PV operation daily question - single log bridge problem

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One 、 Problem description

The car passes the single wooden bridge in the east-west direction , To ensure safety , As long as there is no car on the bridge , Then one side's car is allowed to cross the bridge , When all the cars of the other side have passed , The other side's car is allowed to cross the bridge .

It's not hard to find out , It's a reader writer problem

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Two 、 problem solving

：

int eastCount=0;        // The current number of cars on the bridge from the East 
int westCount=0;        // The current number of cars on the bridge from the West 
semaphore bridge=1;     // The cars on both sides are mutually exclusive 
semaphore eastMutex=1;  // Exclusive access eastCount
semaphore westMutex=1;  // Exclusive access westCount

East()
{
    
    while(1)
    {
    
        P(eastMutex);
        if(eastCount==0)
        {
    
            P(bridge);
        }
        eastCount++;
        V(eastMutex);
         Cross the bridge from the East ;

        P(eastMutex);
        eastCount--;
        if(eastCount==0)
        {
    
            P(bridge);
        }
        V(eastMutex);
    }
}

West()
{
    
    while(1)
    {
    
        P(westMutex);
        if(westCount==0)
        {
    
            P(bridge);
        }
        westCount++;
        V(eastMutex);
         Cross the bridge from the West ;

        P(westMutex);
        westCount--;
        if(westCount==0)
        {
    
            P(bridge);
        }
        V(westMutex);
    }
}

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3、 ... and 、 twitter

There isn't much fresh , No more than is first The car on the bridge bridge resources , The last one to get off the bridge Car release for bridge resources , visit xCount You can use mutually exclusive access .
That's it , Slip away ~