CODEWITHSUNDEEP
csegmentation-fault
vivek_jonam
vivek_jonam

Reputation: 3297

Can I know what causes Segmentation Fault here?

I think I have implemented the pointer correctly in below code. But it causes Segmentation fault. Can someone explain why?

struct list
{
    int index;
    struct list *next;
};
void add(struct list *l,int index)
{
    struct list *temp=l;
    if(l==NULL)
    {
        temp=(struct list *)malloc(sizeof(struct list));
        temp->index=index;
        temp->next=NULL;
        l=temp;
    }
    else
    {
        while(temp->next!=NULL)
        temp=temp->next;
        struct list *nnode=(struct list *)malloc(sizeof(struct list));
        nnode->index=index;
        nnode->next=NULL;
        temp->next=nnode;
    }
}

main()
{
        struct list *l;
        l=NULL;
        int el;
        scanf("%d",&el);
        add(l,el);
        while(l->next!=NULL)    //It causes seg fault
        {
            printf(" %d ",l->index);
            l=l->next;
        }
}

Upvotes: 0

Views: 85

Answers (2)

AndersK
AndersK

Reputation: 36082

This code doesn't do what you think it does:

void add(struct list *l,int index)
{
    struct list *temp=l;
    if(l==NULL)
    {
        temp=(struct list *)malloc(sizeof(struct list));
        temp->index=index;
        temp->next=NULL;
        l=temp;
    }
...

the parameter l will not change since you are just passing it as a pointer, in order to change what l points to you need to pass the address of l

void add(struct list **l,int index)
{
    struct list *temp;
    if(*l==NULL)
    {
        temp=(struct list *)malloc(sizeof(struct list));
        temp->index=index;
        temp->next=NULL;
        *l=temp;
    }
...

otherwise the change will not reach outside the function scope.

Upvotes: 3

iabdalkader
iabdalkader

Reputation: 17312

This code shouldn't segfault if you just call add once, but I assume you call other functions and pass the list which is still NULL you should send a pointer to pointer to the list instead

void add(struct list **l,int index)
{
    if(*l==NULL) {
        *l=(struct list *) malloc(sizeof(struct list));
    }
}

Otherwise it will still be NULL when add returns, because you only initialize a copy of the pointer, or if this confuses you could just initialize the list in main() and avoid initialization in add()

Upvotes: 2

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