pointer malloc fail

Question

I tried to define a node struct which includes a node* next. I write an append(node* n） function to add a new node next to the previous one， but every time I run the code, it gives me seg fault. My code is following:

#include<stdlib.h> #include<stdio.h> typedef struct _log_t { struct _log_t* next; }log_t; void initi(log_t* l) { l = (log_t*)malloc(sizeof(log_t)); l -> next = NULL; } void append(log_t* l){ l->next = (log_t*)malloc(sizeof(log_t)); l->next->next = NULL; l = l->next; }

Thanks in advance for any help!

Yann Ramin · Accepted Answer · 2012-02-21 23:48:47Z

 l = l->next;

That line is not doing what you think it is - in fact, its doing nothing.

Perhaps you want to pass log_t* as a log_t**, or return the new log_t*.

paulsm4 · Accepted Answer · 2012-02-21 23:55:11Z

You're dereference a pointer you never initialized. That's why it's crashing :)

// OK (but you should check for malloc() failing, too!) void initi(log_t* l) { l = (log_t*)malloc(sizeof(log_t)); l -> next = NULL; } void append(log_t* l){ // OK... l->next = (log_t*)malloc(sizeof(log_t)); // BAD!!!! l->next: allocated. l->next->next: *NOT* initialized! l->next->next = NULL; l = l->next; }

Here's what I think you possibly mean:

log_t * append (log_t* l) { // Initialize the head if (l == NULL) { l = (log_t *)malloc(sizeof (log_t)); l->next = NULL; } // Initialize a sub-node else { l->next = (log_t *)malloc(sizeof (log_t)); l->next->next = NULL; } // Always return the head return l; }

hi is just doing the initalization... perfectly fine. there is no dereferencing.
that update is even worse.. append(NULL).. that's just doesn't make any sense.
@yi_H: the point is you can't dereference "l->next->next" until you've both 1) allocated "l", and 2) initialized "l->next". It's perfectly appropriate to initialize l->next either to NULL, or to a newly allocated value. It's NOT OK to use it BEFORE you've initialized it.
you should take a rest, then re-read this thing, because you're talking nonsense. he allocated the memory in the previous line.

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