creating an array of structures and making them as linked lists and traversing them using c

Question

In this program I created a array(node[3]) of structure(struct eg) and made them as linked lists while trying to print the elements of the list I am getting only 1 output that is 3

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

struct eg
{
    struct eg *next;
    int age;
}node[3];

main()
{
    struct eg *head,*temp;
    int i;

    head=temp=node;
    for(i=1;i<3;i++)
    {
       temp->age=i;
       temp->next=temp+i;
    }
    temp->next=0;
    temp->age=3;
    while(head!=0)
    {
       printf("%d",head->age);
       head=head->next;
    } 
}

Please have a look at How to create a Minimal, Complete, and Verifiable example, and review your post. — Hopeful Llama
– Hopeful Llama, Commented Sep 8, 2016 at 16:56
@HopefulLlama I think the code in the question is pretty much MCVE. — Eugene Sh.
– Eugene Sh., Commented Sep 8, 2016 at 16:58

dbush · Accepted Answer · 2016-09-08 16:56:36Z

2

temp->next=temp++;

You're reading and modifying temp in a single expression with no sequence point. This invokes undefined behavior.

You need to separate the increment from the assignment:

temp->next=temp+1;
temp++;

answered Sep 8, 2016 at 16:56

dbush

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1 Comment

Riley Over a year ago

You could even move the temp++ into the increment part of the for to make it clear that it is incremented for each loop (i.e. for(i=1;i<3;i++,temp++))

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