How to make a singly linked list based on inputs - c

Question

I'm just learning linked lists, and want to make one with inputs from the console, but somehow it's not working.
#include <stdio.h>
#include <stdlib.h>
typedef struct datatype Node;
struct datatype {
int data;
Node *next;
};
int main() {
int i,n;
Node *node[50];
printf("Number of intergers: ");
scanf("%d",&n);
for(i=0; i<n; i++) {
printf("Enter interger one by one: ");
scanf("%d",&node[i]->data);
node[i]->next=node[i+1];
}
for(i=0;i<n;i++)
printf("%d->",node[i]->data);
}

Answer 1

Building on Andreas's comment, here is a commented and memory safe implementation. You don't necessarily have to use malloc and free; you could instead have a fixed memory pool:
#define MAX_NODE_SIZE 256
...
Node node[MAX_NODE_SIZE];
followed by a check to make sure the user never puts n greater than MAX_NODE_SIZE. However if you tried to implement list inserts and deletes things would get complicated. Also, using an index "i" to iterate through a linked list kind of defeats the point of having a linked list.
Here is the malloc/free version:
#include <stdio.h>
#include <stdlib.h>
typedef struct datatype Node;
struct datatype {
int data;
Node *next;
};
int main() {
//Allocate the root node
Node *rootnode=malloc(sizeof(Node));
rootnode->data=0; rootnode->next=NULL;
Node *currentnode=rootnode;
//Populate n and check for validity
printf("Number of integers: ");
int n;
scanf("%d",&n);
if(n<=0){
printf("List must have a nonzero/nonnegative number of elements.\n");
return 1;
}
//Populate the list
for(int i=0; i<n; i++) {
printf("Enter integer one by one: ");
scanf("%d",&currentnode->data);
//If there is more data...
if(i!=n-1){
//Allocate room for the data
currentnode->next=malloc(sizeof(Node));
//Initialize things correctly
currentnode->next->data=0;
currentnode->next->next=NULL;
//Step to the next node.
currentnode=currentnode->next;
}
}
//Print the list
currentnode=rootnode;
do {
printf("%d",currentnode->data);
if(currentnode->next!=NULL)
printf(" -> ");
else
printf("\n");
} while((currentnode=currentnode->next)!=NULL);
//Deallocate the list
currentnode=rootnode;
while(currentnode!=NULL){
Node *next=currentnode->next;
free(currentnode);
currentnode=next;
}
return 0;
}
Example output:
user#desktop:~$ ./a.out
Number of integers: 3
Enter integer one by one: 1
Enter integer one by one: 2
Enter integer one by one: 3
1 -> 2 -> 3

Answer 2

You must allocate memory for the individual nodes of the linked list. Instead, you are allocating memory for 50 pointers to individual nodes, but not for the nodes themselves.
I suggest you create a standard linked list and use the function malloc for the individual nodes, like this:
typedef struct Node {
int data;
struct Node *next;
} Node;
int main( void )
{
//this pointer always points to the first element, or NULL if there is no first element
Node *pRoot = NULL;
//this pointer always points to the NULL pointer at the end of the list, which is, when the list is empty, the root pointer
Node **ppNext = &pRoot;
Node *pCurrent;
int retval, n;
//ask user for total number of data elements
printf( "Number of integers: " );
retval = scanf( "%d", &n );
if ( retval != 1)
{
fprintf( stderr, "scanf failed!\n" );
goto cleanup;
}
//build the list from user input
for ( int i = 0; i < n; i++ )
{
//allocate memory for new node
pCurrent = malloc( sizeof( Node ) );
if ( pCurrent == NULL )
{
fprintf( stderr, "malloc failed!\n" );
goto cleanup;
}
//ask user for individual data elements
printf( "Enter integer one by one: " );
retval = scanf( "%d", &pCurrent->data );
if ( retval != 1 )
{
fprintf( stderr, "scanf failed!\n" );
free( pCurrent );
goto cleanup;
}
pCurrent->next = NULL;
//link new node to linked list and update ppNext
*ppNext = pCurrent;
ppNext = &pCurrent->next;
}
//print the list
for ( pCurrent = pRoot; pCurrent != NULL; pCurrent = pCurrent->next )
{
printf( "%d\n", pCurrent->data );
}
cleanup:
//free the linked list
for ( pCurrent = pRoot; pCurrent != NULL; )
{
Node *tmp = pCurrent;
pCurrent = pCurrent->next;
free( tmp );
}
return 0;
}
Please note that in my code, before I use the value written to by scanf, I check the return value of scanf. This is necessary because the function may fail and not write any value, for example when the user enters letters instead of numbers. See this page for further information:
A beginner's guide away from scanf()