我测试过parallel_for_在 OpenCV 中，通过与简单数组求和和乘法的正常操作进行比较。
我有 100 个整数的数组，每个整数分成 10 个，然后使用parallel_for_.
然后我也有正常的0到99的求和和乘法运算。
然后我测量了经过的时间，正常操作比parallel_for_手术。
我的 CPU 是 Intel(R) Core(TM) i7-2600 四核 CPU。parallel_for_求和运算耗时 0.002 秒（耗时 2 个时钟周期），乘法运算耗时 0.003 秒（耗时 3 个时钟周期）。
但正常运行时，求和和乘法都需要 0.0000 秒（小于一键循环）。我缺少什么？我的代码如下。
测试类
#include <opencv2\core\internal.hpp>
#include <opencv2\core\core.hpp>
#include <tbb\tbb.h>
using namespace tbb;
using namespace cv;
template <class type>
class Parallel_clipBufferValues:public cv::ParallelLoopBody
{
private:
type *buffertoClip;
type maxSegment;
char typeOperation;//m = mul, s = summation
static double total;
public:
Parallel_clipBufferValues(){ParallelLoopBody::ParallelLoopBody();};
Parallel_clipBufferValues(type *buffertoprocess, const type max, const char op): buffertoClip(buffertoprocess), maxSegment(max), typeOperation(op){
if(typeOperation == 's')
total = 0;
else if(typeOperation == 'm')
total = 1;
}
~Parallel_clipBufferValues(){ParallelLoopBody::~ParallelLoopBody();};
virtual void operator()(const cv::Range &r) const{
double tot = 0;
type *inputOutputBufferPTR = buffertoClip+(r.start*maxSegment);
for(int i = 0; i < 10; ++i)
{
if(typeOperation == 's')
total += *(inputOutputBufferPTR+i);
else if(typeOperation == 'm')
total *= *(inputOutputBufferPTR+i);
}
}
static double getTotal(){return total;}
void normalOperation(){
//int iteration = sizeof(buffertoClip)/sizeof(type);
if(typeOperation == 'm')
{
for(int i = 0; i < 100; ++i)
{
total *= buffertoClip[i];
}
}
else if(typeOperation == 's')
{
for(int i = 0; i < 100; ++i)
{
total += buffertoClip[i];
}
}
}
};
MAIN
#include "stdafx.h"
#include "TestClass.h"
#include <ctime>
double Parallel_clipBufferValues<int>::total;
int _tmain(int argc, _TCHAR* argv[])
{
const int SIZE=100;
int myTab[SIZE];
double totalSum_by_parallel;
double totalSun_by_normaloperation;
double elapsed_secs_parallel;
double elapsed_secs_normal;
for(int i = 1; i <= SIZE; i++)
{
myTab[i-1] = i;
}
int maxSeg =10;
clock_t begin_parallel = clock();
cv::parallel_for_(cv::Range(0,maxSeg), Parallel_clipBufferValues<int>(myTab, maxSeg, 'm'));
totalSum_by_parallel = Parallel_clipBufferValues<int>::getTotal();
clock_t end_parallel = clock();
elapsed_secs_parallel = double(end_parallel - begin_parallel) / CLOCKS_PER_SEC;
clock_t begin_normal = clock();
Parallel_clipBufferValues<int> norm_op(myTab, maxSeg, 'm');
norm_op.normalOperation();
totalSun_by_normaloperation = norm_op.getTotal();
clock_t end_normal = clock();
elapsed_secs_normal = double(end_normal - begin_normal) / CLOCKS_PER_SEC;
return 0;
}
