操作系統(tǒng)實(shí)驗(yàn)3-銀行家算法

PAGEPAGE4實(shí)驗(yàn)三死鎖

一、實(shí)驗(yàn)題目設(shè)計(jì)一個(gè)n個(gè)并發(fā)進(jìn)程共享m個(gè)系統(tǒng)資源的系統(tǒng)。進(jìn)程可動(dòng)態(tài)申請(qǐng)資源和釋放資源，系統(tǒng)按各進(jìn)程的申請(qǐng)動(dòng)態(tài)的分配資源。要求采用銀行家算法防止死鎖。二、實(shí)驗(yàn)?zāi)康乃梨i會(huì)引起計(jì)算機(jī)工作僵死，造成整個(gè)系統(tǒng)癱瘓。因此，死鎖現(xiàn)象是操作系統(tǒng)特別是大型系統(tǒng)中必須設(shè)法防止的。通過本次實(shí)驗(yàn)，使學(xué)生掌握死鎖的概念和產(chǎn)生死鎖的原因和必要條件，預(yù)防和避免死鎖的方法，死鎖的檢測(cè)與解除。通過本次實(shí)驗(yàn)，使學(xué)生加深了對(duì)死鎖概念的理解和掌握，深刻領(lǐng)會(huì)銀行家算法的實(shí)質(zhì)及實(shí)現(xiàn)過程。三、設(shè)計(jì)思想（本程序中的用到的所有數(shù)據(jù)類型的定義，主程序的流程圖及各程序模塊之間的調(diào)用關(guān)系）程序流程圖安全（safe）函數(shù)框圖如下：2．邏輯設(shè)計(jì)用結(jié)構(gòu)體數(shù)組表示3個(gè)進(jìn)程，其中包括使用數(shù)組形式的線性表表示某個(gè)進(jìn)程申請(qǐng)資源的列表（隊(duì)列），以及進(jìn)程需要的總資源數(shù)、進(jìn)程當(dāng)前已經(jīng)運(yùn)行到的位置等信息。模擬分配的程序部分，采用循環(huán)隊(duì)列對(duì)各進(jìn)程進(jìn)行調(diào)度。3、物理設(shè)計(jì)全局變量的結(jié)構(gòu)定義如下：boolavoid;structinfo//進(jìn)程信息{longtot,n;//最大占用資源,申請(qǐng)次數(shù)longlist[16];//申請(qǐng)資源列表longpre[16];//已占用資源數(shù)longp;//已分配狀況的指針}pro[4];longLeftRes;程序的結(jié)構(gòu)定義如下：intmain()//主函數(shù){init();allot();}voidinit()//函數(shù)功能:輸入和初始化voidallot()//函數(shù)功能:模擬資源分配過程allot()使用以下函數(shù)模擬分配過程：boolrequire(longx)//函數(shù)功能:嘗試分配資源給進(jìn)程xboolsafe(longx)//函數(shù)功能:檢查分配是否安全可以處理3種資源的程序全局變量結(jié)構(gòu)如下：//共有3種資源，3個(gè)進(jìn)程boolavoid;structinfo//進(jìn)程信息{longtot[4];//最大占用資源數(shù)longp[4];//已經(jīng)占有的資源數(shù)}pro[5];longLeftRes[4];//剩余資源數(shù)longqu[4];四、測(cè)試結(jié)果對(duì)于一組會(huì)導(dǎo)致死鎖的數(shù)據(jù)，分別選擇采用和不采用避免算法，結(jié)果如下：Avoiddeadlock?(Y/N)YPleaseinputtheresourse-requirelistsofthese3processes,onelistsinaline:23423331=================================================================Process1:234(Sum=9)Process2:233(Sum=8)Process3:31(Sum=4)=================================================================Process1require2unit(s)resourseSuccess!LeftResourse=8Process2require2unit(s)resourseSuccess!LeftResourse=6Process3require3unit(s)resourseSuccess!LeftResourse=3Process1require3unit(s)resourseDenied!NotsafeProcess2require3unit(s)resourseDenied!NotsafeProcess3require1unit(s)resourseSuccess!LeftResourse=2Process3finish.LeftResourse=6Process1require3unit(s)resourseDenied!NotsafeProcess2require3unit(s)resourseSuccess!LeftResourse=3Process1require3unit(s)resourseDenied!NotsafeProcess2require3unit(s)resourseSuccess!LeftResourse=0Process2finish.LeftResourse=8Process1require3unit(s)resourseSuccess!LeftResourse=5Process1require4unit(s)resourseSuccess!LeftResourse=1Process1finish.LeftResourse=10Finish如果不避免死鎖：Avoiddeadlock?(Y/N)NPleaseinputtheresourse-requirelistsofthese3processes,onelistsinaline:23423331=================================================================Process1:234(Sum=9)Process2:233(Sum=8)Process3:31(Sum=4)=================================================================Process1require2unit(s)resourseSuccess!LeftResourse=8Process2require2unit(s)resourseSuccess!LeftResourse=6Process3require3unit(s)resourseSuccess!LeftResourse=3Process1require3unit(s)resourseSuccess!LeftResourse=0Process2require3unit(s)resourseDenied!NoenoughresourseProcess3require1unit(s)resourseDenied!NoenoughresourseProcess1require4unit(s)resourseDenied!NoenoughresourseAlreadyDeadlockFinish改進(jìn)版本（可以使用3種資源的程序）測(cè)試結(jié)果如下：五、系統(tǒng)不足與經(jīng)驗(yàn)體會(huì)系統(tǒng)不能處理多種不同資源的情況，也缺乏處理大規(guī)模數(shù)據(jù)的能力。safe函數(shù)的編寫是程序成功的關(guān)鍵，其中利用Change標(biāo)記的思鄉(xiāng)值得仔細(xì)體會(huì)。六、附錄：源代碼（帶注釋）//*******************************************************//*操作系統(tǒng)實(shí)驗(yàn)三死鎖*//*實(shí)驗(yàn)學(xué)院一班（0436101）張?jiān)?jìng)(6040310110)*//*2006-10-15*//******************************************************/#include<cstdio>#include<cstring>#include<cstdlib>#include<ctime>//共有3種資源，3個(gè)進(jìn)程boolavoid;structinfo//進(jìn)程信息{longtot[4];//最大占用資源數(shù)longp[4];//已經(jīng)占有的資源數(shù)}pro[5];longLeftRes[4];//剩余資源數(shù)longqu[4];voidinit()//函數(shù)功能:輸入和初始化{chartmp[128];longi,j;//以下由用戶選擇是否采用避免死鎖算法printf("是否采用避免死鎖算法?(Y/N)");gets(tmp);avoid=(tmp[0]=='Y'||tmp[0]=='y');//輸入各進(jìn)程的資源信息printf("請(qǐng)輸入各進(jìn)程的資源信息:\n");printf("依次輸入各進(jìn)程所需要的最大資源數(shù)量:\n");for(i=1;i<=3;i++){for(j=1;j<=3;j++){scanf("%ld",&pro[i].tot[j]);}}printf("依次輸入各進(jìn)程已經(jīng)占據(jù)的資源數(shù)量:\n");for(i=1;i<=3;i++){for(j=1;j<=3;j++){do{scanf("%ld",&pro[i].p[j]);if(pro[i].p[j]>pro[i].tot[j])printf("\n占有資源超過了聲明的最大資源,請(qǐng)重新輸入\n");}while(pro[i].p[j]>pro[i].tot[j]);}}//初始化資源數(shù)量for(i=1;i<=3;i++){LeftRes[i]=10;//初始化總資源數(shù)為10for(j=1;j<=3;j++){LeftRes[i]-=pro[j].p[i];//減去已經(jīng)被占據(jù)的資源}}//以下輸出資源狀況表printf("\n===============================================================\n");printf("%-20s","當(dāng)前剩余資源");for(i=1;i<=3;i++){printf("%8ld",LeftRes[i]);}for(i=1;i<=3;i++){printf("\n%s%ld%-15s","進(jìn)程",i,"(占用/最大)");for(j=1;j<=3;j++){printf("%4ld/%3ld",pro[i].p[j],pro[i].tot[j]);}}printf("\n===============================================================\n\n");}boolsafe(longx)//函數(shù)功能:檢查分配是否安全{longi,j,y[4];boolchange=1;boolT[4];longstack[4],top=0;memset(T,0,sizeof(T));for(i=1;i<=3;i++){y[i]=LeftRes[i]-qu[i];pro[x].p[i]+=qu[i];}while(change){change=0;for(i=1;i<=3;i++){if(T[i])continue;if(y[1]>=pro[i].tot[1]-pro[i].p[1]&&y[2]>=pro[i].tot[2]-pro[i].p[2]&&y[3]>=pro[i].tot[3]-pro[i].p[3]){for(j=1;j<=3;j++){y[j]+=pro[i].p[j];}T[i]=1;change=1;stack[++top]=i;}}}for(i=1;i<=3;i++){if(!T[i]){return0;}}printf("申請(qǐng)成功!安全序列為:");for(i=1;i<=top;i++)printf("%2ld",stack[i]);return1;}voidrequire(longx)//函數(shù)功能:嘗試分配資源給進(jìn)程x{longi,j;if(LeftRes[1]<qu[1]||LeftRes[2]<qu[2]||LeftRes[3]<qu[3]){printf("申請(qǐng)被拒絕!沒有足夠的剩余資源。\n");return;}if(avoid){if(safe(x)){;}else{printf("申請(qǐng)被拒絕!不符合安全規(guī)則。\n");return;}}else{printf("申請(qǐng)成功!\n");}printf("\n===============================================================\n");printf("%-20s","當(dāng)前剩余資源");for(i=1;i<=3;i++){printf("%8ld",LeftRes[i]-qu[i]);}for(i=1;i<=3;i++){printf("\n%s%ld%-15s","進(jìn)程",i,"(占用/最大)");for(j=1;j<=3;j++){if(i==x&&!avoid)pro[i].p[j]+=qu[i];printf("%4ld/%3ld",pro[i].p[j