编程范式 18 冰淇淋商店

clerks
40 cones
10 customers 每人吃1-4个冰淇淋 FIFO
manager不同意要重做 同时只能有一个clerk和manager见面
cashier
需要四类线程

1. manager
2. clerks
3. customers
4. cashier

int main(){    int totalCones=0;    InitThreadPackage();    SetupSemaphores();    for(int i=0;i<10;i++)    {        int numCones=RandomInteger(1,1);        ThreadNew(,Customer,1,numCunstomers);        totalCones+=numCones;    }    ThreadNew(,Cashier,0);    ThreadNew(,Manager,1,totalCones);    RunAllThreads();    FreeSemaphores();    return 0;} 

struct inspection{    bool passed;//false    Semaphore requested;//0    Semaphore finished;//0    Semphore lock;//1}

void Manager(int totalConesNeeded){    int numApproved=0;    int numInsected=0;    while(numApproved<totalConesNeeded)    {        sw(inspection<requested);        numINspected++;        inspection.passed=Random(1,0)        //50%概率通过        if(inspection.passed)        {            numApproved++;        }        ss(inspected.finished);    }}

void clerk(Semaphore semaToSignal){    bool passed=false;    while(!passed)    {        MakeCone();        sw(inspection.lock);//semaphore wait        ss(inspection.requested);//semaphore signal        sw(inspection.finished);        passed=inspection.passed;        ss(inspection.lock);    }    ss(semaToSignal);}

void Customer(int numCones){    Brow X();//等店员做完    Samphore clerkDone;    for(int i=0;i<numCones;i++)    {        TN(,1,clerkDone);//ThreadNew    }    for(int i=0;i<numCones;i++)    {        sw(clerksDone);    }    SemaphoreFree(clerksDone);    WalkToCashier();    ss(line.requested);    sw(line.lock);    int place=line.number++;    ss(line.lock);    ss(line.requested);    sw(line.customers[place]);}   

struct line{    int number;//0    Semaphore requested;//0    Semaphore Customers[10];    Semaphore lock;//1}

void Cashier(){    for(int i=0;i<10;i++)    {        sw(line.requested);        checkout(i);        semaphore line(customers[i]);    }}