Banker's Algorithm

#include 
struct file  { 
  int all[10]; 
  int max[10]; 
  int need[10]; 
  int flag; 
  }; 
  void main() 
  {
    struct file f[10];
    int fl;
    int i, j, k, p, b, n, r, g, cnt=0, id, newr; 
    int avail[10],seq[10]; 
    printf("Enter number of processes -- "); 
    scanf("%d",&n);
    printf("Enter number of resources -- "); 
    scanf("%d",&r);  for(i=0;i    {
      printf("Enter details for P%d",i);
      printf("\nEnter allocation\t -- \t");
      for(j=0;j      scanf("%d",&f[i].all[j]);
      printf("Enter Max\t\t -- \t"); 
      for(j=0;j      scanf("%d",&f[i].max[j]);
      f[i].flag=0; 
     
    } 

    printf("\nEnter Available Resources\t -- \t"); 
    for(i=0;i    scanf("%d",&avail[i]);
    printf("\nEnter New Request Details -- "); 
    printf("\nEnter pid \t -- \t"); 
    scanf("%d",&id); 
    printf("Enter Request for Resources \t -- \t"); 
    for(i=0;i    { 
      scanf("%d",&newr); 
      f[id].all[i] += newr;
      avail[i]=avail[i] - newr; 
          } 
    for(i=0;i    { 
      for(j=0;j      { 
       f[i].need[j] = f[i].max[j] - f[i].all[j];
       if (f[i].need[j] < 0)
       f[i].need[j] = 0;
      }
   }

cnt = 0;
fl = 0;
while (cnt != n)
  {
    g = 0;
    for (j = 0; j < n; j++)
      {
 if (f[j].flag == 0)
   {
     b = 0;
     for (p = 0; p < r; p++)
       {
  if (avail[p] >= f[j].need[p])
    b = b + 1;
  else
    b = b - 1;
       }
     if (b == r)
       {
  printf ("\nP%d is visited", j);
  seq[fl++] = j;
  f[j].flag = 1;
  for (k = 0; k < r; k++)
  avail[k] = avail[k] + f[j].all[k];
  cnt = cnt + 1;
  printf ("(");
  for (k = 0; k < r; k++)
    printf ("%3d", avail[k]);
  printf (")");
  g = 1;
       }
   }
      }
    if (g == 0)
      {
 printf ("\n REQUEST NOT GRANTED -- DEADLOCK OCCURRED");
 printf ("\n SYSTEM IS IN UNSAFE STATE");
 goto y;
      }
  }

printf ("\nSYSTEM IS IN SAFE STATE");
printf ("\nThe Safe Sequence is -- (");
for (i = 0; i < fl; i++)
  printf ("P%d ", seq[i]);
printf (")");
y:printf ("\nProcess\t\tAllocation\t\tMax\t\t\tNeed\n");
for (i = 0; i < n; i++)
  {
    printf ("P%d\t", i);
    for (j = 0; j < r; j++)
      printf ("%6d", f[i].all[j]);
    for (j = 0; j < r; j++)
     
printf ("%6d", f[i].max[j]);
    for (j = 0; j < r; j++)
      printf ("%6d", f[i].need[j]);
    printf ("\n");
  }
}

INPUT 

Enter number of processes – 5

Enter number of resources -- 3 

Enter details for P0 

Enter allocation -- 0 1 0 

Enter Max -- 7 5 3

Enter details for P1 

Enter allocation -- 2 0 0 

Enter Max -- 3 2 2 

Enter details for P2

Enter allocation -- 3 0 2 

Enter Max -- 9 0 2 

Enter details for P3 

Enter allocation -- 2 1 1 

Enter Max -- 2 2 2

Enter details for P4 

Enter allocation -- 0 0 2 

Enter Max -- 4 3 3 

Enter Available Resources -- 3 3 2 

Enter New Request Details -- Enter pid -- 1 

Enter Request for Resources -- 1 0 2 

OUTPUT

P1 is visited( 5 3 2) 

P3 is visited( 7 4 3) 

P4 is visited( 7 4 5) 

P0 is visited( 7 5 5) 

P2 is visited( 10 5 7)

 

SYSTEM IS IN SAFE STATE

The Safe Sequence is -- (P1 P3 P4 P0 P2 ) 

Process       Allocation         Max            Need 

P0              0 1 0                  7 5 3           7 4 3 

P1              3 0 2                  3 2 2           0 2 0 

P2              3 0 2                  9 0 2           6 0 0

P3              2 1 1                  2 2 2           0 1 1

P4              0 0 2                  4 3 3           4 31