我已经实现了避免死锁的银行家算法。。。。。但我没有得到一个安全的序列。。。。。。有人能告诉我我的代码有什么问题吗。。。。。？??？请引导我。。。。。
程序代码如下：
#include <stdio.h>
#include <stdlib.h>
int main()
{
int Max[10][10], need[10][10], alloc[10][10], avail[10], completed[10], safeSequence[10];
int p, r, i, j, process, count;
count = 0;
printf("Enter the no of processes : ");
scanf("%d", &p);
for(i = 0; i< p; i++)
completed[i] = 0;
printf("\n\nEnter the no of resources : ");
scanf("%d", &r);
printf("\n\nEnter the Max Matrix for each process : ");
for(i = 0; i < p; i++)
{
printf("\nFor process %d : ", i + 1);
for(j = 0; j < r; j++)
scanf("%d", &Max[i][j]);
}
printf("\n\nEnter the allocation for each process : ");
for(i = 0; i < p; i++)
{
printf("\nFor process %d : ",i + 1);
for(j = 0; j < r; j++)
scanf("%d", &alloc[i][j]);
}
printf("\n\nEnter the Available Resources : ");
for(i = 0; i < r; i++)
scanf("%d", &avail[i]);
for(i = 0; i < p; i++)
for(j = 0; j < r; j++)
need[i][j] = Max[i][j] - alloc[i][j];
do
{
printf("\n Max matrix:\tAllocation matrix:\n");
for(i = 0; i < p; i++)
{
for( j = 0; j < r; j++)
printf("%d ", Max[i][j]);
printf("\t\t");
for( j = 0; j < r; j++)
printf("%d ", alloc[i][j]);
printf("\n");
}
process = -1;
for(i = 0; i < p; i++)
{
if(completed[i] == 0)//if not completed
{
process = i ;
for(j = 0; j < r; j++)
{
if(avail[j] < need[i][j])
{
process = -1;
break;
}
}
}
if(process != -1)
break;
}
if(process != -1)
{
printf("\nProcess %d runs to completion!", process + 1);
safeSequence[count] = process + 1;
count++;
for(j = 0; j < r; j++)
{
avail[j] += alloc[process][j];
alloc[process][j] = 0;
Max[process][j] = 0;
completed[process] = 1;
}
}
}
while(count != p && process != -1);
if(count == p)
{
printf("\nThe system is in a safe state!!\n");
printf("Safe Sequence : < ");
for( i = 0; i < p; i++)
printf("%d ", safeSequence[i]);
printf(">\n");
}
else
printf("\nThe system is in an unsafe state!!");
}
程序输出如下：
[root@comps 111a1059]# gcc bankerssafesequence.c
[root@comps 111a1059]# ./a.out
Enter the no of processes : 5
Enter the no of resources : 3
Enter the Max Matrix for each process :
For process 1 : 7
5
3
For process 2 : 3
2
2
For process 3 : 7
0
2
For process 4 : 2
2
2
For process 5 : 4
3
3
Enter the allocation for each process :
For process 1 : 0
1
0
For process 2 : 2
0
0
For process 3 : 3
0
2
For process 4 : 2
1
1
For process 5 : 0
0
2
Enter the Available Resources : 3
3
2
Max matrix: Allocation matrix:
7 5 3 0 1 0
3 2 2 2 0 0
7 0 2 3 0 2
2 2 2 2 1 1
4 3 3 0 0 2
Process 2 runs to completion!
Max matrix: Allocation matrix:
7 5 3 0 1 0
0 0 0 0 0 0
7 0 2 3 0 2
2 2 2 2 1 1
4 3 3 0 0 2
Process 3 runs to completion!
Max matrix: Allocation matrix:
7 5 3 0 1 0
0 0 0 0 0 0
0 0 0 0 0 0
2 2 2 2 1 1
4 3 3 0 0 2
Process 4 runs to completion!
Max matrix: Allocation matrix:
7 5 3 0 1 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
4 3 3 0 0 2
Process 1 runs to completion!
Max matrix: Allocation matrix:
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
4 3 3 0 0 2
Process 5 runs to completion!
The system is in a safe state!!
Safe Sequence : < 2 3 4 1 5 >
最佳答案
可能这段代码对你有帮助。
#include<stdio.h>
#include<conio.h>
void main()
{
int process,resource,i,j,instanc,k=0,count1=0,count2=0; //count,k variables are taken for counting purpose
printf("\n\t Enter No. of Process:-\n");
printf("\t\t");
scanf("%d",&process); //Entering No. of Processes
printf("\n\tEnter No. of Resources:-\n");
printf("\t\t");
scanf("%d",&resource); //No. of Resources
int avail[resource],max[process][resource],allot[process][resource],need[process][resource],completed[process];
for(i=0;i<process;i++)
completed[i]=0; //Setting Flag for uncompleted Process
printf("\n\tEnter No. of Available Instances\n");
for(i=0;i<resource;i++)
{
printf("\t\t");
scanf("%d",&instanc);
avail[i]=instanc; // Storing Available instances
}
printf("\n\tEnter Maximum No. of instances of resources that a Process need:\n");
for(i=0;i<process;i++)
{
printf("\n\t For P[%d]",i);
for(j=0;j<resource;j++)
{
printf("\t");
scanf("%d",&instanc);
max[i][j]=instanc;
}
}
printf("\n\t Enter no. of instances already allocated to process of a resource:\n");
for(i=0;i<process;i++)
{
printf("\n\t For P[%d]\t",i);
for(j=0;j<resource;j++)
{
printf("\t\t");
scanf("%d",&instanc);
allot[i][j]=instanc;
need[i][j]=max[i][j]-allot[i][j]; //calculating Need of each process
}
}
printf("\n\t Safe Sequence is:- \t");
while(count1!=process)
{
count2=count1;
for(i=0;i<process;i++)
{
for(j=0;j<resource;j++)
{
if(need[i][j]<=avail[j])
{
k++;
}
}
if(k==resource && completed[i]==0 )
{
printf("P[%d]\t",i);
completed[i]=1;
for(j=0;j<resource;j++)
{
avail[j]=avail[j]+allot[i][j];
}
count1++;
}
k=0;
}
if(count1==count2)
{
printf("\t\t Stop ..After this.....Deadlock \n");
break;
}
}
getch();
}
关于c - Bankers中避免死锁的算法，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/15501861/