building a tree

Hi,
I'm trying to make a predictive text program, so I'm going to build a tree with pointers. Every node must have 9 branches since the input from the user is through the numeric keypad (like the T9).
I have found some codes to build trees but they're usually for binary trees. How do I define that I want 9 branches coming out of every node (instead of a left and right)?

Code:

---

typedef struct node
{

int number;
struct node *left;
struct node *right;

};

---

struct node *branches[9];

I have changed the code like you said, I'm trying it with 3 branches at the moment. I'm trying to build 3 branches coming out of the root node, 1st branch having a value of 2, 2nd branch having a value of 3 and the 3rd having a value of 4. My code won't compile can you please tell me what's wrong?

Code:

---


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

typedef struct node
{
#define BRANCHES    3
struct node*    branch[ BRANCHES ];
}node;

struct node *root = 0;

main(struct node **leaf)
{

                        if( *leaf == 0 )
                {
                        *leaf = malloc( sizeof( struct node ) );
               
                       
                        leaf->branch[1] = 2;
                        leaf->branch[2] = 3;
                        leaf->branch[3] = 4;
                       
                }
               
               
        return 0;
}

---

> main(struct node **leaf)
New and interesting ways of declaring main()
Try one of the normal forms.

> leaf->branch[1]
Arrays start at 0, not 1

If you'd figured out how to allocate *left and *right pointers, then there really isn't anything more difficult than that.

> branches[ BRANCHES ] rather than branch[ BRANCHES ]. What do you guys feel is right?
I often use the plural form when declaring arrays of things, if that makes the code more readable.

As for putting the #define inside the struct, I wouldn't do that. It seems to imply a scope which simply isn't there.

This still doesn't compile, there's something wrong with the lines:
I consider myself a beginner in C i'm sorry if it's a very silly question.

*leaf->branches[0]->value = 2;
*leaf->branches[1]->value = 3;
*leaf->branches[2]->value = 4;

Code:

---

#include <stdio.h>
#include <stdlib.h>
#define BRANCHES    3

typedef struct node
{
struct node *branches[ BRANCHES ];
struct node *value;
};



void main()
{ struct node *root = 0;
struct node **leaf;
               
               
                if( *leaf == 0 )
                {
                        *leaf = malloc( sizeof( struct node ) );
                       
                        *leaf->branches[0]->value = 2;
                        *leaf->branches[1]->value = 3;
                        *leaf->branches[2]->value = 4;
                }
               
}

---

See, that's where it all falls apart.
I thought you actually knew something about what you were trying to do.

There's at least a few things that my compiler gakked on....

My comments are in red in the code block below.... Just a
couple of pointers (no pun intended) to get you going in the
right direction.....

Code:

---

#include <stdio.h>
#include <stdlib.h>
#define BRANCHES    3

typedef struct node
{
struct node *branches[ BRANCHES ];
struct node *value;
};      /* My compiler says: "
          useless keyword or type name in empty
          declaration."  I think you forgot "node" here
        between the closing brace and semilcolon */



void main()
{ struct node *root = 0;
struct node **leaf;
               
               
                if( *leaf == 0 )
                {
                        *leaf = malloc( sizeof( struct node ) );

/* You've got some precedence issues.  For starters, it should be
 (*leaf)->branches[0]->value
But, even with that, it still won't compile.  You've declared
value as a pointer to struct node, and you're trying to set it to an
integer value here.
My compiler says:
assignment makes integer from pointer without a cast
And THAT will usually cause a seg fault when it is run.
I'm not sure if you want value to be an int or not, but it
does not seem that you'd want it to be a pointer to another
struct node.
*/                       
                        *leaf->branches[0]->value = 2;
                        *leaf->branches[1]->value = 3;
                        *leaf->branches[2]->value = 4;
                }
               
}

---

[/QUOTE]

Selent, here's a bit of code to show you some ideas of what to do.
But you should really study a bit more C, and how binary trees work in particular before trying this for real.

Code:

---

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

#define BRANCHES    9
typedef struct node
{
  struct node *branches[ BRANCHES ];
  int          value;
} node ;

node *makeNode ( int value ) {
  node *result = malloc ( sizeof *result );
  if ( result != NULL ) {
    int i;
    for ( i = 0 ; i < BRANCHES ; i++ ) {
      result->branches[i] = NULL;
    }
    result->value = value;
  }
  return result;
}

node *addTree ( node *tree, int value, int branchNum ) {
  node *newNode = makeNode ( value );
  if ( tree == NULL ) {
    return newNode;
  } else {
    tree->branches[branchNum] = newNode;
    return tree;
  }
}

int main ( ) {
  node *tree = NULL;
  tree = addTree ( tree, 0, 0 );
  tree = addTree ( tree, 1, 1 );
  tree = addTree ( tree, 2, 2 );
  return 0;
}

---

Everyone else who wants to chew the fat over preprocessor and naming semantics should play here
http://cboard.cprogramming.com/showthread.php?t=74051