Segmentation fault while implementing linked lists in C

general
, ,
1

i am getting a segmentation fault error while running the following code to implement linked list and the display them. your help would be appreciated.

#include<stdio.h>
#include<stdlib.h>

typedef struct sl
{
    int num;
    struct sl * next;
}node;

void ins(node *, int);
void printll(node *);

int main(void)
{
    node * start;
    start=NULL;
    int n;
    char ch;
    do
    {
        printf("enter an integer \n");
        scanf(" %i",&n);
        ins(start,n);
        printf("do you want to insert more elements \n");
        scanf(" %c",&ch);
    }while(ch=='y');

    //print the linked list

    printll(start);
}

void ins(node *start, int n)
{
    if(start==NULL)
    {
        start=malloc(sizeof(node));
        start->num=n;
        start->next=NULL;
    }

    else
    {
        node * np;
        np=malloc(sizeof(node));
        np->num=n;
        np->next=start;
        start=np;
    }

}

void printll(node * start)
{
    node * nt=NULL;
    nt=start;
    printf("[[");
    while(nt->next!=NULL)
    {
        printf("%i>>",nt->num);
        nt=nt->next;
    }
    printf("]]");
}
2

You call function ‘ins’ with argument ‘int* start’.
This send the pointer down into the function.
In the function, if you change the value contained in the pointer, for example by
*start = something
the calling function will see the new value.

You change the pointer itself in the function. This changed pointer is not passed back up to the calling function when the argument is like this.

To see the changed pointer at a higher level, you need to call ‘ins’ with a pointer to a pointer
‘int** start’. The first few lines of ‘ins’ then look like this

void ins(node **start, int n)
{
	if (*start == NULL)
	{
		*start = malloc(sizeof(node));

with a call from ‘main’

	ins(&start, n);

I have not looked at whether your code now does what you want.

3

My answer did not come out quite right in the browser, although it looked right in the preview window. You should be able to understand what I meant to write.