Linked Lists in C, Random output? - c

Question

#include <stdio.h>
struct list
{
int data;
struct list *next;
};
struct list *start, *end;
void add(struct list *head, struct list *list, int data);
void delete(struct list *head, struct list *tail);
int main(void)
{
start=end=NULL;
add(start, end, NULL);
add(start, end, NULL);
printf("First element: %d");
delete(start, end);
return 0;
}
void add(struct list *head, struct list *tail, int data)
{
if(tail==NULL)
{
head=tail=malloc(sizeof(struct list));
head->data=data; head->next=NULL;
} else {
tail->next=malloc(sizeof(struct list));
tail=tail->next;
tail->data=data;
tail->next=NULL;
}
}
void delete(struct list *head, struct list *tail)
{
struct list *temp;
if(head==tail)
{
free(head);
head=tail=NULL;
} else {
temp=head->next;
free(head);
head=temp;
}
}
I am aiming to return an output of 3 but keep getting random results. Any insight is greatly appreciated

Answer 1

As you want to modify head and tail you need to pass a pointer to them.
i.e.
void add(struct list **head, struct list **tail, int data)
{
if(*tail==NULL)
{
*head = *tail = malloc(sizeof(struct list));
(*head)->data = data;
(*head)->next = NULL;
} else {
(*tail)->next = malloc(sizeof(struct list));
*tail = (*tail)->next;
(*tail)->data = data;
(*tail)->next = NULL;
}
}
Do similarly for the other function. Then head and tail will also change outside the function.
The line
printf("First element: %d");
It requires an integer to print - supply it - see the manual page for printf
Avoid using keywords for C++ in C programs - such as delete

Answer 2

Parameters of C functions are passed by value. Thus, the changes that you make to the values of head and tail in the add() function will not be reflected in main() from which add() is called.
And, it doesn't seem that you've supplied a second parameter for printf(), so the %d format is not going to get the integer value that it will be looking for.

Answer 3

In your program start and end are global variables, so there is no need to pass it as arguments to other functions, because other methods can directly access it.
//Structure to store data
struct list
{
int data;
struct list *next;
};
// global variables
struct list *start, *end;
void add(int);
void delete();
// start of program
int main(void)
{
start=end=NULL;
add(5);
add(6);
printf("\nFirst element: %d",start->data);
delete();
printf("\nFirst element: %d",start->data);
return 0;
}
//add node to list
void add(int data)
{
if(end==NULL)
{
start=end=malloc(sizeof(struct list));
start->data=data; start->next=NULL;
} else {
end->next=malloc(sizeof(struct list));
end=end->next;
end->data=data;
end->next=NULL;
}
}
// delete node from list
void delete()
{
struct list *temp;
if(start==end)
{
free(start);
start=end=NULL;
} else {
temp=start->next;
free(start);
start=temp;
}
}
OUTPUT:
First element: 5
First element: 6
Note:
If you don't want your start and end to be global than it can be a local variable of main function. Here you have to either work on double pointer mechanism or return memory address to retain the modification.

Answer 4

There are a number of problems here.
First, when you pass a pointer to a function, the pointer is passed by value. Any changes to the pointer in the called function will not be reflected in the calling function. To change what the pointer in the calling function is pointing to, you need to pass a pointer-to-pointer. So your add()function needs to be:
void add(struct list **head, struct list **tail, int data) {
if(*tail == NULL) {
*head = *tail = malloc(sizeof(struct list));
(*head)->data = data;
(*head)->next = NULL;
}
else {
(*tail)->next = malloc(sizeof(struct list));
(*tail) = (*tail)->next;
(*tail)->data = data;
(*tail)->next = NULL;
}
return;
}
And your delete() function needs to be changed similarly.
Secondly, you are passing NULL as your data value to the add function. NULL is a pointer; it is typically defined as a macro, and could be (void*) 0 in the implementation. It should not be used as an integer. Pass 0 as the integer, not NULL.
Third, you have this statement:
printf("First element: %d");
Your format string has the conversion specifier %d but there is no argument that matches the %d. This is undefined behavior. What integer exactly are you trying to print?
Where do you expect the output 3, and for what reason?