UNIX C Program for Banker's Algorithm | CS1254-Operating Systems Lab

AIM:
To write a LINUX/UNIX C Program for the Implementation of Banker's Algorithm in CS1254 - Operating Systems Laboratory.

SOURCE CODE:
#include<stdio.h>
struct process
{
   int allocation[3];
   int max[3];
   int need[3];
   int finish;
}p[10];
int main()
{
  int n,i,j,avail[3],work[3],flag,count=0,sequence[10],k=0;
  printf(“\n Enter the number of process:”);
  scanf(“%d”,&n);
  for(i=0;i<n;i++)
  {
     printf(“\nEnter the %dth  process allocated resources”,i);
     scanf(“%d%d%d”,&p[i].allocation[0],&p[i].allocation[1],&p[i].allocation[2]);
     printf(“\n Enter the %d process maxium resources:”,i);
     scanf(“%d%d%d”,&p[i].max[0],&p[i].max[1],&p[i].max[2]);
     p[i].finish=0;
     p[i].need[0]=p[i].max[0]-p[i].allocation[0];
     p[i].need[0]=p[i].max[1]-p[i].allocation[1];
     p[i].need[0]=p[i].max[2]-p[i].allocation[2];
}
printf(“\n Enter the available vector:”);
scanf(“%d%d%d”,&avail[1],&avail[2]);
for(i=0;i<3;i++);
work[i]=avail[i];
while(count!=n)
{
   count=0
   for(i=0;i<n;i++)
   {
      flag=1;
      if(p[i].finish==0)
      if(p[i].need[0]<=work[0])
      if(p[i].need[1]<=work[1])
      if(p[i].need[2]<=work[2])
        {
           for(j=0;j<3;j++)
           work[j]+=p[i].allocation[j]
           p[i].finish=1;
           sequence[k++]=I;
           flag=0;
        }
       if(flag==1)
       count++;
   }
}
for(i=0;i<n;i++)
if(p]i].finish==1)
count++;
printf(“\n The safe sequence is:\t”);
if(count++n)
for(i=0;i<k;i++)
printf(“ %d\t”,sequence[i]);
else
printf(“SYSTEM IS NOT IN A SAFE STATE \n\n”);
return 0;
}

OUTPUT:
[examuser35@localhost Jebastin]$ cc ban.c
[examuser35@localhost Jebastin]$ ./a.out

Enter the number of  process:3
Enter the 0th process allocated resources:1 2 3
Enter the 0th process maximum resources:4 5 6
Enter the 1th process allocated resources:3 4 5
Enter the 1th process maximum resources:6 7 8
Enter the 2th process allocated resources:1 2 3
Enter the 2th process maximum resources:3 4 5
Enter the available vector:10 12 11

The state sequence :SYSTEM IS NOT IN A SAFE STATE
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