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文件名称: 银行家算法(c语言)
  所属分类: 网络攻防
  开发工具:
  文件大小: 883kb
  下载次数: 0
  上传时间: 2009-01-18
  提 供 者: baidu_3*******
 详细说明: 操作系统课的实验(银行家算法)#include "malloc.h"   #include "stdio.h"   #include "stdlib.h"   #define alloclen sizeof(struct allocation)   #define maxlen sizeof(struct max)   #define avalen sizeof(struct available)   #define needlen sizeof(struct need)   #define finilen sizeof(struct finish)   #define pathlen sizeof(struct path)   struct allocation   {   int value;   struct allocation *next;   };   struct max   {   int value;   struct max *next;   };   struct available /*可用资源数*/   {   int value;   struct availabl e *next;   };   struct need /*需求资源数*/   {   int value;   struct need *next;   };   struct path   {   int value;   struct path *next;   };   struct finish   {   int stat;   struct finish *next;   };   int main()   {   int row,colum,status=0,i,j,t,temp,processtest;   struct allocation *allochead,*alloc1,*alloc2,*alloctemp;   struct max *maxhead,*maxium1,*maxium2,*maxtemp;   struct available *avahead,*available1,*available2,*workhead,*work1,*work2,*worktemp,*worktemp1;   struct need *needhead,*need1,*need2,*needtemp;   struct finish *finihead,*finish1,*finish2,*finishtemp;   struct path *pathhead,*path1,*path2;   printf("\n请输入系统资源的种类数:");   scanf("%d",&colum);   printf("请输入现时内存中的进程数:");   scanf("%d",&row);   printf("请输入已分配资源矩阵:\n");   for(i=0;inext=alloc2->next=NULL;   scanf("%d",&allochead->value);   status++;   }   else   {   alloc2=(struct allocation *)malloc(alloclen);   scanf("%d,%d",&alloc2->value);   if(status==1)   {   allochead->next=alloc2;   status++;   }   alloc1->next=alloc2;   alloc1=alloc2;   }   }   }   alloc2->next=NULL;   status=0;   printf("请输入最大需求矩阵:\n");   for(i=0;inext=maxium2->next=NULL;   scanf("%d",&maxium1->value);   status++;   }   else   {   maxium2=(struct max *)malloc(maxlen);   scanf("%d,%d",&maxium2->value);   if(status==1)   {   maxhead->next=maxium2;   status++;   }   maxium1->next=maxium2;   maxium1=maxium2;   }   }   }   maxium2->next=NULL;   status=0;   printf("请输入现时系统剩余的资源矩阵:\n");   for (j=0;jnext=available2->next=NULL;   work1->next=work2->next=NULL;   scanf("%d",&available1->value);   work1->value=available1->value;   status++;   }   else   {   available2=(struct available*)malloc(avalen);   work2=(struct available*)malloc(avalen);   scanf("%d,%d",&available2->value);   work2->value=available2->value;   if(status==1)   {   avahead->next=available2;   workhead->next=work2;   status++;   }   available1->next=available2;   available1=available2;   work1->next=work2;   work1=work2;   }   }   available2->next=NULL;   work2->next=NULL;   status=0;   alloctemp=allochead;   maxtemp=maxhead;   for(i=0;inext=need2->next=NULL;   need1->value=maxtemp->value-alloctemp->value;   status++;   }   else   {   need2=(struct need *)malloc(needlen);   need2->value=(maxtemp->value)-(alloctemp->value);   if(status==1)   {   needhead->next=need2;   status++;   }   need1->next=need2;   need1=need2;   }   maxtemp=maxtemp->next;   alloctemp=alloctemp->next;   }   need2->next=NULL;   status=0;   for(i=0;inext=finish2->next=NULL;   finish1->stat=0;   status++;   }   else   {   finish2=(struct finish*)malloc(finilen);   finish2->stat=0;   if(status==1)   {   finihead->next=finish2;   status++;   }   finish1->next=finish2;   finish1=finish2;   }   }   finish2->next=NULL; /*Initialization compleated*/   status=0;   processtest=0;   for(temp=0;tempstat==0)   {   for(j=0;jnext,worktemp=worktemp->next)   if(needtemp->value<=worktemp->value)   processtest++;   if(processtest==colum)   {   for(j=0;jvalue+=alloctemp->value;   worktemp1=worktemp1->next;   alloctemp=alloctemp->next;   }   if(status==0)   {   pathhead=path1=path2=(struct path*)malloc(pathlen);   path1->next=path2->next=NULL;   path1->value=i;   status++;   }   else   {   path2=(struct path*)malloc(pathlen);   path2->value=i;   if(status==1)   {   pathhead->next=path2;   status++;   }   path1->next=path2;   path1=path2;   }   finishtemp->stat=1;   }   else   {   for(t=0;tnext;   finishtemp->stat=0;   }   }   else   for(t=0;tnext;   alloctemp=alloctemp->next;   }   processtest=0;   worktemp=workhead;   finishtemp=finishtemp->next;   }   }   path2->next=NULL;   finishtemp=finihead;   for(temp=0;tempstat==0)   {   printf("\n系统处于非安全状态!\n");   exit(0);   }   finishtemp=finishtemp->next;   }   printf("\n系统处于安全状态.\n");   printf("\n安全序列为: \n");   do   {   printf("p%d ",pathhead->value);   }   while(pathhead=pathhead->next);   printf("\n");   return 0;   } #include "string.h" #include #include #define M 5 #define N 3 #define FALSE 0 #define TRUE 1 /*M个进程对N类资源最大资源需求量*/ int MAX[M][N]={{7,5,3},{3,2,2},{9,0,2},{2,2,2},{4,3,3}}; /*系统可用资源数*/ int AVAILABLE[N]={10,5,7}; /*M个进程对N类资源最大资源需求量*/ int ALLOCATION[M][N]={{0,0,0},{0,0,0},{0,0,0},{0,0,0},{0,0,0}}; /*M个进程已经得到N类资源的资源量 */ int NEED[M][N]={{7,5,3},{3,2,2},{9,0,2},{2,2,2},{4,3,3}}; /*M个进程还需要N类资源的资源量*/ int Request[N]={0,0,0}; void main() { int i=0,j=0; char flag='Y'; void showdata(); void changdata(int); void rstordata(int); int chkerr(int); showdata(); while(flag=='Y'||flag=='y') { i=-1; while(i<0||i>=M) { printf("请输入需申请资源的进程号(从0到"); printf("%d",M-1); printf(",否则重输入!):"); scanf("%d",&i); if(i<0||i>=M)printf("输入的进程号不存在,重新输入!\n"); } printf("请输入进程"); printf("%d",i); printf("申请的资源数\n"); for (j=0;jNEED[i][j]) { printf("进程"); printf("%d",i); printf("申请的资源数大于进程"); printf("%d",i); printf("还需要"); printf("%d",j); printf("类资源的资源量!申请不合理,出错!请重新选择!\n"); /*printf("申请不合理,出错!请重新选择!\n");*/ flag='N'; break; } else { if(Request[j]>AVAILABLE[j]) { printf("进程"); printf("%d",i); printf("申请的资源数大于系统可用"); printf("%d",j); printf("类资源的资源量!申请不合理,出错!请重新选择!\n"); /*printf("申请不合理,出错!请重新选择!\n");*/ flag='N'; break; } } } if(flag=='Y'||flag=='y') { changdata(i); if(chkerr(i)) { rstordata(i); showdata(); } else showdata(); } else showdata(); printf("\n"); printf("是否继续银行家算法演示,按'Y'或'y'键继续,按'N'或'n'键退出演示: "); scanf("%c",&flag); } } void showdata() { int i,j; printf("系统可用的资源数为:\n"); printf(" "); for (j=0;j"); } printf("\n"); return 0; } ...展开收缩
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