Linked List pointer printing memory leak - c

Question

The error is that it prints out the memory instead the values of each node. I tried all combinations of pointers and different printing styles but they all show up as memory leaks.
Here is my code:
#include <stdio.h>//headers
#include <stdlib.h>
struct node{
int val;
struct node *next;
};//linked list
struct node *curr = NULL;//list pointers
struct node *prev = NULL;
struct node *head = NULL;
int main(){
int i;
struct node *curr = (struct node*) malloc(sizeof(struct node));
head=curr;//sets head node
for (i=1;i<=5;i++){
curr->val=i;//sets data
struct node *prev = (struct node*) malloc(sizeof(struct node));
curr->next=prev;
curr=prev;//links to previous node
printf("%d\n", *curr);//prints out data
}
return 0;
}

Answer 1

Well, you have no calls to free in your program, so yes, every call to malloc is technically a leak. That said, I think you are confusing your terms here.
If you want to print out the value of the node, use cur->val, not *cur. *cur returns a struct node, which you are telling printf to interpret as an int.

Answer 2

You should print the value accessing it directly like this
printf("%d\n", curr->val);

Answer 3

Plus you are not setting the value of prev at any point so curr = prev; and then printf("%d\n", curr->val); will just print rubbish.

Answer 4

There are several things wrong apart from the way you access the val field, which as has been commented should be curr->val.
Your program, when thus corrected, prints rubbish because you only print it after pointing curr to prev, and val is uninitialised.
curr->val = i;
...
curr = prev;
printf ("%d\n", curr->val);
Also, you have declared curr and prev both globally and within the function.
I have simplified this, printing the list after it has been created. If you want a different sequence, reverse the i loop. Note I have removed prev.
#include <stdio.h>
#include <stdlib.h>
struct node{
int val;
struct node *next;
};
int main(){
int i;
struct node *curr, *head = NULL;
// build the list
for (i=1; i<=5; i++) {
curr = malloc(sizeof(struct node));
curr->val = i;
curr->next = head;
head = curr;
}
// print the list
curr = head;
while (curr) {
printf ("%d\n", curr->val);
curr = curr->next;
}
return 0;
}
You should check the return value from malloc() before using it.

Answer 5

the following code compiles cleanly
caveat: I have not run this code
#include <stdio.h>//headers
#include <stdlib.h>
struct node{
int val;
struct node *next;
};//linked list
struct node *curr = NULL;//list pointers
struct node *prev = NULL;
struct node *head = NULL;
int main()
{
int i;
for (i=1;i<6;i++)
{
// get block
if( NULL == (curr = malloc(sizeof(struct node)) ) )
{ // then, malloc failed
perror( "malloc failed" );
exit( EXIT_FAILURE );
}
// implied else, malloc successful
// fill in fields
curr->val = i;
curr->next = NULL;
// display val
printf("%d\n", curr->val);//prints out data
if( NULL == head)
{ // then, first node
head = curr;
prev = curr;
}
else
{ // else, after first node
// link in new node
prev->next = curr; // link in new node
} // end if
} // end while
prev = head;
curr = prev;
while( curr )
{ // then, node to free
curr = prev->next;
free(prev);
prev = NULL;
} // end while
return 0;
} // end function: main