Question

This is a debugging question. Please write comments where it is fixed. Thanks.

q7.4 Fix the errors in the code (in C)

//This program is supposed to scan 5 ints from the user

//Using those 5 ints, it should construct a linked list of 5 elements

//Then it prints the elements of the list using the PrintList function

#include <stdio.h>

struct Node{

int data;

Node* next;

};

int main(void){

struct Node first = {0, 0};

struct Node* second = {0, 0};

Node third = {0, 0};

struct Node fourth = {0, 0};

struct Node fifth = {0, &first};

int i;

scanf(" %d", &i);

first.data = i;

scanf(" %d", &i);

second.data = i

first.next = &second;

scanf(" %d", &i);

third.data = i;

second.next = third;

scanf(" %d", &i);

data = i;

third.next = &fourth;

scanf(" %d", &i);

fifth.data = i;

fourth->next = &fifth;

PrintList(first);

}

PrintList(struct Node* n){

while(n != 0){

printf("%d ", n.data);

n = n.next;

}

printf("\n");

}

Answer 1

Fixed all the errors, the code is working fine now.

//This program is supposed to scan 5 ints from the user

//Using those 5 ints, it should construct a linked list of 5 elements

//Then it prints the elements of the list using the PrintList function

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

struct Node{

int data;

struct Node* next; // fixed the error here by adding 'struct'

};

// Moved PrintList upwards to prevent any conflicts/warnings
void PrintList(struct Node* n){

while(n != NULL){

printf("%d ", n->data); // Changed n.data to n->data

n = n->next; //Changed n.next to n->next

}

printf("\n");

}

int main(void){

// Initialised the nodes to avoid incompatible pointer type error
struct Node* first;
first = (struct Node *) malloc( sizeof(struct Node) );
struct Node* second;
second = (struct Node *) malloc( sizeof(struct Node) );
struct Node* third;
third = (struct Node *) malloc( sizeof(struct Node) );
struct Node* fourth;
fourth = (struct Node *) malloc( sizeof(struct Node) );
struct Node* fifth;
fifth = (struct Node *) malloc( sizeof(struct Node) );

int i=0;
// Syntax node->data = i is used because node is a pointer here
// Syntax node->next = nextnode is used to assign the next node in the linked list
scanf("%d", &i);

first->data = i;

scanf("%d", &i);

second->data = i;

first->next = second;

scanf(" %d", &i);

third->data = i;

second->next = third;

scanf(" %d", &i);

fourth->data = i;

third->next = fourth;

scanf(" %d", &i);

fifth->data = i;

fourth->next = fifth;

PrintList(first);

}