C - 带指针的函数出现分段错误

Question

I have a list of 6 elements of a struct type:

struct entry
{
    int value;
    struct entry *next;
};

I'm trying to create a function to initialize these six elements by assigning a random int to each value member.

void initializeList(struct entry *l_p)
{
    while(l_p!=_END)
    {
        l_p->value=rand()%999;
        l_p=l_p->next;
    }
}

When this function is called in the main, *l_p points at the very first element of the list. _END is a global constant defined as it follows:

struct entry const *_END=(struct entry *)0;

Now, every time I run my code I get this:

Segmentation fault (core dumped)

Process returned 139 (0x8B)

I know this means I'm trying to access a part of memory I'm not allowed to, but I can't figure out how to fix my code. Also, I'm pretty sure that the problem is caused by initializeList because if I remove it and manually initialize every element of the list, the program runs smoothly.

Answer 1

Sorry everybody, my fault. I've been trying to figure it out for so long and now, after a 5 minutes break, I can see that since my list wasn't initialized and the elements weren't linked to each other, I couldn't run a sequential scan over it. Again, my fault, i'm still a noob, got plenty to learn :)

Answer 2

Actually there's an error in this:

    l_p->value=rand()%999;
    l_p=l_p->next;

You haven't allocated memory for l_p & then tried to assign a value in a restricted area leading to a Segmentation fault.

You may try this

void initializeList(struct entry *l_p)
{
    l_p=(struct entry *)malloc(sizeof(struct entry));
    l_p->next=NULL;
    struct entry *pointer,* back=l_p;
    for(int i=0;i<5;i++)
    {
        pointer=(struct entry *)malloc(sizeof(struct entry));
        pointer->next=NULL;
        back->next=pointer;
    }
    for(int i=0;i<6;i++)
    {
        l_p->value=rand()%999;
        l_p=l_p->next;
    }
}

I've created 6 nodes and connected them in the first 9 lines.

Then assigned values to each of them in the next lines.