Thread: Link list function call

hello im having some trouble in my function call here.. what i am trying to accomplish is to build a linked list in a function and send the link back to main. Can anyone sort out my problem?(i think it might be syntax with how i am calling the function)

Code:

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

struct info_node
{
    char fname[20] ;
    char lname[20] ;
    int rank ;
    float score ;
    struct info_node * next ;
} ;

int build_list(struct info_node*, int)

int main()
{
	FILE* infile ;
	FILE* infile2 ;
	struct info_node* link1, * link2 ;
 	struct info_node* top, * top2 ;	
	struct info_node* move_pntr = NULL, * move_pntr2 = NULL ;
	struct info_node* pre_pntr ;
	struct info_node* current_pntr ;
	int num = 0 ;
	

    infile2 = fopen("p3data.txt", "r") ;
	infile = fopen("p3purge_data.txt", "r") ;

	build_list(&link1, &top) ;




fclose(infile);
fclose(infile2);
return (0) ;

} //end of func main()
int build_list(info_node * lk1, int tp)
{
	int pre_pntr = NULL ;
	int current_pntr = NULL ;

	while( !feof(infile) )
    {

		lk1 = (info_node *)malloc(sizeof(info_node) ) ;
		fscanf( infile, "%s%s%d%f", lk1->fname , lk1->lname, &lk1->rank, &lk1->score ) ;

		lk1->next = NULL ;
		pre_pntr = lk1 ;
		current_pntr = lk1 ;
		tp = lk1 ;

		lk1->next = tp ;
		tp = lk1 ;
	}
	return(*tp) ;
}

"%s %s %d %f"

kk i think i get what you mean there but lets say i declare the FILE* infile inside of my function would that still be a problem?

What happens in the function, stays in the function. Namely the value of lk1 can never change (what it points to can, but not the variable itself). That is to say, the malloc into lk1 is not going to be seen in the main function. Either make the first parameter a **, or don't pass it in at all but return a pointer to the head node.

Also, it would appear that you don't make a linked list, so much as a bunch of disconnected nodes that are linked to themselves (i.e., every time through the list you set lk1->next equal to lk1, instead of waiting to connect it to the next one).

so basically send a pointer to malloc from main into my function and work from there? Will the infile scan work inside of the function if i do that?

You're sending a pointer. You need to send a pointer to a pointer (you can only change what is pointed to, and you need to change a pointer, so you need to point to a pointer).

im still having trouble understanding how to get this function to work properly, so instead of passing a pointer to a pointer i can remove the struct info node pointer, build the list and return a pointer to the head, or top in this case

That ought to work. Either/or.

right. i get the idea but how am i supposed to build the list if i dont have a pointer to the structure?

You are allowed to declare variables inside a function.

so basically just shove this into my function and that works?

Code:

struct info_node
{
    char fname[20] ;
    char lname[20] ;
    int rank ;
    float score ;
    struct info_node * next ;
} ;

Sometimes its far easier to write your own code than it is to copy and paste someone else's then alter it to your specifications.

well so far this is the addition ive made im just using test code to see if i can get it to print out before i implement my official list. Can someone give me a hint as to how i can return the value of the node to main

Code:

int build_list(struct info_node* pnode)
{

	FILE* infile ;
	struct info_node* link1 ;
	struct info_node* top ;
	int result ;


		link1 = (info_node *)malloc(sizeof(info_node) ) ;
		fscanf( infile, "%s%s%d%f", link1->fname , link1->lname, &link1->rank, &link1->score ) ;

		top = link1 ;
	
	return(top) ;
}

I would do it in much the same way as you have it there, except for the "reading from a file you didn't open" part.