实验三《使用动态优先权的进程调度算法的模拟》.docx

使用动态优先权的进程调度算法的模拟实验三 使用动态优先权的进程调度算法的模拟 1、实验目的 通过动态优先权算法的模拟加深对进程概念和进程调度过程的理解。 2、实验内容 (1)用C语言来实现对N个进程采用动态优先算法的进程调度; (2)每个用来标识进程的进程控制块 PCB用结构来描述，包括以下字段: , 进程标识符id , 进程优先数priority，并规定优先数越大的进程，其优先权越高; , 进程已占用的CPU时间cputime ; , 进程还需占用的CPU时间alltime，当进程运行完毕时，alltime变为0; , 进程的阻塞时间startblock，表示当进程再运行startblock个时间片后，进程将进入阻塞状态; , 进程被阻塞的时间blocktime，表示已阻塞的进程再等待blocktime个时间片后，将转换成就绪态 , 进程状态state; ，用来将PCB排成队列 , 队列指针next(3)优先数改变的原则: , 进程在就绪队列中呆一个时间片，优先数增加1 , 进程每运行一个时间片，优先数减3。 (4)假设在调度前，系统中有5个进程，它们的初始状态如下: 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)为了清楚地观察诸进程的调度过程，程序应将每个时间片内的进程的情况显示出来，参照的具体格式如下: 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、思考题 (1)在实际的调度中，除了按调度算法选择下一个执行的进程外，还应处理哪些工作, 队列实现: #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; 运行结果: rootlocalhost root# gcc -o a shiyan4.c rootlocalhost root# ./a 数组实现: 实验目的 通过动态优先权算法的模拟加深对进程概念和进程调度过程的理解。 2、 实验内容: (1) 用C语言来实现对N个进程采用动态优先权优先算法的进程调度。 (2) 每个用来标识进程的进程控制块PCB用结构来描述，包括以下字段: 进程标识数ID 进程优先数PRIORITY，并规定优先数越大的进程，其优先权越高 进程已占用的CPU时间CUPTIME 进程还需占用的CPU时间ALLTIME。当进程运行完毕时，ALLTIME变为0 进程的阻塞时间STARTBLOCK，表示当进程再运行STARTBLOCK个时间片后，进程将进入阻塞状态。 进程被阻塞的时间BLOCKTIME，表示已阻塞的进程再等待BLOCKTIME个时间片后，将转换成就绪状态。 进程状态STATE. 队列指针NEXT，用来将 PCB排成队列。 (3) 优先数改变的原则: 进程在就绪队列中呆一个时间片，优先数增加1。 进程每运行一个时间片，优先数减3。 (4) 假设在调度前，系统中有 5个进程，它们的初始状态如下: 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)为了清楚地观察诸进程的调度过程，程序将每个时间片内的进程的情况显示出来，参照的具体格式如下: 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;/循环值 int j;/还在阻塞或就绪队列中的进程数 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现在的数组编号 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;/刚开始有5个进程 while(j!=0) for(i=0;ij;i+) if(proi.state=1) for(;ij;i+) proi=proi+1; j=j-1; display(); getchar(); 运行结果: 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