基于windows线程的并行前缀求和

#include <Windows.h>
#include <process.h>
#include <stdio.h>
#include <time.h>

#define NUM_THREADS 4

int N = 10000, *A;
int intTotals[NUM_THREADS], outTotals[NUM_THREADS];
HANDLE doneStep1[NUM_THREADS];
HANDLE doneStep2;

unsigned __stdcall prefixScan(LPVOID pArg)
{
	if(NULL != pArg)
	{
		int tNum = *((int*)pArg);
		int start, end, i;
		int lPrefixTotal;

		free(pArg);
		pArg = NULL;

		start = (N/NUM_THREADS)*tNum;
		end = (N/NUM_THREADS)*(tNum +1);

		if (tNum == (NUM_THREADS - 1))
		{
			end = N;
		}

		for (i = start + 1; i < end; i++)
		{
			//		printf(" %d, %d\r\n", A[i-1],A[i]);
			A[i] = A[i-1] + A[i];		
			Sleep(1);
		}

		intTotals[tNum] = A[end - 1];

		//	printf("sub = %d, from %d to %d result = %d\r\n", tNum, start, end, intTotals[tNum]);

		SetEvent(doneStep1[tNum]);

		WaitForSingleObject(doneStep2, INFINITE);

		lPrefixTotal = outTotals[tNum];
		for (i = start; i < end; i++)
		{
			A[i] = lPrefixTotal + A[i];
		}
	}
	return 0;
}

int main(int argc, char * argv[])
{
	int i, j;
	HANDLE tHandles[NUM_THREADS];

	time_t begin;
	time_t end;

	begin = time(NULL);

	A = (int*)malloc(sizeof(int)*N);

	int * p = A;

	for (i = 0; i < N; i++)
	{
		(*p++) = i;
	}

	for (i = 0; i < NUM_THREADS; i++)
	{
		doneStep1[i] = CreateEvent(NULL, TRUE, FALSE, NULL);
	}
	doneStep2 = CreateEvent(NULL, TRUE, FALSE, NULL);

	for (i = 0; i < NUM_THREADS; i++)
	{
		int *tnum = new int;
		*tnum = i;
		tHandles[i] = (HANDLE)_beginthreadex(NULL, 0, prefixScan, (LPVOID)tnum, 0, NULL);
	}

	WaitForMultipleObjects(NUM_THREADS, doneStep1, TRUE, INFINITE);

	outTotals[0] = 0;
	for (j = 1; j < NUM_THREADS; j++)
	{
		outTotals[j] = outTotals[j - 1] + intTotals[j -1];
	}

	SetEvent(doneStep2);

	WaitForMultipleObjects(NUM_THREADS, tHandles, TRUE, INFINITE);

	end = time(NULL);

	printf("total = %ld cost = %d\r\n", outTotals[NUM_THREADS -1], (end - begin));

	if(NULL != A)
	{
		free(A);
		A = NULL;
	}

	return 0;
}