Banker's algorithm implementation

/*implementation of Banker's Algorithm

#include<stdio.h>
#include<conio.h>
struct process
{
int all[6],max[6],need[6],finished,request[6];
}p[10];
int avail[6],sseq[10],ss=0,check1=0,check2=0,n,pid,work[6];
int nor,nori;

void main()
{
int safeseq(void);
int ch,i=0,j=0,k,pid,ch1;
int violationcheck=0,waitcheck=0;

do
{
clrscr();
printf(“\n\n\t 1. Input”);
printf(“\n\n\t 2. New Request”);
printf(“\n\n\t 3. Safe State or Not”);
printf(“\n\n\t 4. print”);
printf(“\n\n\t 5. Exit”);
printf(“\n\n\t Enter ur choice : “);
scanf(“%d”,&ch);
switch(ch)
{
case 1:
printf(“\n\n\t Enter number of processes : “);
scanf(“%d”,&n);
printf(“\n\n\t Enter the Number of Resources : “);
scanf(“%d”,&nor);
printf(“\n\n\t Enter the Available Resouces : “);

for(j=0;j<n;j++)
{
for(k=0;k<nor;k++)
{
if(j==0)
{
printf(“\n\n\t For Resource type %d : “,k);
scanf(“%d”,&avail[k]);
}
p[j].max[k]=0;
p[j].all[k]=0;
p[j].need[k]=0;
p[j].finished=0;
p[j].request[k]=0;
}
}
for(i=0;i<n;i++)
{
printf(“\n\n\t Enter Max and Allocated resources for P%d : “,i);       for(j=0;j<nor;j++)
{
printf(“\n\n\t Enter the Max of resource %d : “,j);
scanf(“%d”,&p[i].max[j]);
printf(“\n\n\t Allocation of resource %d    : “,j);
scanf(“%d”,&p[i].all[j]);
if(p[i].all[j]>p[i].max[j])
{
printf(“\n\n\t Allocation should be less < or == max”);
j–;
}
else
p[i].need[j]=p[i].max[j]-p[i].all[j];
avail[j]=avail[j]-p[i].all[j];
}
}
break;
case 2:
violationcheck=0;
waitcheck=0;
printf(“\n\n\t Requesting process id : “);
scanf(“%d”,&pid);
for(j=0;j<nor;j++)
{
printf(“\n\n\t Number of Request for resource %d : “,j);                scanf(“%d”,&p[pid].request[j]);
if(p[pid].request[j]>p[pid].need[j])
violationcheck=1;
if(p[pid].request[j]>avail[j])
waitcheck=1;
}
if (violationcheck==1)
printf(“\n\n\t The Process Exceeds it’s Max Need: Terminated”);
else if(waitcheck==1)
printf(“\n\n\t Lack of Resourcess : Process State – Wait”);
else
{
for(j=0;j<nor;j++)
{
avail[j]=avail[j]-p[pid].request[j];
p[pid].all[j]=p[pid].all[j]+p[pid].request[j];
p[pid].need[j]=p[pid].need[j]-p[pid].request[j];
}
ch1=safeseq();
if(ch1==0)
{
printf(“\n\n\t Granting leads to Unsafe state : “);                                                                                                                     printf(“\n\n\t Request Denied “);
for(j=0;j<nor;j++)
{
avail[j]=avail[j]+p[pid].request[j];
p[pid].all[j]=p[pid].all[j]-p[pid].request[j];
p[pid].need[j]=p[pid].need[j]+p[pid].request[j];
}
}
else if(ch1==1)
printf(“\n\n\t Request Committed “);
}
break;
case 3:
if(safeseq()==1)
printf(“\n\n\t The System is in safe state “);
else
printf(“\n\n\t The System is Not in safe state “);
break;
case 4:
printf(“\n\n\t Number of processes : %d”,n);
printf(“\n\n\t Number of Resources : %d”,nor);
printf(“\n\n\t Pid \t   Max \t   Allocated \t Need “);
for(i=0;i<n;i++)
{
printf(“\n\n\t  P%d : “,i);
for(j=0;j<nor;j++)
printf(” %d “,p[i].max[j]);
printf(“\t”);
for(j=0;j<nor;j++)
printf(” %d “,p[i].all[j]);
printf(“\t”);
for(j=0;j<nor;j++)
printf(” %d “,p[i].need[j]);
}
printf(“\n\n\t Available : “);
for(i=0;i<nor;i++)
printf(” %d “,avail[i]);
break;
case 5:
break;
}
getch();
}while(ch!=5);
}
int safeseq()
{
int i,j,k;
ss=0;
for(j=0;j<nor;j++)
work[j]=avail[j];
for(j=0;j<n;j++)
p[j].finished=0;
for(int tk=0;tk<nor;tk++)
{
for(j=0;j<n;j++)
{
if(p[j].finished==0)
{
check1=0;
for(k=0;k<nor;k++)
if(p[j].need[k]<=work[k])
check1++;
if(check1==nor)
{
for(k=0;k<nor;k++)
{
work[k]=work[k]+p[j].all[k];
p[j].finished=1;
}
sseq[ss]=j;
ss++;
}
}
}
}
check2=0;
for(i=0;i<n;i++)
if(p[i].finished==1)
check2++;
printf(“\n\n\t”);
if(check2>=n)
{
printf(“\n\n\t The system is in safe state”);
for(int tj=0;tj<n;tj++)
printf(“P%d, “,sseq[tj]);
return 1;
}
else
printf(“\n\n\t The system is Not in safe state”);
return 0;
}