Re: in c++ I need a pointer to a pointer ?????

"Doug Harrison [MVP]" <>
Wed, 03 May 2006 09:00:31 -0500
On Wed, 03 May 2006 12:42:12 GMT, sparks <> wrote:

Ok I am not the greatest in the world at pointers.

but I can build a linked list with no problems.

New problem is this I was parsing out some xml data
and I was trying my hand at recursion.

so to make it simple I took out all the parsing and just
did a little something to try and see if it works.

I did this

struct tree
        string data;
        tree *child;
        tree *sibling;
    tree *Root_node;

What about initializing child and sibling?

     tree* add_node(tree*);

tree* tree::add_node(tree *root)
    int found;
    char str[120];
    string ReadData();

The line above declares a global function named ReadData that returns a
string. This language "feature" should not be used.

         string lineinput;
        static ifstream
input_line("c:\\testproject3.txt",ios::in);//input file
    while (!input_line.eof())
    found = lineinput.find("</");
    tree *node = *root;

The line above attempts to copy a tree to a tree*, which can't work.

     root->data = lineinput;
    if (found >0)

    if (found <=0)

You shouldn't loop around an eof test; see this message for more:

Also, I don't see where your loop reads anything. Nor do I see where it
creates any nodes. Finally, the two return statements don't return
anything, which is illegal.

try as I might this worked but didn't.

Then I was told oh this wont work unless you use a pointer to a
pointer...something like **root.

Well I am lost and dont' know what to do this makes no since(to me)
and not sure what they are saying or how to do it.

can someone tell me what a pointer to a pointer does and if there is
something I can find to explain the syntax behind it?

A pointer is an object that contains the address of something. So a pointer
to a pointer is just an object that holds the address of a pointer. In C,
pointers are used (among other things) to simulate pass-by-reference. As
you know, for a function in C to modify the caller's object through a
function parameter, that parameter must be a pointer:

void f(int* p)
   *p = 2;

void g()
   int x;
   assert(x == 2);

You can do the same thing with pointers, but you need a pointer to a

int i;

void f(int** p)
   *p = &i;

void g()
   int* p;
   assert(p == &i);

On the other hand, C++ has references, so you can use pass-by-reference:

void f(int& r)
   r = 2;

void g()
   int x;
   assert(x == 2);

And the pointer version looks like this:

int i;

void f(int*& r) // This & declares a reference
   r = &i; // This & is the address-of operator

void g()
   int* p;
   assert(p == &i);

Relating this to your example, the function add_node should probably be
declared like one of these:

tree* add_node();
void add_node(tree*& p);

You would call it like this:

t->child = add_node();



Of course, someone will have had to create t... Concerning your class
design, it would be more typical for a tree class to define a nested class
to represent its nodes.

Doug Harrison
Visual C++ MVP

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