float *pBufMag = (float *) malloc(bufferSize * sizeof(float));
clock_t start = clock();
float temp_stop = SearchRegion->AngleStop;
SearchRegion->AngleStop = -30;
thread t1(&CShapeMatch::match, this, ShapeInfoVec, *SearchRegion, MinScore, Greediness, width, height, pBufGradX ,pBufGradY,pBufMag, corr);
SearchRegion->AngleStart = -30;
SearchRegion->AngleStop=0;
thread t2(&CShapeMatch::match, this, ShapeInfoVec, *SearchRegion, MinScore, Greediness, width, height, pBufGradX ,pBufGradY,pBufMag, corr);
SearchRegion->AngleStart = 0;
SearchRegion->AngleStop=30;
thread t3(&CShapeMatch::match, this, ShapeInfoVec, *SearchRegion, MinScore, Greediness,width, height, pBufGradX ,pBufGradY,pBufMag, corr);
SearchRegion->AngleStart = 30;
SearchRegion->AngleStop=temp_stop;
thread t4(&CShapeMatch::match, this, ShapeInfoVec, *SearchRegion, MinScore, Greediness,width, height, pBufGradX ,pBufGradY,pBufMag, corr);
t1.join();
t2.join();
t3.join();
t4.join();
clock_t end = clock();
cout << 1000*(double)(end-start)/CLOCKS_PER_SEC << endl;
}
我们可以看到有很多堆访问权限,但是它们只是只读的。只有totalResultTemp
和totalResultNum
是共享的全局资源,在其上执行写操作。
我的PC配置是
i5-7200U CPU @ 2.50GHz 4 cores
4 Gig RAM
Ubuntu 18
答案 0 :(得分:0)
for(int i = 0; i < resultsNumPerDegree; i++)
{
mtx.lock();
totalResultsTemp[totalResultsNum] = resultsPerDeg[i];
totalResultsNum++;
mtx.unlock();
}
您写入静态数组,互斥锁实际上是time consuming。与其创建锁,不如尝试使用std::atomic_int,或者我认为更好的方法是,将其传递给存储结果的确切位置,所以同步问题不再是您的问题了
答案 1 :(得分:-1)
c / c ++中的POSIX线程不是并发的,因为操作系统必须将分配给每个父进程的时间分成其拥有的线程数。因此,您的算法仅执行核心。要利用多核技术,必须使用OpenMP。该接口库使您可以将算法拆分为不同的物理核心。这是一个很好的OpenMP tutorial