生产者与消费者得问题描述
生产者＿消费者问题（参考资料；计算机操作系统）
1：利用记录型信号量解决生产者与消费者问题：
Var mutext,empty,full:semaphore=1,n,0;
buffer ：array[0,…,n-1] ofitem;
in out :interge:=0,0;
begin
parbegin
producer:begin
repeat
proceduceran item nextp;
wait(empty);
wait(mutex);
buffer(in):=nextp;
in=(in+1)mod n;
signal(mutex);
signal(full);
until false;
end
consumer:begin
repeat
wait(full);
wait(mutex);
nextc:=buffer(out);
out:=(out+1)mod n;
signal(empty);
signal(mutex);
consumerthe item in nextc;
until false;
end
parend
end
2：利用AND信号量解决生产者与消费者问题：
Var mutext,empty,full:semaphore=1,n,0;
buffer ：array[0,…,n-1] ofitem;
in out :interge:=0,0;
begin
parbegin
producer:begin
repeat
proceduceran item nextp;
Swait(empty,mutex);
buffer(in):=nextp;
in=(in+1)mod n;
Ssignal(mutex.mutex);
utilfalse;
end
consumer:begin
repeat
Swait(full,mutex);
nextc:=buffer(out);
out:=(out+1)mod n;
Ssignal(empty,mutex);
consumerthe item in nextc;
until false;
end
parend
end
3：利用管程解决生产者与消费者问题：
(1):Pc管道可以描述为：
type producer-consumer=monitor
Varint,out,count :integer;
buffer:array[0,…,n-1] of item;
notfull,notempty:condition;
proceducerentry put(item);
begin
if(count>=n)then notfull.wait;
buffer(in):=nextp;
in:=(in+1)mod n;
count:=count+1;
ifnotempty.queue then notempty.signal;
end
procedure entry get(item)
begin
if count<=0 then notempty.wait;
nextc:=buffer(out);
out:=(out+1)mod n;
count:=count-1;
ifnotfull.queue then notfull.signal;
end
(2):利用管程解决生产者与消费者问题是，可以描述为：
producer: begin
repeat
producean item in nextp;
PC.put(item);
utilfalse;
end
consumer:begin
repeat
PC.get(item);
Consumer the item in nextc;
util false;
end