Re: One more question

"Jacky" <>
Sun, 18 Feb 2007 15:36:58 +0800
"Jacky" <> ???g???l???s?D:ejj7u6yUHHA.3592@TK2MSFTNGP06.phx.gbl...

"Arnaud Debaene" <> ???g???l???s?D:%23RGBCwyUHHA.192@TK2MSFTNGP04.phx.gbl...

"Jacky" <> a ?crit dans le message de news:

If I have the following code:

template<class I, class T>
I find (I first, I last, const T& value)
       while (first != last && *first != value)
       return first;

struct int_node {
    int val;
    int_node *next;

// Wrapper
template <class Node>
struct node_wrap {
        Node *ptr;

        node_wrap(Node* p = 0) : ptr(p) { }
        Node& operator *() const { return *ptr } <<<<< what is this? and
more coming up

This is the star operator, which allow youto dereference a node_wrap as
if it were a pointer. This operation returns a Node&

In the scenrio, what is the difference between Node, Node *, Node&?

        Node* operator->() const { return ptr } <<<<< what does cont
mean here?

It means that the operation is a const-functin, which doesn't modify the
current object, and therefore the function can be called on a const

So there would be a compilation error when I apply this function on a
non-const object?

As a side note : you miss the ending ";" for both of those functions

        // pre-increment operator
        node_wrap& operator++ () { ptr = ptr->next; return
*this; }
        // post-increment operator
        node_wrap operator++(int) { node_wrap tmp = *this;
++*this; return tmp; }

        bool operator == (const node_wrap& i) const { return ptr ==
i.ptr; }
        bool operator != (const node_wrap& i) const { return ptr !=
i.ptr; }

bool operator == (const int_node& node, int n) { return node.val == n; }
<<< what are these extra lines for?
bool operator != (const int_node& node, int n) { return node.val != n; }

They allow to compare for equality an int_node and an int.


void main()
       int_node *list_head, *list_tail;
       int_node *in = new(int_node);
       in->val = 100;
       in->next = 0;
       list_head = list_tail = in;

       for (int i = 0; i < 10; i++)
            int_node* in = new(int_node);
            in->val = i;
            in->next = 0;

           list_tail->next = in;
           list_tail = in;

      // 100, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
      node_wrap<int_node> r;

      r = find(node_wrap<int_node>(list_head), node_wrap<int_node>(),
10); <<<<< don't understand this as well

First, it builds a node_wrap<int_node> object around list_head. It then
creates an empty node_wrap<int_node>. The both objects form a range (that
is, by aplying operator++ several times on the 1st, you should reach the
Since they form a range, they can be used as an input for the find
function. Since node_wrap defines equality operators against int, an int
(10) can be passed directly as predicate to the find fnction (as the
object to search).

Are there any good reference books that I can grab on (as mine is in
another language) over this topic?

So there is the C++ programming language which was proposed in the previous
Just forgot. Sorry about it

      if ( r != node_wrap<int_node>())
           std::cout << (*r).val << std::endl;

      r = find(node_wrap<int_node>(list_head), node_wrap<int_node>(),
      if (r != node_wrap<int_node>())
           std::cout << (*r).val << std::endl;

it says that find(list_head, null, 5); won't work, why?

"It" says? Who say what? The compiler, the program, your teacher? Also,
what is "null" ????

Whoops. The reference material does.

Anyway, the first call to find fails because 10 is not in the list. But
the second call succeeds because 3 is in the list.


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