处理器调度 1 内容 选择一个调度算法实现处理器调度 2 目的 在采用.doc

处理器调度1 内容选择一个调度算法，实现处理器调度。2 目的在采用多道程序设计的系统中，往往有若干个进程同时处于就绪状态。当就绪状态进程个数大于处理器数时，就必须依照某种策略来决定那些进程优先占有处理器。本题目模拟在单处理器情况下处理器器调度，帮助学生加深了解处理器调度的工作。3 设计思路设计一个结构体数组，来存放个进程的名称,要求运行时间，优先级和当前状态，把优先级最高的进程放在队列的首位，执行一次，就将队列首位的进程的运行时间减一，优先级减一，再重新排序，直到所有进程的运行时间为零。4 程序代码如下：#include#includeint count=5;int head=0;int time=1;int start=1;typedef struct process /定义进程结构,包含信息进程名,进程要求运行的时间,优先及和状态 char pname; int ptime; int prior; char apper;jin;/定义一个结构体变量void creat(jin a)/创建进程队列 int i; jin *q; int k; for(i=0;i5;i+) q=(jin*)malloc(sizeof(jin); ai=*q; printf(please input the %d processs info:n,i); printf(run time and prior are:n); scanf(%d %d,&ai.ptime,&ai.prior); if(i=0) ai.pname=a;/将五个进程命名为a,b,c,d,e. else if(i=1) ai.pname=b; else if(i=2) ai.pname=c; else if(i=3) ai.pname=d; else ai.pname=e; ai.apper=R; for(k=0;k5;k+) /输出所有进程信息printf(%c(%d,%d,%c)n,ak.pname,ak.ptime,ak.prior,ak.apper); void paixu(jin a)/对所有进程按优先及由大至小排序int k; jin temp;int i,j; for(j=0;jcount;j+)for(i=0;icount-1;i+)if(ai.priorai+1.prior) temp=ai; ai=ai+1; ai+1=temp; printf(after order:n); for(k=0;kcount;k+) printf(%c(%d,%d,%c)n,ak.pname,ak.ptime,ak.prior,ak.apper);void run(jin a)/运行进程,操作:使优先数减一,要求运行时间减一int k,i=1;ahead.ptime-; ahead.prior-;printf(run %d times and ,time); time+;if(ahead.ptime=0) & (count=1)start=0; printf(every jin is End:n);if(ahead.ptime=0)&(count!=1) printf(jin %c is End:n,ahead.pname);ahead.apper=E; /输出所有进程信息 printf(%c(%d,%d,%c)n,ahead.pname,ahead.ptime,ahead.prior,ahead.apper); for(k=0;kcount-1;k+) ak=ak+1; count-; main() struct process jin5; creat(jin);/调用创建进程队列函数 paixu(jin);/调用排序函数 while(1)/当标志为1时,运行并排序 run(jin);/运行程序 paixu(jin);/排序程序 if(start=0)/标志为0则跳出 break; 5 程序运行结果：rootlocalhost root# gcc 1.c rootlocalhost root# ./a.out please input the 0 processs infor: run time and prior are: 2 1 please input the 1 processs infor: run time and prior are: 3 5 please input the 2 processs infor: run time and prior are: 1 3 please input the 3 processs infor: run time and prior are: 2 4 please input the 4 processs infor: run time and prior are: 4 2 a(2,1,R) b(3,5,R) c(1,3,R) d(2,4,R) e(4,2,R) after order: b(3,5,R) d(2,4,R) c(1,3,R) e(4,2,R) a(2,1,R) run 1 times and after paixu: b(2,4,R) d(2,4,R) c(1,3,R) e(4,2,R) a(2,1,R) run 2 times and after paixu: d(2,4,R) b(1,3,R) c(1,3,R) e(4,2,R) a(2,1,R) run 3 times and after paixu: d(1,3,R) b(1,3,R) c(1,3,R) e(4,2,R) a(2,1,R) run 4 times and jin d is End: d(0,2,E) after paixu: b(1,3,R) c(1,3,R) e(4,2,R) a(2,1,R) run 5 times and jin b is End: b(0,2,E) after paixu: c(1,3,R) e(4,2,R) a(2,1,R) run 6 times and jin c is End: c(0,2,E) after paixu: e(4,2,R) a(2,1,R) run 7 times and after paixu: e(3,1,R) a(2,1,R) run 8 times and after paixu: a(2,1,R) e(2,0,R) run 9 times and after paixu: a(1,0,R) e(2,0,R) run 10 times and jin a is E: a(0,