操作系统课程设计-模拟银行家算法
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)
|
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 |
|
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(" 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");
scanf("%d %d %d %d",&id,&Request[0],&Request[1],&Request[2]);
Bank(id,Request,Need,Avialable,Allocation);
}