實驗三《使用動態(tài)優(yōu)先權(quán)的進(jìn)程調(diào)度算法的模擬》.docx

使用動態(tài)優(yōu)先權(quán)的進(jìn)程調(diào)度算法的模擬實驗三 使用動態(tài)優(yōu)先權(quán)的進(jìn)程調(diào)度算法的模擬 1、實驗?zāi)康?通過動態(tài)優(yōu)先權(quán)算法的模擬加深對進(jìn)程概念和進(jìn)程調(diào)度過程的理解。 2、實驗內(nèi)容 (1)用C語言來實現(xiàn)對N個進(jìn)程采用動態(tài)優(yōu)先算法的進(jìn)程調(diào)度; (2)每個用來標(biāo)識進(jìn)程的進(jìn)程控制塊 PCB用結(jié)構(gòu)來描述，包括以下字段: , 進(jìn)程標(biāo)識符id , 進(jìn)程優(yōu)先數(shù)priority，并規(guī)定優(yōu)先數(shù)越大的進(jìn)程，其優(yōu)先權(quán)越高; , 進(jìn)程已占用的CPU時間cputime ; , 進(jìn)程還需占用的CPU時間alltime，當(dāng)進(jìn)程運行完畢時，alltime變?yōu)?; , 進(jìn)程的阻塞時間startblock，表示當(dāng)進(jìn)程再運行startblock個時間片后，進(jìn)程將進(jìn)入阻塞狀態(tài); , 進(jìn)程被阻塞的時間blocktime，表示已阻塞的進(jìn)程再等待blocktime個時間片后，將轉(zhuǎn)換成就緒態(tài) , 進(jìn)程狀態(tài)state; ，用來將PCB排成隊列 , 隊列指針next(3)優(yōu)先數(shù)改變的原則: , 進(jìn)程在就緒隊列中呆一個時間片，優(yōu)先數(shù)增加1 , 進(jìn)程每運行一個時間片，優(yōu)先數(shù)減3。 (4)假設(shè)在調(diào)度前，系統(tǒng)中有5個進(jìn)程，它們的初始狀態(tài)如下: ID 0 1 2 3 4 PRIORITY 9 38 30 29 0 CPUTIME 0 0 0 0 0 ALLTIME 3 3 6 3 4 STARTBLOCK 2 -1 -1 -1 -1 BLOCKTIME 3 0 0 0 0 STATE READY READY READY READY READY (5)為了清楚地觀察諸進(jìn)程的調(diào)度過程，程序應(yīng)將每個時間片內(nèi)的進(jìn)程的情況顯示出來，參照的具體格式如下: RUNNING PROG: i READY_QUEUE:-id1-id2 BLOCK_QUEUE:-id3-id4 = ID 0 1 2 3 4 PRIORITY P0 P1 P2 P3 P4 CPUTIME C0 C1 C2 C3 C4 ALLTIME A0 A1 A2 A3 A4 STARTBLOCK T0 T1 T2 T3 T4 BLOCKTIME B0 B1 B2 B3 B4 STATE S0 S1 S2 S3 S4 3、思考題 (1)在實際的調(diào)度中，除了按調(diào)度算法選擇下一個執(zhí)行的進(jìn)程外，還應(yīng)處理哪些工作, 隊列實現(xiàn): #include #include /#define NULL 0 #define M 10 typedef struct node int id; int pr; int ct; int at; int bt; int sb; struct node *next; jd; jd *max(jd *p) jd *maxnode=NULL,*p1,*p2,*p3=p; int maxnum; p1=p; p2=p; if(p-next=NULL) return NULL; maxnum=p-next-pr; while(p1-next!=NULL) p2=p1-next; if(maxnum pr) maxnode=p2; p3=p1; maxnum=p2-pr; p1=p1-next; p3-next=maxnode-next; maxnode-next=NULL; return maxnode; void blocktoready(jd *pblock,jd *pready) jd *p1=pblock,*p3; while(p1-next!=NULL) p3=p1-next; if(p3-bt=0) p1-next=p3-next; p3-next=pready-next; pready-next=p3; p1=p1-next; if(p1=NULL) break; void ready(jd *p) jd *p1=p-next; while(p1!=NULL) p1-pr+; p1=p1-next; void run(jd *p) jd *p1; if(p-next!=NULL) p1=p-next; p1-pr=p1-pr-3; p1-at-; p1-ct+; void block(jd *p) jd *p1=p-next; while(p1!=NULL) p1-bt-; p1=p1-next; void runtoreadyorblock(jd *prun,jd *pready,jd *pblock) jd *p; if(prun-next=NULL) return; p=prun-next; if(p-at=0) prun-next=NULL; else if(p-ct=p-sb) p-next=pblock-next; pblock-next=p; prun-next=NULL; else p-next=pready-next; pready-next=p; prun-next=NULL; jd *head(jd pcb,int L) int i; for(i=0;iL;i+) printf(input pcd%d informationn,i); printf(PRIORITY:); scanf(%d,&pcbi.pr); printf(ALLTIME:); scanf(%d,&pcbi.at); printf(STARTBLOCK:); scanf(%d,&pcbi.sb); printf(BLOCKTIME:); scanf(%d,&pcbi.bt); pcbi.id=i; pcbi.ct=0; for(i=0;inext=NULL) printf(ttthe queue are emptyn); while(p-next!=NULL) p1=p-next; printf(tt%dt%dt%dt%dt%dt%dn,p1-id,p1-pr,p1-ct,p1-at,p1-sb,p1-bt); p=p-next; if(p=NULL) break; int main() jd pcbM; jd *pready=(jd *)malloc(sizeof(jd); jd *prun=(jd *)malloc(sizeof(jd); jd *pblock=(jd *)malloc(sizeof(jd); int L,i,n=1; pready-next=NULL; prun-next=NULL; pblock-next=NULL; printf(please input the number of pcb :n); scanf(%d,&L); pready-ne #include #include /#define NULL 0 #define M 10 typedef struct node int id; int pr; int ct; int at; int bt; int sb; struct node *next; jd; jd *max(jd *p) jd *maxnode=NULL,*p1,*p2,*p3=p; int maxnum; p1=p; p2=p; if(p-next=NULL) return NULL; maxnum=p-next-pr; while(p1-next!=NULL) p2=p1-next; if(maxnum pr) maxnode=p2; p3=p1; maxnum=p2-pr; p1=p1-next; p3-next=maxnode-next; maxnode-next=NULL; return maxnode; void blocktoready(jd *pblock,jd *pready) jd *p1=pblock,*p3; while(p1-next!=NULL) p3=p1-next; if(p3-bt=0) p1-next=p3-next; p3-next=pready-next; pready-next=p3; p1=p1-next; if(p1=NULL) break; void ready(jd *p) jd *p1=p-next; while(p1!=NULL) p1-pr+; p1=p1-next; void run(jd *p) jd *p1; if(p-next!=NULL) p1=p-next; p1-pr=p1-pr-3; p1-at-; p1-ct+; void block(jd *p) jd *p1=p-next; while(p1!=NULL) p1-bt-; p1=p1-next; void runtoreadyorblock(jd *prun,jd *pready,jd *pblock) jd *p; if(prun-next=NULL) return; p=prun-next; if(p-at=0) prun-next=NULL; else if(p-ct=p-sb) p-next=pblock-next; pblock-next=p; prun-next=NULL; else p-next=pready-next; pready-next=p; prun-next=NULL; jd *head(jd pcb,int L) int i; for(i=0;iL;i+) printf(input pcd%d informationn,i); printf(PRIORITY:); scanf(%d,&pcbi.pr); printf(ALLTIME:); scanf(%d,&pcbi.at); printf(STARTBLOCK:); scanf(%d,&pcbi.sb); printf(BLOCKTIME:); scanf(%d,&pcbi.bt); pcbi.id=i; pcbi.ct=0; for(i=0;inext=NULL) printf(ttthe queue are emptyn); while(p-next!=NULL) p1=p-next; printf(tt%dt%dt%dt%dt%dt%dn,p1-id,p1-pr,p1-ct,p1-at,p1-sb,p1-bt); p=p-next; if(p=NULL) break; int main() jd pcbM; jd *pready=(jd *)malloc(sizeof(jd); jd *prun=(jd *)malloc(sizeof(jd); jd *pblock=(jd *)malloc(sizeof(jd); int L,i,n=1; pready-next=NULL; prun-next=NULL; pblock-next=NULL; printf(please input the number of pcb :n); scanf(%d,&L); pready-ne #include #include /#define NULL 0 #define M 10 typedef struct node int id; int pr; int ct; int at; int bt; int sb; struct node *next; jd; jd *max(jd *p) jd *maxnode=NULL,*p1,*p2,*p3=p; int maxnum; p1=p; p2=p; if(p-next=NULL) return NULL; maxnum=p-next-pr; while(p1-next!=NULL) p2=p1-next; if(maxnum pr) maxnode=p2; p3=p1; maxnum=p2-pr; p1=p1-next; p3-next=maxnode-next; maxnode-next=NULL; return maxnode; void blocktoready(jd *pblock,jd *pready) jd *p1=pblock,*p3; while(p1-next!=NULL) p3=p1-next; if(p3-bt=0) p1-next=p3-next; p3-next=pready-next; pready-next=p3; p1=p1-next; if(p1=NULL) break; void ready(jd *p) jd *p1=p-next; while(p1!=NULL) p1-pr+; p1=p1-next; void run(jd *p) jd *p1; if(p-next!=NULL) p1=p-next; p1-pr=p1-pr-3; p1-at-; p1-ct+; void block(jd *p) jd *p1=p-next; while(p1!=NULL) p1-bt-; p1=p1-next; void runtoreadyorblock(jd *prun,jd *pready,jd *pblock) jd *p; if(prun-next=NULL) return; p=prun-next; if(p-at=0) prun-next=NULL; else if(p-ct=p-sb) p-next=pblock-next; pblock-next=p; prun-next=NULL; else p-next=pready-next; pready-next=p; prun-next=NULL; jd *head(jd pcb,int L) int i; for(i=0;iL;i+) printf(input pcd%d informationn,i); printf(PRIORITY:); scanf(%d,&pcbi.pr); printf(ALLTIME:); scanf(%d,&pcbi.at); printf(STARTBLOCK:); scanf(%d,&pcbi.sb); printf(BLOCKTIME:); scanf(%d,&pcbi.bt); pcbi.id=i; pcbi.ct=0; for(i=0;inext=NULL) printf(ttthe queue are emptyn); while(p-next!=NULL) p1=p-next; printf(tt%dt%dt%dt%dt%dt%dn,p1-id,p1-pr,p1-ct,p1-at,p1-sb,p1-bt); p=p-next; if(p=NULL) break; int main() jd pcbM; jd *pready=(jd *)malloc(sizeof(jd); jd *prun=(jd *)malloc(sizeof(jd); jd *pblock=(jd *)malloc(sizeof(jd); int L,i,n=1; pready-next=NULL; prun-next=NULL; pblock-next=NULL; printf(please input the number of pcb :n); scanf(%d,&L); pready-next=head(pcb,L); while(1) prun-next=max(pready); run(prun); ready(pready); block(pblock); printf(running%d every pcb information:n,n); printf(ttidtprtcttattsbtbtn); printf(the ready pcb:n); print(pready); printf(the run pcb:n); print(prun); printf(the block pcb:n); print(pblock); printf(the all pcb:n); for(i=0;inext=NULL & pblock-next=NULL) break; xt=head(pcb,L); while(1) prun-next=max(pready); run(prun); ready(pready); block(pblock); printf(running%d every pcb information:n,n); printf(ttidtprtcttattsbtbtn); printf(the ready pcb:n); print(pready); printf(the run pcb:n); print(prun); printf(the block pcb:n); print(pblock); printf(the all pcb:n); for(i=0;inext=NULL & pblock-next=NULL) break; xt=head(pcb,L); while(1) prun-next=max(pready); run(prun); ready(pready); block(pblock); printf(running%d every pcb information:n,n); printf(ttidtprtcttattsbtbtn); printf(the ready pcb:n); print(pready); printf(the run pcb:n); print(prun); printf(the block pcb:n); print(pblock); printf(the all pcb:n); for(i=0;inext=NULL & pblock-next=NULL) break; 運行結(jié)果: rootlocalhost root# gcc -o a shiyan4.c rootlocalhost root# ./a 數(shù)組實現(xiàn): 實驗?zāi)康?通過動態(tài)優(yōu)先權(quán)算法的模擬加深對進(jìn)程概念和進(jìn)程調(diào)度過程的理解。 2、 實驗內(nèi)容: (1) 用C語言來實現(xiàn)對N個進(jìn)程采用動態(tài)優(yōu)先權(quán)優(yōu)先算法的進(jìn)程調(diào)度。 (2) 每個用來標(biāo)識進(jìn)程的進(jìn)程控制塊PCB用結(jié)構(gòu)來描述，包括以下字段: 進(jìn)程標(biāo)識數(shù)ID 進(jìn)程優(yōu)先數(shù)PRIORITY，并規(guī)定優(yōu)先數(shù)越大的進(jìn)程，其優(yōu)先權(quán)越高 進(jìn)程已占用的CPU時間CUPTIME 進(jìn)程還需占用的CPU時間ALLTIME。當(dāng)進(jìn)程運行完畢時，ALLTIME變?yōu)? 進(jìn)程的阻塞時間STARTBLOCK，表示當(dāng)進(jìn)程再運行STARTBLOCK個時間片后，進(jìn)程將進(jìn)入阻塞狀態(tài)。 進(jìn)程被阻塞的時間BLOCKTIME，表示已阻塞的進(jìn)程再等待BLOCKTIME個時間片后，將轉(zhuǎn)換成就緒狀態(tài)。 進(jìn)程狀態(tài)STATE. 隊列指針NEXT，用來將 PCB排成隊列。 (3) 優(yōu)先數(shù)改變的原則: 進(jìn)程在就緒隊列中呆一個時間片，優(yōu)先數(shù)增加1。 進(jìn)程每運行一個時間片，優(yōu)先數(shù)減3。 (4) 假設(shè)在調(diào)度前，系統(tǒng)中有 5個進(jìn)程，它們的初始狀態(tài)如下: ID 0 1 2 3 4 PRIORITY 9 38 30 29 0 CPUTIME 0 0 0 0 0 ALLTIME 3 3 6 3 4 STARTBLOCK 2 -1 -1 -1 -1 BLOCKTIME 3 0 0 0 0 STATE READY READY READY READY READY (5)為了清楚地觀察諸進(jìn)程的調(diào)度過程，程序?qū)⒚總€時間片內(nèi)的進(jìn)程的情況顯示出來，參照的具體格式如下: RUNING PROG:i READY_QUEUE: -id1-id2 BLOCK_QUEUE: -id3-id4 ID 0 1 2 3 4 PRIORITY P0 P1 P2 P3 P4 CPUTIME C0 C1 C2 C3 C4 ALLTIME A0 A1 A2 A3 A4 STARTBLOCK T0 T1 T2 T3 T4 BLOCKTIME B0 B1 B2 B3 B4 STATE S0 S1 S2 S3 S4 #include #include using namespace std; int i;/循環(huán)值 int j;/還在阻塞或就緒隊列中的進(jìn)程數(shù) int s; int m;/最大priority的id struct pcb int id; int p; /priority int cputime; int alltime; int startblock; int blocktime; int state; /0 表示ready 1表示end -1表示block ; struct pcb pro5= 0,9,0,3,2,3,0, 1,38,0,3,-1,0,0, 2,30,0,6,-1,0,0, 3,29,0,3,-1,0,0, 4,0,0,4,-1,0,0 ; int changestate0() if(pro0.startblock=0) pro0.state=-1; pro0.startblock-; return 1; if(pro0.blocktime=0) pro0.state=0; return 1; if(pro0.state=0&pro0.startblock!=-1) pro0.startblock-;return 1; if(pro0.state=-1&pro0.blocktime!=0) pro0.blocktime-;return 1; int state0() changestate0(); s=pro0.p; if(pro0.state=-1) s=-100; return s; int maxp()/求出最大priority state0(); int max=s; m=pro0.id; for(i=0;iproi.p) max=proi+1.p; m=proi+1.id; return m; void change() maxp(); int x;/得到m現(xiàn)在的數(shù)組編號 for(i=0;ij;i+) proi.p+; for(i=0;ij;i+) if(proi.id=m) x=i; prox.cputime+; prox.p=prox.p-4; prox.alltime-; if(prox.alltime=0) prox.state=1; void display() change(); coutRUNNING PROG:mendl; cout=n; coutID ; for(i=0;ij;i+) cout.width(10); coutproi.id; coutendlPRIORITY ; for(i=0;ij;i+) cout.width(10); coutproi.p; coutendlCPUTIME ; for(i=0;ij;i+) cout.width(10); coutproi.cputime; coutendlALLTIME ; for(i=0;ij;i+) cout.width(10); coutproi.alltime; coutendlSTARTBLOCK; for(i=0;ij;i+) cout.width(10); coutproi.startblock; coutendlBLOCKTIME ; for(i=0;ij;i+) cout.width(10); coutproi.blocktime; coutendlSTATE ; for(i=0;ij;i+) cout.width(10); coutproi.state; coutendl; int main() j=5;/剛開始有5個進(jìn)程 while(j!=0) for(i=0;ij;i+) if(proi.state=1) for(;ij;i+) proi=proi+1; j=j-1; display(); getchar(); 運行結(jié)果: rootlocalhost root# g+ -o c c.c rootlocalhost root# ./c RUNNING PROG:1 = ID 0 1 2 3 4 PRIORITY 10 35 31 30 1 CPUTIME 0 1 0 0 0 ALLTIME 3 2 6 3 4 STARTBLOCK 1 -1 -1 -1 -1 BLOCKTIME 3 0 0 0 0 STATE 0 0 0 0 0 RUNNING PROG:1 = ID 0 1 2 3 4 PRIORITY 11 32 32 31 2 CPUTIME 0 2 0 0 0 ALLTIME 3 1 6 3 4 STARTBLOCK 0 -1 -1 -1 -1 BLOCKTIME 3 0 0 0 0 STATE 0 0 0 0 0 RUNNING PROG:1 = ID 0 1 2 3 4 PRIORITY 12 29 33 32 3 CPUTIME 0 3 0 0 0 ALLTIME 3 0 6 3 4 STARTBLOCK -1 -1 -1 -1 -1 BLOCKTIME 3 0 0 0 0 STATE -1 1 0 0 0 RUNNING PROG:2 = ID 0 2 3 4 PRIORITY 13 30 33 4 CPUTIME 0 1 0 0 ALLTIME 3 5 3 4 STARTBLOCK -1 -1 -1 -1 BLOCKTIME 2 0 0 0 STATE -1 0 0 0 RUNNING PROG:3 = ID 0 2 3 4 PRIORITY 14 31 30 5 CPUTIME 0 1 1 0 ALLTIME 3 5 2 4 STARTBLOCK -1 -1 -1 -1 BLOCKTIME 1 0 0 0 STATE -1 0 0 0 RUNNING PROG:2 = ID 0 2 3 4 PRIORITY 15 28 31 6 CPUTIME 0 2 1 0 ALLTIME 3 4 2 4 STARTBLOCK -1 -1 -1 -1 BLOCKTIME 0 0 0 0 STATE -1 0 0 0 RUNNING PROG:3 = ID 0 2 3 4 PRIORITY 16 29 28 7 CPUTIME 0 2 2 0 ALLTIME 3 4 1 4 STARTBLOCK -1 -1 -1 -1 BLOCKTIME 0 0 0 0 STATE 0 0 0 0 RUNNING PROG:2 = ID 0 2 3 4 PRIORITY 17 26 29 8 CPUTIME 0 3 2 0 ALLTIME 3 3 1 4 STARTBLOCK -1 -1 -1 -1 BLOCKTIME 0 0 0 0 STATE 0 0 0 0 RUNNING PROG:3 = ID 0 2 3 4 PRIORITY 18 27 26 9 CPUTIME 0 3 3 0 ALLTIME 3 3 0 4 STARTBLOCK -1 -1 -1 -1 BLOCKTIME 0 0 0 0 STATE 0 0 1 0 RUNNING PROG:2 = ID 0 2 4 PRIORITY 19 24 10 CPUTIME 0 4 0 ALLTIME 3 2 4 STARTBLOCK -1