多线程锁竞争造成的开销

这是执行结果先看实例代码:#include "stdafx.h"#include <Windows.h>#include <string>#include <set>#include <io.h>#include <fstream>#include <Windows.h>#include<time.h>#include <stdio.h>#include <process.h>using namespace std;CRITICAL_SECTION cs;unsigned __stdcall thread1(LPVOID lpVoid){clock_t t1,t2;t1 = clock();for ( int i = 0; i < 5000000; i++ ){EnterCriticalSection(&cs);LeaveCriticalSection(&cs);}t2 = clock();char buf[100]={0};sprintf(buf, "thread1 t2-t1=%ld", t2-t1);printf(buf);printf("\n");return 1;}unsigned __stdcall thread2(LPVOID lpVoid){clock_t t1,t2;t1 = clock();for ( int i = 0; i < 5000000; i++ ){EnterCriticalSection(&cs);LeaveCriticalSection(&cs);}t2 = clock();char buf[100]={0};sprintf(buf, "thread2 t2-t1=%ld", t2-t1);printf(buf);printf("\n");return 1;}int _tmain(int argc, _TCHAR* argv[]){InitializeCriticalSection(&cs);clock_t t1,t2;t1 = clock();for ( int i = 0; i < 10000000; i++ ){EnterCriticalSection(&cs);LeaveCriticalSection(&cs);}t2 = clock();char buf[100]={0};sprintf(buf, "main thread t2-t1=%ld", t2-t1);printf(buf);printf("\n");_beginthreadex(NULL, 0, thread1,NULL,0,NULL);_beginthreadex(NULL, 0,thread2,NULL,0,NULL);Sleep(100000);return 0;}

上面的代码，愿意是想通过多核优势，分两个线程来计算主线程中的1000W次循环。理论上来说，thread1和thread2分别花费的时间应该是main thread的一半，结果分别是main thread的一倍还要多，原因就是:两个线程因为锁竞争成了串行化操作，没有充分利用多核优势，并且平凡的大量锁竞争和锁竞争在某些情况下造成的线程切换，使得时间倍增。所以，尽量减少线程，减少锁竞争，而且这样还少些BUG，好调试。