Getting input into a linked list using a loop - c

Question

I made this program to learn about linked lists as I am just starting out with them. The program terminates immediately after the statement "Enter the amount of pesticide"(it is a school assignment). Also, I am unsure how do I keep the length of the list limited to the size input by the user.
#include<stdio.h>
struct plants{
int val;
struct plants *next;
};
void printlist();
int main(){
struct plants* head = NULL;
struct plants* current= head;
head = malloc(sizeof(struct plants));
int counter,size;
printf("Enter the number of plants\n");
scanf("%d",&size);
printf("Enter the amount of pesticide each plant has.\n");
while(current!=NULL){
scanf("%d",current->val);
current= current->next;
}
return 0;
}

Answer 1

#include<stdio.h>
#include<malloc.h>
int main()
{
int count = 0;
int size = 0;
printf("Enter the number of plants\n");
scanf("%d",&size);
printf("Enter the amount of pesticide each plant has.\n");
You have to allocate the memory for each node inside the while loop. If you like to add the new node at the end of the list notice the end of the list by a pointer to the pointer at the end of the list. Apart form this you have to pass the addres of the value to be read to scanf:
struct plants * head = NULL;
struct plants ** current = &head; // current refers there, where next node has to be placed
while( count < size ) // do it for "size" nodes
{
*current = malloc(sizeof(struct plants)); // allocate memory for the node right to target
scanf( "%d", &((*current)->val)); // read the data
(*current)->next = NULL; // node is last node in list, so its successor is NULL
current = &((*current)->next); // step on forward
count ++; // increment number of nodes
}
Note since the type of current is struct plants ** this codes puts the new node to head for the first element of the list and to (*current)->next for all further nodes of the list.
It would be easier to add the new node at the head of the list:
struct plants * head = NULL; // init head with NULL (this becomes end of the list)
while( count < size ) // do it for "size" nodes
{
struct plants * current = malloc(sizeof(struct plants)); // allocate memory for the node
scanf( "%d", &current->val); // read the data
current->next = head; // successor of node is head
head = current; // new node is head of list
count ++; // increment number of nodes
}
This will print your list:
struct plants *temp = head;
while( temp != NULL )
{
printf( "%d ", temp->val );
temp = temp->next;
}
Don't forget to free the list at the end of your program:
while ( head != NULL )
{
struct plants *next = head->next;
free( head );
head = next;
}
return 0;
}

Answer 2

struct plants* head = malloc(sizeof(struct plants));//declare an initialize in memory
struct plants* current= head;
int i = 0;
struct plants* aux = NULL;
while(i ++ < size)
{
aux = malloc(sizeof(struct plants)); // used for new values
scanf("%d", aux->val);
aux->next = NULL;
current->next = aux;
current = aux;
}
Loop while you have less than the quantity required by user. Use an auxiliar node. Read the value, set the current next position to the aux node, the aux node to null and set the current node to aux to be sure you are at the last node from the list.

Answer 3

Lot of things to be corrected, check for the code below. and should be pretty easy for a beginner to understand. I haven't changed the format of your code for ease of understanding
#include<stdio.h>
#include<stdlib.h>
struct plants{
int val;
struct plants *next;
};
void printlist();
int main(){
int i;
struct plants* head = NULL,*temp=NULL;
struct plants* current= head;
int counter,size;
printf("Enter the number of plants\n");
scanf("%d",&size);
printf("Enter the amount of pesticide each plant has.\n");
for(i=0;i<size;i++){
temp = malloc(sizeof(struct plants));
scanf("%d",&temp->val);
temp->next=NULL;
if(head==NULL)
{
head=temp;
current=head;
}
else
{
current->next=temp;
current=temp;
}
}
current = head;
while(current!=NULL){
printf("%d\t",current->val);
current= current->next;
}
}