操作系统课程设计-模拟银行家算法|课程设计|计算机数据库课程设计
模拟银行家算法，用银行家算法实现资源分配。
Output() 输出某时刻的资源分配情况；
Bank() 银行家算法的步骤 1、2、3、及4调用安全性算法；
Security() 进行安全性检查；
Main() 各种数据结构的定义。
假设 有0-4 五个进程(假设系统的当前状态是安全的)，三类资源 即：A、B、C
(以下各种数据均取自课本Ｐ97例子)
T0 时刻的资源分配表(各种资源的数量分别为：10、5、7)
资源情况
进程
Max
A B C
Allocation
A B C
Need
A B C
Available
A B C
P0
7 5 3
0 1 0
7 4 3
3 3 2
P1
3 2 2
2 0 0
1 2 2
P2
9 0 2
3 0 2
6 0 0
P3
2 2 2
2 1 1
0 1 1
P4
4 3 3
0 0 2
4 3 1
#include output(int arr[5][3]){ int i,j; printf(" A B C\n"); for(i=0;i<5;i++) { printf("P%d",i); for(j=0;j<3;j++) { printf(" %d",arr[i][j]); } printf("\n"); }}int Security(int avialable[3],int need[5][3],int allocation[5][3]){ int j,i; int work[3]; int finish[5]={0,0,0,0,0}; for(j=0;j<3;j++) work[j]=avialable[j]; for(i=0;i<5;i++) { if(finish[i]==0) { for(j=0;j<3;j++) if(need[i][j]<=work[i]&need[i][j]<=work[i]&need[i][j]<=work[i]) work[j]=work[j]+allocation[i][j]; finish[j]=1; } } for(i=0;i<5;i++) { if(finish[i]==0) return 0; else return 1; }} */Bank(int i,int request[3],int need[5][3],int avialable[3],int allocation[5][3]){ int int j; printf("The process id: %d\n",i); printf("The Process Request:"); for(j=0;j<3;j++) printf(" %d",request[j]); printf("\n"); if(request[0]<=need[i][0]&request[1]<=need[i][1]&request[2]<=need[i][2]) { if(request[0]<=avialable[0]&request[1]<=avialable[1]&request[2]<=avialable[2]) { for(j=0;j<3;j++) { avialable[j]=avialable[j]-request[j]; allocation[i][j]=allocation[i][j]+request[j]; need[i][j]=need[i][j]-request[j]; } printf("The source after request:\n"); printf(" Allocation\n"); output(allocation); printf(" Need\n"); output(need); printf("The Avialable after request:"); for(j=0;j<3;j++) printf(" %d",avialable[j]); printf("\n"); } else printf("RequestError!"); } else printf("RequestError!"); /*Security(avialable,need,allocation);*/}main(){ int id,i; int Avialable[3]={3,3,2}; int Max[5][3]={{7,5,3},{3,2,2},{9,0,2},{2,2,2},{4,3,3}}; int Allocation[5][3]={{0,1,0},{2,0,0},{3,0,2},{2,1,1},{0,0,2}}; int Need[5][3]={{7,4,3},{1,2,2},{6,0,0},{0,1,1},{4,3,1}}; int Request[3]; /*i Id of Process*/
printf("The source before request:\n"); printf(" MAX\n"); output(Max); printf(" Allocation\n"); output(Allocation); printf(" Need\n"); output(Need); printf("The Avialable before request:");for(i=0;i<3;i++) printf(" %d",Avialable[i]); printf("\n");
printf("Input the id and request of Process:"); scanf("%d %d %d %d",&id,&Request[0],&Request[1],&Request[2]); Bank(id,Request,Need,Avialable,Allocation);}
