SilverMage - Viewing Profile

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Last Active:: Apr 10 2013 07:05 AM
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Posts I've Made

In Topic: C thread accepting same connection twice, results in failure

Posted 6 Mar 2013

Never mind, it seems I have fixed my problem
In Topic: How to find out if a client has disconnected from server.

Posted 5 Mar 2013

Yes, that one precisely
In Topic: How to find out if a client has disconnected from server.

Posted 5 Mar 2013

Well, a server that is capable of handling multiple clients using the select() function. Basically, think of a chat server.
In Topic: How come an insert to a binary tree struct doesn't work this way

Posted 8 Oct 2012
Salem_c, on 08 October 2012 - 08:20 AM, said:
```
void foo ( int a ) {
  a = 42;
}
int bar ( void ) {
  return 42;
}
int main ( ) {
  int a = 0;
  foo(a);
  printf("%d\n", a );
  a = bar();
  printf("%d\n", a );
  return 0;
}
```
Do you understand why foo() doesn't work, and bar() does?

One way is to declare
```
struct node* insertNode(struct node* n, int x) {
  if ( n == NULL ) {
    n = malloc.....
  }
  return n;
}
```
Then do
n->left = insertNode(n->left, x);

Or you pass a pointer to the node itself (see your other thread).

Or you switch to C++, and use a reference instead.
Hmmm, Now I see why that tutorial told me to use a pointer to the node. Now that I see, the root is a pointer itself, so the method makes sense now. Thanks, Salem, I just wanted to understand the tutorial better
In Topic: How come an insert to a binary tree struct doesn't work this way

Posted 8 Oct 2012
Skydiver, on 08 October 2012 - 07:58 AM, said:
Your line 9 crashes because no where in the code you've post are you setting/updating the pointer to the root node.

Are you setting the root pointer elsewhere, or are you assuming a code call like this will update the root?
```
struct node * root = NULL;
insertNode(root, 25);
```
Given your current insertNode() method, root will not be updated.
Salem_c, on 08 October 2012 - 08:04 AM, said:
http://www.dreaminco...to-binary-tree/
Note the extra level of indirection in this example.

> insertNode(n->left, x);
This does NOT change n->left.

Skydiver, I have my root set that way, but why won't my method update the root?

Salem, I was able to get the method working that way, but it's got me curious as to why I have to use pointers to pointers instead of an intuitive way involving just the node or a pointer to it.