设有一缓冲池 P ， P 中含有 20 个可用缓冲区，一个输

解：

semaphore mutex=1;

semaphore empty=20;

semaphore full=0;

int in,out = 0;

item p [20];

void Producer(){

while(ture){

producer an item in nextp;

wait(empty);

wait(mutex);

p[in] := nextp;

in := (in+1) mod 20;

signal(mutex);

signal(full);

}

}

void Consumer(){

while(ture){

wait(full);

wait(mutex);

nextc := p[out];

out := (out+1) mod 20;

signal(mutex);

signal(empty);

}

}