number multiply by 2 with out using * operator

Posted by prawin
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2. number multiply by 2 with out using * operator

#include<stdio.h>

main()

{

int num = 5;

num = num<<1;

printf("%d",num);

}

Op : 10

Exp : num number left shift by the 1 position we can get.

1^st---------------0101

2^nd--------------1010

Author

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#include
struct node
{
int data;
struct node* next;
};
void addatbeg(struct node,int);
void append(struct node,int);
void display(struct node*);
void delete(struct node**,int num);
void count(struct node);
void insert(struct node,int,int);
void reverse(struct node);
void reversing(struct node);
void mid(struct node);
main()
{
struct node head;
head = NULL;
addatbeg(&head,10);
addatbeg(&head,20);
addatbeg(&head,30);
display(head);
append(&head,60);
append(&head,70);
printf("\n");
display(head);
delete(&head,60);
printf("\n");
printf("after deletion the linked list be a\n");
display(head);
printf("\n");
printf("the number of nodes will be\n");
count(head);
insert(&head,333,20);
printf("\n");
display(head);
reverse(&head);
printf("\n");
printf("this is reverse the nodes\n");
display(head);
reversing(head);
printf("\n");
printf("this is reverse printing\n");
display(head);
printf("\n");
printf("the mid element is\n");
mid(head);

}
void addatbeg(struct node **q,int num)
{
struct node temp;
temp = (struct node)malloc(sizeof(struct node));
temp->data = num;
temp->next = q;
q = temp;
}
void display(struct node *q)
{
while(q!=NULL)
{
printf("%d ",q->data);
q = q->next;
}
}
void append(struct node**q,int num)
{
struct node temp,r;
if(q==NULL)
{
temp = (struct node)malloc(sizeof(struct node));
temp->data = num;
temp->next = q;
q = temp;
}
else
{
temp = q;
while(temp->next!=NULL)
temp = temp->next;
r = (struct node*)malloc(sizeof(struct node));
r->data = num;
r->next = NULL;
temp->next = r;
}
}
void delete(struct node **q,int num)
{
struct node* temp,old;
temp = q;
while(temp!=NULL)
{
if(temp->data == num)
{
if(temp == q)
{
q = NULL;
free(temp);
}
else
{
old->next = temp->next;
free(temp);
}
}
else
{

old = temp;
temp = temp->next;
}

}
}
void count(struct node*q)
{
int c=0;
while(q!=NULL)
{
c++;
q = q->next;
}
printf("the nuber of nodes = %d\n",c);

}
void insert(struct node**q,int num,int n)
{
struct node temp,r;
temp = q;
while(temp!=NULL)
{
if(temp->data == n)
{
r = (struct node)malloc(sizeof(struct node));
r->data = num;
r->next = temp->next;
temp->next = r;
}
temp = temp->next;
}
}
void reverse(struct node**q)
{
struct node temp,prev=NULL,rev;
temp = q;
while(temp!=NULL)
{
rev = temp;
temp = temp->next;
rev->next = prev;
prev = rev;
}
q = rev;
}
void reversing(struct nodeq)
{
if(q==NULL)
{
reversing(q->next);
printf("%d ",q->data);
}
}
void mid(struct node*q)
{
int m=0;
if(q->next==NULL)
{
printf("only one node is there\n");
}
if(q->next->next==NULL)
{
printf(" two nodes is there\n");
}
while( (q->next!=NULL) && (q->next->next!=NULL))
{
m++;
q = q->next;
}
printf("%d\n",m);
}

-------------------------------------------------------------------------------------------------------------------------------------------------

30 20 10
30 20 10 60 70
after deletion the linked list be a
30 20 10 70
the number of nodes will be
the nuber of nodes = 4

30 20 333 10 70
this is reverse the nodes
70 10 333 20 30
this is reverse printing
70 10 333 20 30
the mid element is
3