CODEWITHSUNDEEP
c++listpointers
user3447353
user3447353

Reputation: 1

How to return Pointer to a object from List of objects using iterators

When I traverse a list of objects and I have to return the pointer to one of my object when I have a iterator pointing to it, how I do so ? In short can one tell what should I put in the place of ???

#include <iostream>
#include <list>

using namespace std;

class Node
{
private:
    int name;
public:
    Node() : name(0) {}
    Node(int n) : name(n) {}
    int getName() {return name;}
    void setName(int n) {name = n;}
};

list<Node> mylist;

Node* getNodeObjectPointer(int n)
{
    Node* toBeReturned = NULL;
    for(list<Node>::iterator it=mylist.begin(); it != mylist.end(); it++)
    {
        if(((*it).getName()) == n)
            toBeReturned = ???
    }
    return toBeReturned;
} 

int main()
{
    Node* a = new Node(1);
    mylist.push_back(*a);
    Node* b = new Node(2);
    mylist.push_back(*b);

    Node* myPointer = getNodeObjectPointer(1);
    return 0;
}

Upvotes: 0

Views: 4918

Answers (5)

Vlad from Moscow
Vlad from Moscow

Reputation: 311078

Node* getNodeObjectPointer( int n )
{
    list<Node>::iterator it = mylist.begin();

    while ( it != mylist.end() && it->getName() != n ) ++it;

    return ( it == mylist.end() ? NULL : &*it );
}

It would be much better and simpler if you wrote in main

int main()
{
    mylist.push_back( 1 );
    mylist.push_back( 2 );

    Node* myPointer = getNodeObjectPointer(1);

    return 0;
}

Also member function getName should be declared as

int getName() const {return name;}

Upvotes: 0

Neil Kirk
Neil Kirk

Reputation: 21803

Your code to add objects to the list leaks memory and it is needlessly inefficient. Use this instead

mylist.push_back(Node(1));
mylist.push_back(Node(2));

Your function should be

Node* getNodeObjectPointer(int n)
{
    Node* toBeReturned = NULL;
    for(list<Node>::iterator it=mylist.begin(); it != mylist.end(); it++)
    {
        if(((*it).getName()) == n)
        {
            toBeReturned = &*it;
            break;
        }
    }
    return toBeReturned;
}

If you remove that node from the list, the pointer becomes invalid.

Upvotes: 1

Lightness Races in Orbit
Lightness Races in Orbit

Reputation: 385295

You can obtain a pointer from an iterator like this:

Node* getNodeObject(int n)
{
    for (list<Node>::iterator it = mylist.begin(); it != mylist.end(); it++) {
        if ((it->getName()) == n)
            return &*it;   // this is the key, if you really want a pointer
    }

    return NULL;
}

But I'd return a reference instead (throwing an exception on failure to find). You could even just return the iterator. Or use std::find…

Related, this leaks like a sieve:

Node* a = new Node(1);
mylist.push_back(*a);

Just write:

Node a(1);
mylist.push_back(a);

or even:

mylist.push_back(Node(1));

Upvotes: 6

Marius Bancila
Marius Bancila

Reputation: 16338

Your list should be a list of pointers to Node.

list<Node*> mylist;

It makes no sense to do things like creating objects on the heap, dereferencing them and putting values on stack and then returning pointers to those objects.

Upvotes: -4

pippin1289
pippin1289

Reputation: 4939

What you should do is dereference the iterator and find its address.

return &(*itr);

If you can you could also alter the code to return the iterator instead of a raw pointer.

Upvotes: 1

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