Using a for loop to create a linked list
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This might help..
#include<stdio.h>
#include<stdlib.h>
// Prototypes
void InitList(struct list *sList);
void push(struct list *sList, int data);
void pop(struct list *sList);
void print(struct list *sList);
/* Node Structure */
struct node {
int data;
struct node *next;
};
/* List Structure */
struct list {
struct node *start;
};
int main(int argc, char** argv)
{
int x;
struct list MyList;
InitList(&MyList);
for(x = 0; x < 100; x++) push(&MyList, x);
print(&MyList);
printf("\n");
for(x = 0; x < 25; x++) pop(&MyList);
print(&MyList);
printf("\n");
for(x = 0; x < 80; x++) pop(&MyList);
print(&MyList);
printf("\n");
return 0;
}
/* Initializes the list structure */
void InitList(struct list *sList)
{
sList->start = NULL;
}
/* Adds a value to the front of the list */
void push(struct list *sList, int data)
{
struct node *p;
p = malloc(sizeof(struct node));
p->data = data;
p->next = sList->start;
sList->start = p;
}
/* Prints the list */
void print(struct list *sList)
{
struct node *p = sList->start;
while(p != NULL) {
printf("%d ", p->data);
p = p->next;
}
}
/* Removes the first value of the list */
void pop(struct list *sList)
{
if(sList->start != NULL) {
struct node *p = sList->start;
sList->start = sList->start->next;
free(p);
}
}
Author by
user2917692
Updated on June 04, 2022Comments
-
user2917692 almost 2 years
Newb C programmer here, assuming I have a struct for a node as follows
struct node{ int data; struct node *next; };
How would I use a loop to make a linked list where the first node's data is 0, and a pointer to the next node whose data is 1. etc.
EDIT:
int main(int argc, char* argv[]){ struct node a; a.data = 0; struct node * tempnode = &a; for (int i = 1; i < 5; i++){ struct node * next; next->data = i; tempnode->next = next; tempnode = next; }
Heres what i tried but it doesn't work