带有指针的C错误中的Malloc函数

Question

I've create this function that's supposed to create a randomly generated binary tree, it works fine but at the end of the function the root == NULL, i can't understand why!

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

#define MAX_B 7

typedef struct _ramo{
    int nbanane;
    struct _ramo *dx;
    struct _ramo *sx;
}ramo;

void creaAlbero(ramo *root, int n){
    printf("%d\n",n);
    root = malloc(sizeof(ramo));
    root->nbanane=rand()%MAX_B;
    printf("BANANA! %d\n",root->nbanane);
    root->dx=NULL;
    root->sx=NULL;
    if ((int)(rand()%n)==0)
        creaAlbero(root->dx, n+1);
    if ((int)(rand()%n)==0)
        creaAlbero(root->sx, n+1);
 }

int main(){
    srand((unsigned int)time(NULL));
    ramo *root=NULL;
    creaAlbero(root, 1);
    if (root==NULL) {
        printf("EMPTY!!");
    }
    return 0;
}

Answer 1

You set root to NULL :

ramo *root=NULL;

then pass a copy of it to creaAlbero() :

creaAlbero(root, 1);

which modifies the copy

root = malloc(sizeof(ramo));

then returns. The original root is still NULL , because nothing changed it.

Consider returning root from creaAlbero() :

ramo * creaAlbero(int n){
  printf("%d\n",n);

  ramo *root = malloc(sizeof(ramo));
  root->nbanane=rand()%MAX_B;
  printf("BANANA! %d\n",root->nbanane);
  root->dx=NULL;
  root->sx=NULL;

  if ((int)(rand()%n)==0)
    root->dx = creaAlbero(n+1);
  if ((int)(rand()%n)==0)
    root->sx = creaAlbero(n+1);

  return root;
}

int main(){
  srand((unsigned int)time(NULL));
  ramo *root=NULL;
  root = creaAlbero(1);
  if (root==NULL) {
    printf("EMPTY!!");
  }
  return 0;
}

Example: https://ideone.com/dXiv8A

Answer 2

creaAlbero(ramo *root, int n) is a function that takes a copy of a pointer to a ramo . It then proceeds to do stuff with this copy of the pointer, and then returns. main then looks a the value of the origonal root variable, which was (obviously) never changed.

If you want a function to modify a value that's passed in, you must pass the object by pointer. To clarify: if you want a function to modify a pointer, the function must take as a parameter a pointer to a pointer to a thing:

void creaAlbero(ramo **rootptr, int n){     //pass rootptr by pointer     
    *rootptr = malloc(sizeof(ramo)); //modify pointer pointed at by rootptr
    ramo* root = *rootptr; //make local copy of value for ease of use
    //rest of your code here
}
int main(){
    ramo *root=NULL;
    creaAlbero(&root, 1);  //pass by pointer

Paul Roub's answer also suggests another excellent idea: return the ramo* from the function instead of taking it as a mutable parameter. It's simpler and more intuitive by far.

Answer 3

root is passed to creaAlbero by value. Any changes made to root in creaAlbero are only local modifications. They don't change the value of root in main. A better alternative would be to change the signature of creaAlbero to:

ramo* creaAlbero(int n){
   printf("%d\n",n);
   ramo* root = malloc(sizeof(ramo));
   root->nbanane=rand()%MAX_B;
   printf("BANANA! %d\n",root->nbanane);
   root->dx=NULL;
   root->sx=NULL;
   if ((int)(rand()%n)==0)
      root->dx = creaAlbero(n+1);
   if ((int)(rand()%n)==0)
      root->sx = creaAlbero(n+1);

   return root;
}

and change the usage to:

int main(){
   srand((unsigned int)time(NULL));
   ramo *root = creaAlbero(1);
   if (root==NULL) {
      printf("EMPTY!!");
   }
   return 0;
}