生产者与消费者得问题描述

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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

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