一、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，待一方的汽車全部過完后，另一方的汽車才允許過橋。

semaphore wait,mutex1,mutex2; mutex1=mutex2=1;wait=1; bridge=1; int counter1,counter2; counter1=0;counter2=0; semaphore S1,S2;S1=3;S2=0; process P左() { while(true) {P(mutex1); count1++; if (count1==1) P(wait); V(mutex1); 過獨木橋; P(mutex1); count1--;if(count1==0) V(wait); V(mutex1); } } process P右() { while(true) {P(mutex2); count2++; if (count2==1) P(wait); V(mutex2); 過獨木橋; P(mutex1); count2--; if(count2==0) V(wait); V(mutex2); } } <strong> </strong>

二、在獨木橋問題1中，限制橋面上最多可以有k輛汽車通過。

答1：? semaphore wait,mutex1,mutex2,bridge;mutex1=mutex2=1;bridge=k;wait=1;int counter1,counter2; counter1=0;counter2=0; cobegin process P東( ) { process P西( ) {P(mutex1); P(mutex2);count1++; count2++;if (count1==1) P(wait); if (count2==1) P(wait);V(mutex1); V(mutex2);P(bridge); P(bridge);{過橋}; {過橋}; V(bridge); V(bridge); P(mutex1); P(mutex2);count1--; count2--; if (count1==0) V(wait); if (count2==0) V(wait);V(mutex1); V(mutex2);} } coend 答2： cobegin process P東( ) { process P西( ) {P(bridge1); P(bridge2);P(mutex1); P(mutex2);count1++; count2++;if (count1==1 ) P(wait); if (count2==1) P(wait);V(mutex1); V(mutex2);{過橋}; {過橋}; V(bridge1); V(bridge2); P(mutex1); P(mutex2); count1--; count2--; if (count1==0) V(wait) if (count2==0 ) V(wait); V(mutex1); V(mutex2);} } coend
三、 在獨木橋問題1中，以叁輛汽車為一組，要求保證左方和右方以組為單位交替通過汽車。

semaphore 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左() { while(true) {P(S1) P(mutex1); countu1 ++; if (countu1==1) & (countd1==0) 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右() { while(true) {P(S2) P(mutex2); countu2++; if (countu2==1) & (countd2==0) P(wait); V(mutex2); 過獨木橋; V(S1) P(mutex2); countu2--;countd2++ if ((countu2==0)&(countd2==3)) {countd2=0; V(wait); }V(mutex2); } }

左邊過橋分為兩種典型(不完全)：

(1)PL1上橋, PL1下橋, PL2上橋, PL2下橋, PL3上橋, PL3下橋

如果PL1上橋, PL1下橋，countu1會減到0，但是countd1計數還是1，可以避免這時就由左邊執行V(wait)；PL2上橋, PL2下橋, countu1會減到0，countd1加到2，不執行V(wait)；PL2上橋, PL2下橋, countu1會減到0，countd1累計到3，因此只有左邊累計有3個下橋之后，才一次性將countd1改為0，并執行V(wait)喚醒右邊。

(2)PL1上橋, PL2上橋, PL3上橋, PL1下橋, PL2下橋, PL3下橋

PL1,PL2, PL3依次上橋，此時countu1加到3；然后PL1下橋，countu1減到2，countd1加到1，不執行V(wait)；PL2下橋，countu1減到1，countd1加到2，不執行V(wait)；PL3下橋，countu1減到0，countd1加到3，執行V(wait) 喚醒右邊。

四、 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。

stop用于當另一方提出過橋時，應阻止對方未上橋的后繼車輛。semaphore stop,wait,mutex1,mutex2;stop=mutex1=mutex2=1;wait=1; int counter1,counter2; counter1=0;counter2=0; cobegin process P東( ) { process P西( ) {P(stop); P(stop);P(mutex1); P(mutex2);count1++; count2++;if (count1==1) P(wait); if (count2==1) P(wait); V(mutex1); V(mutex2); V(stop); V(stop); {過橋}; {過橋}; P(mutex1); P(mutex2);Count1--; count2--; if (count1==0) V(wait); if (count2==0) V(wait);V(mutex1); V(mutex2);} } coend

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