Print a linked list in reverse order: program not outputting anything-CodePudding

I am trying to print a linked list in reverse order using recursion. i.e. : if i added 1,2,3,4,5 to the linked list , the program should output 5,4,3,2,1. Here's the code:

#include <stdio.h>
#include <stdlib.h>

struct Node{
    int data;
    struct Node* next;
};

void Print(struct Node* n)
{
    if(n->next == NULL)
    {
        return;
    }
    Print(n->next);
    printf("%d",n->data);
    
}

void Insert(struct Node** head_ref,int data)
{
    struct Node* new_node = (struct Node*)malloc(sizeof(struct Node));
    new_node->data = data;
    new_node->next = NULL;
    if((*head_ref)==NULL)
    {
        (*head_ref) = new_node;
    }
    else
    {
        struct Node* temp = (*head_ref);
        while(temp!=NULL)
        {
            temp = temp->next;
        }
        temp->next = new_node;
    }
}

int main()
{
    struct Node* head = NULL;
    Insert(&head,1);
    Insert(&head,2);
    Insert(&head,3);
    Insert(&head,4);
    
    Print(head);
}

I have also tried declaring the head pointer as a global variable but still is is not outputting anything. Any help would be appreciated. Also if you think there something missing in my code or something that i don't understand correctly, if possible please point me to some resources where i can learn the concept.

Thanks

CodePudding user response：

Two mistakes:

The loop condition int the loop that finds the last node is wrong.

It is:

while(temp!=NULL)

while it should be:

while(temp->next!=NULL)

Otherwise, the update at:

while (temp != NULL) { ... }
temp->next = new_node; // segmentation fault

will explode because temp is NULL at that place.

Second error:

When printing the early check if tail is reached should be:

if(n == NULL) // not `if (n->next == NULL)`

Otherwise, the tail will not be printed.

After those fixes the program prints:

If you want to reverse the order just swap printf and recursive Print() calls:

    Print(n->next);
    printf("%d",n->data);