CODEWITHSUNDEEP
c++pointers
Snowman
Snowman

Reputation: 32071

Pointer in copy constructor

I have a copy constructor that looks like this:

Qtreenode(const QtreeNode * & n) {

    x=n->x;
    y=n->y;
    height=n->height;
    width=n->width;
    element=n->element;

}

I wrote this a week ago, and I looked back at it today and I was surprised by the line that when called the copy constructor with say swChild=new QtreeNode(*(n->swChild)); The arguement to the cc is a pointer by reference, right? But when I call it, I do (*(n->swChild)), which means the value of that child right? Why does my code work?

Upvotes: 2

Views: 367

Answers (2)

James McNellis
James McNellis

Reputation: 355069

That is not a copy constructor. A copy constructor for a class is a constructor that takes a reference of the class's type. Any of the following would be copy constructors, though the second, taking a const reference, is by far the most commonly used:

QTreeNode(QTreeNode&);
QTreeNode(const QTreeNode&);
QTreeNode(volatile QTreeNode&);
QTreeNode(const volatile QTreeNode&);

Since this isn't a copy constructor, the implicitly declared copy constructor is still provided. It is declared as QTreeNode(const QTreeNode& n) and basically just copies each member of the class (i.e., it's a "shallow" copy, not a "deep" copy). It is this implicitly declared copy constructor that is used in your code.

Upvotes: 5

dfan
dfan

Reputation: 5814

That's not a copy constructor, it's just a regular old constructor. A copy constructor would take a const QtreeNode&, not a const QtreeNode*&.

Since you didn't define a copy constructor, the compiler made one for you. That's the one that is being called in the line swChild=new QtreeNode(*(n->swChild));, since you are passing a QtreeNode and not a pointer to one.

Upvotes: 1

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