簡單線程的順序

我的VC++ 2010快速安裝的時間在工作，已經過期，所以我不能再使用它，而不改變從右下的日期。我不想這樣做，所以我問你這些問題：簡單線程的順序

你認爲，在多線程程序的線程下面的順序工作？

如果我把類似的3多個線程計算下一個1000粒子，情景開關會耗盡還是睡眠（10）正在用盡？

注意：每個計算（。的力的每一個，VEL，POS）約需9毫秒。

Something like this: 
core1:first 1000 particles forces  // 
core2:first 1000 particles velocities //===>these 3 are connected 
core3:first 1000 particles positions // ------------------------------ 
                     | 
core4:next 1000 particles forces  //        ====these 2 will be connected 
core5:next 1000 particles velocities //===>these 3 are connected  | 
core6:next 1000 particles positions // ------------------------------ 

boolean locker1; 
boolean locker2; 
boolean locker3; 

boolean worker1; 
boolean worker2; 
boolean worker3; 

void core1(void * x) 
{ 
    while(worker1) 
    { 
     while(!locker1){Sleep(10);} 
     for(int i=0;i<1000;i++) 
     { 
       //calculate 1000 particles forces 
     } 
     locker2=true; //starts core2 thread 
     while(locker2){Sleep(10);} //core2 must be working 
     while(locker3){Sleep(10);} //core3 must be working 
    } 
    _endthread(); 
} 

void core2(void * y) 
{ 
    while(worker2) 
    { 
     if(!locker2){Sleep(10);} 
     for(int i=0;i<1000;i++) 
     { 
       //calculate 1000 particles velocities 
     } 
     locker3=true; //starts core3 thread 
     while(locker3){Sleep(10);} //core3 must be working 
     while(locker1){Sleep(10);} //core1 must be working 
    } 
    _endthread(); 

} 

void core3(void * z) 
{ 
    while(worker3) 
    { 
     if(!locker3){Sleep(10);} 
     for(int i=0;i<1000;i++) 
     { 
       //calculate 1000 particles positions 
     } 
     locker1=true; //starts core1 thread 
     while(locker1){Sleep(10);} //core1 must be working 
     while(locker2){Sleep(10);} //core2 must be working 
    } 
    _endthread(); 

} 

int main() 
{ 
    locker1=false; 
    locker2=false; 
    locker3=false; 
    worker1=true; 
    worker2=true; 
    worker3=true; 
    _beginthread(core1,0,(void *)0); 
    _beginthread(core2,0,(void *)0); 
    _beginthread(core3,0,(void *)0); 

    locker1=true; //gets the waiting core1-thread working 
        //so i think when it finishes, it releases core2 to work 
        //when core2 finishes, core3 starts working 

    Sleep(100); 
    worker1=false; 
    worker2=false; 
    worker3=false; //after a while i shut them down 

}

請如果有任何人有VC++可以給我一些提示或建議我欣賞。

Or should i just forget the 3+3-->2 system and do this(6)? : 

core1 first 233 particles for computing forces ----->all for velocity ----->all for psoition 
core2 next 233 particles for computing forces ----->all for velocity ----->all for psoition 
core3 next 233 particles for computing forces ----->all for velocity ----->all for psoition 
core4 next 233 particles for computing forces ----->all for velocity ----->all for psoition 
core5 next 233 particles for computing forces ----->all for velocity ----->all for psoition 
core6 last 233 particles for computing forces ----->all for velocity ----->all for psoition 

and just wait all of them finish to get to next calculation?

來源

2012-09-03 huseyin tugrul buyukisik

你是要我們寫和分析你的應用程序？ –

您是否知道只需註冊Visual C++ 2010 Express即可在30天后繼續使用它？我的安裝已經超過兩年了，工作正常... – Blastfurnace

：O我不知道。我以爲需要錢 –