lec11 并發(fā)程序設計3 獨木橋問題

1、計算機與操作系統(tǒng)第十一講并發(fā)程序設計習題講解南京大學軟件學院13:278、獨木橋問題1 獨木橋問題1：東西向汽車過獨木橋，為了保證安全，只要橋上無車，則允許一方的汽車過橋，待一方的車全部過完后， 另一方的車才允許過橋。請用信號量和PV操作寫出過獨木橋問題的同步算法。獨木橋問題1var wait,mutex1,mutex2:semaphore; mutex1:=mutex2:=1;wait:=1; counter1,counter2:integer;process P東() P(mutex1); count1+;if (count1=1) P(wait); V(mutex1);過獨木橋; P(m

2、utex1);count1-;if (count1=0) V(wait); V(mutex1);process P西() P(mutex2); count2+;if (count1=1) P(wait); V(mutex2);過獨木橋; P(mutex1); count2-;if (count1=0) V(wait); V(mutex2);獨木橋問題 2 獨木橋問題2：在獨木橋問題1中，限制橋面上最多可以有k輛汽車通過。試用信號量和P，V操作寫出過獨木橋問題的同步算法獨木橋問題2semaphore wait,mutex1,mutex2,bridge;mutex1=mutex2=1;bridge

3、=k;wait=1; int count1,count2; count1=0;count2=0;process P東( ) P(mutex1);count1+;if (count1=1) P(wait); V(mutex1);P(bridge); 過 橋 ; V(bridge); P(mutex1);count1-;if (count1=0) V(wait); V(mutex1);process P西( ) P(mutex2);count2+;if (count2=1) P(wait); V(mutex2);P(bridge); 過 橋 ; V(bridge); P(mutex2);count

4、2-;if (count2=0) V(wait); V(mutex2);獨木橋問題3 獨木橋問題3：在獨木橋問題1中，以叁輛汽車為一組，要求保證東方和西方以組為單位交替通過汽車。試用信號量和P，V操作寫出汽車過獨木橋問題的同步算法獨木橋問題3semaphore wait,mutex1,mutex2; mutex1=mutex2=1;wait=1;int counter1,counter2; counteru1=0; countd1=0; counteru2=0; counterd2=0; semaphore S1,S2;S1=3;S2=0;Process P東() P(S1)P(mutex1)

5、; countu1+;if (countu1=1) P(wait); V(mutex1); 過獨木橋; V(S2)P(mutex1);countu1-;countd1+if (countu1=0)&(countd1=3)countd1=0;V(wait); V(mutex1);Process P西() P(S2)P(mutex2); countu2+;if (countu2=1) P(wait); V(mutex2);過獨木橋; V(S1)P(mutex2);countu2-; countd2+if (countu2=0)&(countd2=3)countd2=0;V(wait); V(mut

6、ex2);獨木橋問題4 獨木橋問題4：在獨木橋問題1中，要求各方向的汽車串行過橋，但當另一方提出過橋時，應能阻止對方未上橋的后繼車輛，待橋面上的汽車過完橋后，另一方的汽車開始過橋。試用信號量和P，V操作寫出過獨木橋問題的同步算法獨木橋問題4semaphore stop,wait,mutex1,mutex2; stop=mutex1=mutex2=1;wait=1;int count1,count2; count1=0;count2=0;process P東( ) P(stop);P(mutex1); count1+;if (count1=1) P(wait); V(mutex1);V(stop); 過 橋 ; P(mutex1);Count1-;if (count1=0) V(wait); V(mutex1);process P西( ) P(stop);P(mutex2); count2+