Thread: Nodes

It looks like it should work to me from what I see here. Can you paste more code so that we can try to replicate the issue more fully?

Is it a runtime error, or compilation error? For example, this compiles and works for me:

Code:

struct node {
  int i; 
};
struct randomstruct {   /*codes*/
    struct node *start; 
};

int access(struct randomstruct *point) {    
  struct node *trial;
  trial = (struct node *)malloc( sizeof( struct randomstruct ) );
  trial = point->start;
  printf( "From access: %d\n", point->start->i );
  return 0; 
}

int main() {
  struct randomstruct *root = (randomstruct*)malloc( sizeof( struct randomstruct ) );
  root->start = (node*) malloc( sizeof( struct node ) ); 
  root->start->i = 10; 

  access( root ); 

  printf( "From main: %d\n", root->start->i );
  free( root ); 

  return 0;
}

Code:

int access(struct randomstruct *point)
{
    struct node *trial;
    trial = malloc(sizeof(*trial));
    trial = point->start;

That last line is a memory leak BTW. You allocate memory to the pointer trial with the malloc call and then the very next line you throw that away when you reassign trial to point to something else.

---

The first thing you should do before assigning point->start to trial is test that point is a valid (non-NULL) pointer. Then, only if that is true can you safely dereference point to get at start. The malloc call is probably unnecessary, unless start itself is NULL and you're attempting to assign it some memory.

---

For the incompatible thing, do both of those bits of code have the same definition for node available? Where is node defined?

Code:

trial =(struct randomstruct *) malloc(sizeof(struct node)); 

malloc() always return void pointer so you should type cast it in to your struct pointer.....

Originally Posted by amzc

Code:

trial =(struct randomstruct *) malloc(sizeof(struct node)); 

malloc() always return void pointer so you should type cast it in to your struct pointer.....

You should avoid casting the return value of the malloc call. The void pointer returned can be safely assigned to other pointer types without need for a cast. From a maintenance standpoint the malloc call for pointer trial (if it were to be used) should be as follows:

Code:

trial = malloc(sizeof(*trial));

That way, if the type of variable trial changes then you don't need to go hunting down every sizeof(struct node) in your program to match what you just changed it to. Using sizeof(*trial) instead means you won't need to change a thing if/when the type of trial changes.