Explain how this code is working

#1

sir i am not able to understand the working of this code please help

#2

hello @we_kaash
pls share the code that u want me to explain .

#3

Screenshot (130)
Screenshot (130)1920×1080 251 KB

actually sir last time i was not able to drag the image so i posted it empty and thank you for responding to this post this is the code in the image

#4

@we_kaash

its just reversing the doubly linked list.
focus on these lines->

  while(current!=NULL){
        temp=current->prev;
        current->prev=current->next;
        current->next=temp;
        current=current->prev;
    }

here they are swapping previous node with next node.
......
temp=current->prev;//storing previous node address in temp
current->prev=current->next //putting next node in previous
current->next=temp;// since temp has address of previous node, this line will put previous node in next.
current=current->prev ; //going to next node, since address of next node is now in prev , we are doing current->prev

they are just repeating the same on evry node and assining the lead node as head,as a result u will get reversed linked list.

for clarity run this code ->

#5

sir what does this line exactly means :
node* current = *head_ref ;
what is the compiler actually doing while compiling this statement of code?

#6

in function they have sent pointer of pointer so to get the head node we need to dereference the pointer they have sent in function.

so in this line
node *current=*head_ref ; // we are putting address of head node to current pointer.

look at the code that i have shared that will avoid such confusion, as i have used familar syntaxes

1 Like
#7

yes sir i saw the code

it means that
node* current = head;
is same as
node current =head;?

#8

no they are not same ,

infact this is not possible at all. how u can store head (a pointer) to current (an object).

#9

sir i dont know today my pc is not working properly so typing errors are occuring i am sorry for that, i was asking that
node* current =head ;
is same as
node* current = *head; ?

#10

no bro. . . .they are not, first look at the argumetnts of the function.

in quiz they are giving pointer of a pointer of list head (**head)
whereas in the code(that i shared) has a pointer of list head(*head).

they are differnet right?
in first case ,to get the head pointer u need to derefernce the given pointer.
whereas in second pointer u already have head pointer so no needof dereference

1 Like
#11

ok sir, i got it
thanks for explaining this so smoothly : )

#12

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