Skip to main content

Circular Queue using Array in C Programming

#include< stdio.h>
#include< conio.h>
#define MAX 3
#define NULL -1
void main()
{
 int Q[MAX],i,j,f=NULL,r=NULL,val,ch,item;
 clrscr();
 while(1)
 {
  printf("\n1.Insert \n2.Delete \n3.Display \n4.Exit");
  printf("\n Enter Your choice:");
  scanf("%d",&val);
  switch(val)
  {
   case 1:
    if((f==0 && r==(MAX-1))||(f==r+1))
     printf("\nCircular Queue Overflow.");
    else
    {
     if(r==MAX-1)
      r=0;
     else
      r++;
     printf("\n Enter value for circular queue:");
     scanf("%d",&item);
     Q[r]=item;

     if(f==NULL)
      f=0;

             }
    break;

   case 2:
    if(f==NULL)
     printf("\n circular queue is empty.");
    else
    {
     printf("\n Deleted element is :%d",Q[f]);

     if(f==r)
      f=r=NULL;
     else if(f==(MAX-1))
      f=0;
     else
      f++;
    }
    break;
   case 3:
    if(f==NULL)
     printf("\n Circular queue is empty.");
    else
    {
     if(f<=r)
     {
      for(i=f;i<=r;i++)
         printf("\t  %d",Q[i]);
     }
     if(f>r)
     {
      for(i=f;i<=MAX-1;i++)
       printf("\t %d",Q[i]);
      for(i=0;i<=r;i++)
       printf("\t %d",Q[i]);
     }
    }
    break;
   case 4:
    exit(0);
  }
 }
getch();
}

OUTPUT
1.Insert
2.Delete
3.Display
4.Exit
 Enter Your choice:2

 circular queue is empty.
1.Insert
2.Delete
3.Display
4.Exit
 Enter Your choice:1

 Enter value for circular queue:11

1.Insert
2.Delete
3.Display
4.Exit
 Enter Your choice:1
 
 Enter value for circular queue:22

1.Insert
2.Delete
3.Display
4.Exit
 Enter Your choice:1

 Enter value for circular queue:33

1.Insert
2.Delete
3.Display
4.Exit
 Enter Your choice:3
          11      22      33
1.Insert
2.Delete
3.Display
4.Exit
 Enter Your choice:1

Circular Queue Overflow.
1.Insert
2.Delete
3.Display
4.Exit
 Enter Your choice:2

 Deleted element is :11
1.Insert
2.Delete
3.Display
4.Exit
 Enter Your choice:1

 Enter value for circular queue:44

1.Insert
2.Delete
3.Display
4.Exit
 Enter Your choice:3
         22      33      44


Comments

Popular posts from this blog

MVT (Multiprogramming Variable Task) in C Programming

#include #include void main() { int i,os_m,nPage,total,pg[25]; clrscr(); printf("\nEnter total memory size:"); scanf("%d",&total); printf("\nEnter memory for OS:"); scanf("%d",&os_m); printf("\nEnter no. of pages:"); scanf("%d",&nPage); for(i=0;i =pg[i]) { printf("\n Allocate page %d",i+1); total=total-pg[i]; } else printf("\n page %d is not allocated due to insufficient memory.",i+1); } printf("\n External Fragmentation is:%d",total); getch(); } OUTPUT Enter total memory size:1024 Enter memory for OS:256 Enter no. of pages:4 Enter size of page[1]:128 Enter size of page[2]:512 Enter size of page[3]:64 Enter size of page[4]:512 Allocate page 1 Allocate page 2 Allocate page 3 page 4 is not allocated due to insufficient memory. External Fragmentation is:64

First Come First Serve (FCFS) Page replacement algorithm in C Programming

#include #include int fsize; int frm[15]; void display(); void main() { int pg[100],nPage,i,j,pf=0,top=-1,temp,flag=0; clrscr(); printf("\n Enter frame size:"); scanf("%d",&fsize); printf("\n Enter number of pages:"); scanf("%d",&nPage); for(i=0;i OUTPUT Enter frame size:3 Enter number of pages:12 Enter page[1]:1 Enter page[2]:2 Enter page[3]:3 Enter page[4]:4 Enter page[5]:1 Enter page[6]:2 Enter page[7]:5 Enter page[8]:1 Enter page[9]:2 Enter page[10]:3 Enter page[11]:4 Enter page[12]:5 page | Frame content -------------------------------------- 1 | 1 -1 -1 2 | 1 2 -1 3 | 1 2 3 4 | 4 2 3 1 | 4 1 3 2 | 4 1 2 5 | 5 1 2 1 | 5 1 2 2 | 5 1 2 3 | 5 3 2 4 | 5 3 4 5 | 5 3 4 ---------------------------...

Deadlock Prevention using Banker’s Algorithm in C Programming

#include #include void main() { int allocated[15][15],max[15][15],need[15][15],avail[15],tres[15],work[15],flag[15]; int pno,rno,i,j,prc,count,t,total; count=0; clrscr(); printf("\n Enter number of process:"); scanf("%d",&pno); printf("\n Enter number of resources:"); scanf("%d",&rno); for(i=1;i OUTPUT Enter number of process:5 Enter number of resources:3 Enter total numbers of each resources:10 5 7 Enter Max resources for each process: for process 1:7 5 3 for process 2:3 2 2 for process 3:9 0 2 for process 4:2 2 2 for process 5:4 3 3 Enter allocated resources for each process: for process 1:0 1 0 for process 2:3 0 2 for process 3:3 0 2 for process 4:2 1 1 for process 5:0 0 2 available resources: 2 3 0 Allocated matrix Max need 0 1 0| 7 5 3| 7 4 3 3 0 2| 3 2 2| 0 2 0 3 0 2| 9 0 2| 6 0 0 2 1 1| ...