scanf to struct pointer

This is the framework of my program (without the code of functions). It is meant to build a binary tree. My question i about the scanf with pointer to struct. In current code, though it is compiled, it is crushed in the scanf stage when debug it.

where is my mistake?

Code:

struct tree_el {
   int val;
   struct tree_el * right, * left;
};

typedef struct tree_el node;

void main() {
   node *curr=NULL, *root= NULL;
 
   printf("please enter a number\n");
   scanf ("%d", curr->val);

   while (curr->val!=0)
   {
      curr = (node *)malloc(sizeof(node));
      curr->left = curr->right = NULL;
      printf("please enter next number\n");
      scanf ("%d", curr->val);
      insert(&root, curr);
  }

TIA.

Code:

 scanf ("%d", &curr->val);

scanf() functions require an address to write to.

> void main() {
This is your first mistake

> scanf ("%d", curr->val);
This is your second - you haven't malloc'ed the space yet

Gotta wonder how many compiler errors this generated.

Ah yes, everything Salem said are of more importance since your code has some fundamental errors that should prevent it from really doing anything other than seg faulting. I just saw scanf() in the title and CTRL+F'ed scanf() with the standard address error. I am curious, why do you even have the initial scanf() that resides outside the loop?

This is your first mistake

To go further into it, main should always return an int. It may work for you, but the standard is very clear about it: main returns an int. It will cause problems on other implementations, and also on this board.

by Salem
> scanf ("%d", curr->val);
This is your second - you haven't malloc'ed the space yet

this solved my original issue

by sean_mackrory
To go further into it, main should always return an int. It may work for you, but the standard is very clear about it: main returns an int. It will cause problems on other implementations, and also on this board.

thanx for notice. I know it, but keep forget...

by master5001
I am curious, why do you even have the initial scanf() that resides outside the loop?

thanx, again for notice. It was a mattar of habit.
fixed that too.

and now to the insert function:

Code:

void insert(node ** tree, node * item) 
{
   if(!(*tree)) {
      *tree = item;
      return;
   }
   if(item->val<(*tree)->val)
      insert(&(*tree)->left, item);
   else if(item->val>(*tree)->val)
      insert(&(*tree)->right, item);
}

it doesn't build the tree...

> curr->left = curr->right = NULL;
Did you remember to do this on your first malloc (which you just added) ?

curr->left = curr->right = NULL;

now I did...
but this doesn't seem to make a difference.
I have noticed that
(*tree)->val
and
item->val
have the same value during the function. it doesn't seem healthy, but I think it is the right code. isn't it?

Dunno - post the WHOLE code as it currently stands

This is my full code:

Code:

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

struct tree_el {
   int val;
   struct tree_el * right, * left;
};

typedef struct tree_el node;

void insert(node ** tree, node * item) 
{
   if(!(*tree)) {
      *tree = item;
      return;
   }
   if(item->val<(*tree)->val)
      insert(&(*tree)->left, item);
   else if(item->val>(*tree)->val)
      insert(&(*tree)->right, item);
}

void printout(node * tree) {
   if(tree->left) printout(tree->left);
   printf("%d\n",tree->val);
   if(tree->right) printout(tree->right);
}

int main(void) 
{
   node *curr=NULL, *root= NULL;

   curr = (node *)malloc(sizeof(node));
   curr->left = curr->right = NULL;

   while (curr->val!=0)
   {
      curr->left = curr->right = NULL;
	  printf("please enter next number\n");
	  scanf ("%d", &curr->val);
      insert(&root, curr);
  }

   printout(root);

   return 1;
}

New nodes need malloc to be called each time. You were using the same node for all inserts.

Code:

int main(void) 
{
  node *curr=NULL, *root= NULL;

  do
  {
    curr = (node *)malloc(sizeof(node));
    curr->left = curr->right = NULL;
    printf("please enter next number\n");
    scanf ("%d", &curr->val);
    insert(&root, curr);
  } while (curr->val != 0);

  printout(root);

  return 0;
}

1 as the return for main means failure, 0 means success.

Thank you!