Segmentation fault on initializing pointer (malloc) in different function

Question

I am trying to initialize a tree by doing this:

typedef struct {
    char *value;
    struct children_list *children;
} tree;

typedef struct t_children_list {
    tree *child;
    struct t_children_list *next;
} children_list;

void initializeTree(tree *root, char *input)
{
  if((root = malloc(sizeof(tree))) == NULL) { abort(); }
  root->value = input;
}

void main()
{
  // Create the tree
  char *input = "aaaaaa";
  tree *my_tree = NULL;

  initializeTree(my_tree, input);
}

But I am getting a segmentation fault. Why is that happening? I am passing a pointer to a function and I am reserving memory inside of it. Is that wrong?

Duplicate hundreds of times over: http://c-faq.com/ptrs/passptrinit.html — Carl Norum, Feb 23 '13 at 15:53
["Initializing" the pointer in the separate function in C](http://stackoverflow.com/questions/2486235/initializing-the-pointer-in-the-separate-function-in-c), for example. — Carl Norum, Feb 23 '13 at 15:54
Are you getting a segmentation fault with this exact program? At what line? — n. m. could be an AI, Feb 23 '13 at 15:55
'my_tree' is being passed by copy, and you are initialising the copy of the pointer. Then that copy is discarded. — Bingo, Feb 23 '13 at 15:55
Why you guys **don't learn C before trying to write an entire OS in it...** — , Feb 23 '13 at 15:55
@H2CO3 fyi, I'm upvoting your comment until my mouse breaks. — Andreas Grapentin, Feb 23 '13 at 16:08

Daij-Djan · Accepted Answer · 2013-02-24T11:20:21.003

the pointer 'my_tree' is passed by value (as is the only way it is done in C)

so my_tree is basically COPIED and no assignment of 'root' has any effect on the 'my_tree' variable.

you want to get a pointer back so pass a pointer to a pointer (**) and then init *root to actually modify my tree

void initializeTree(tree **pRoot, char *input)
{
  if((*pRoot = malloc(sizeof(tree))) == NULL) { abort(); }
  *pRroot->value = input;
}

void main()
{
  // Create the tree
  char *input = "aaaaaa";
  tree *my_tree = NULL;

  initializeTree(&my_tree, input);
}

OR don't pass it at all but return it:

tree *initializeTree(char *input)
{
  tree *root = NULL;
  if((root = malloc(sizeof(tree))) == NULL) { abort(); }
  root->value = input;
  return root;
}

void main()
{
  // Create the tree
  char *input = "aaaaaa";
  tree *my_tree = initializeTree(input);
}

Byval passing is not only the default, it's the **only** way in C to pass arguments. — , Feb 23 '13 at 15:57
you're right the . & thingy in the param is C++ .. illl edit my answer — Daij-Djan, Feb 23 '13 at 15:58

Segmentation fault on initializing pointer (malloc) in different function

1 Answers1