Testbook-ds

Question

Assume that p & q are pointers. What will be the output after performing following sets of operations on a given linked list?

Struct node {

char info;

struct node *link;

};

Operations are:

q = p → link → link ;

q → link → link = p → link;

printf (“%c”,  q → link → link → link → info);

---

the given ans is b but I am getting c.. can anyone plz verify?

Comments

answer is b, because in the second operation the second last node (d) is made to point to the head and then in the last operation the element to print is (c->d->a->b).

---

I think that the second last node points to the second node i.e b because it says p->link which means b .. @Jaideep Bankoti can u plz elaborate

Answer 1

q = p → link → link 

means q is pointing to node $b$

q → link → link = p → link

means node $c$ is now pointing to node $a$

q → link → link → link → info

| b | link | → | c | link | →| a | info | → $b$

$so$ $answer$ $ is $ $b$

Comments

Thanks @kumar.dilip for notifying my blunder(but node with data item e will detach)! and we all are wrong except @ anjali007.

The answer is indeed 'c'. The illustration and the simple code snippet will clarify the doubts.

*note that you can change the printf statement and modify output (also the list will loop a,b,c,d,b,c...).☻

https://onlinegdb.com/SkWDGZskE

---

thanks @Jaideep Bankoti

---

Jaideep Bankoti

i don't think so node e will be disconnected. 

q = p → link → link ; ( q points to node c )

q → link → link  ( it will node e because already q is pointing to node c  q-> link  point to node d  and q -> link the it will point to node e) =  p → link ( here node b )

then e pointing to node b.

q → link → link → link → info

q → link( node d) → link   (node e)→ link  ( node b) → info (node b data)

The answer will be b.