how to make a function return a structure pointer in c++

Question

I want to make the function return the address of a particular node .but the compiler is not detecting the node datatype structure i created.

struct node
{
int data;

node *link;
};

node *header,*current;
node traverse(int pos);


node *Linkedlist::traverse(int pos)
{
    int location = 0;  
    current->link = header->link;
    node *address = new node;
    address->data = NULL;
    address->link = NULL;


    while(current->link != NULL)
    {

        if(location == pos)
        {
            cout <<current->link->data <<" "<< endl; 
            address->link=current->link;
        }
        location ++;
        current->link = current->link->link;

    }


    return  address->link;
}

Since this is a C++ question, consider std::unique_ptr and/or std::shared_ptr. If you find yourself writing new and delete often then you're probably doing something wrong. — Ed Swangren
– Ed Swangren, Commented Jan 13, 2013 at 4:28
I did not understand anything from your question. Please, be more concise about what you are asking. Also, your code seems to have too much faults and bad practices. — mualloc
– mualloc, Commented Oct 22, 2017 at 12:47

Ed Heal · Accepted Answer · 2013-01-13 04:09:16Z

5

Change

return  *address;

to

return  address;

answered Jan 13, 2013 at 4:09

Ed Heal

60.3k18 gold badges91 silver badges137 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Abhijit · Accepted Answer · 2013-01-13 04:24:41Z

As address is a pointer variable to node, you need to simply return the variable.

A * preceding a pointer variable is an explicit referencing, which means to get the value pointed by the pointer variable address. This is antagonistic to what the operator & would have done, fetch the address of a variable.

return  address;

So its logical that you should return the variable instead of returning the value pointed by the variable.

Note, care needs to be taken by the caller of traverse to free the memory explicitly by calling delete else this would result in memory leak. This is because of a potential design issue, where you have allocated a heap object inside a local scope and returned the address.

node * foo = Linkedlist::traverse(n);
...............
delete foo;

You could have simply created the object in the heap or add it as a class member, where in the former case, you can easily transfer the ownership of the object from one scope to another, where in the second case, the lifetime of the object would have been controlled by the object Linkedlist

Collectives™ on Stack Overflow

how to make a function return a structure pointer in c++

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related