Answer / ruchi

#include<stdio.h>
#include<conio.h>
void insert();
void del();
void trav();
int f=-1,r=-1,a[20],n=5;
int main()
{
int ch;
char ch1='y';
clrscr();
while(ch1=='y')
{
printf("\nMEnu ");
printf("\n1.Insertion ");
printf("\n2.Deletion ");
printf("\n3.Traversal ");
printf("\n4.Exit ");
printf("\nEnter your chioce ");
scanf("%d",&ch);
switch(ch)
{
case 1:
insert();
break;
case 2:
del();
break;
case 3:
trav();
break;
case 4:
exit(0);
}
printf("\nDo you want to repeat ");
ch1=getche();
}
}
void insert()
{
int item;
if(r<n-1)
{
printf("\nEnter the element ");
scanf("%d",&item);
if(f==-1)
{
f=0;
r=0;
}
else

r=r+1;
a[r]=item;
}
}
void del()
{
if((f>=0)&&(f<=r))
{
f=f+1;
printf("\nItem is deleted ");
}
else
{
printf("\nUnderflow ");
}
}
void trav()
{
int i;
if((f==-1)||(f>r))
{
printf("\nQueue is empty ");
}
for(i=f;i<=r;i++)
{
printf("%d\n",a[i]);
}
}

write the output of following code .. main() { static int a[]={10,20,30,40,50}; int *ptr=a; static int arr[2][2]={1,2,3,4}; char str[]="ABCD * 4#"; char *s=str+2; int i,j; for(i=0;i<5,i++) printf("%d",*ptr++); for(i=0;i<2;i++) for(j=0;j<2;j++) printf("%d\n",*(*(n+i)+j)); printf("%c\n%c\n%c\n",*(str+2),*s++,*--s); }

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