Tree with std::unique_ptr. Reference cannot be bound to dereferenced null pointer in well-defined C++ code

Question

I am trying to write a tree data structure with unique_ptr but I ma getting following warning while printing the Tree:

struct Node {
    int data_{0};
    std::unique_ptr<Node> left_{nullptr};
    std::unique_ptr<Node> right_{nullptr};
};

void print(Node &root) {
    if (&root == nullptr) return;
    print(*root.left_);
    std::cout << root.data_ << " ";
    print(*root.right_);
}

I get following warning from compiler:

Reference cannot be bound to dereferenced null pointer in well-defined C++ code; comparison may be assumed to always evaluate to false. In print(...) function.

How to fix this warning?

I want solve this using unique_ptr.

Answer 1

You test if the address of root is equal to nullptr. The address of a reference variable can not be nullptr.

void print(Node &root) {
    if (&root == nullptr) return; // will never be true

What you probably want is to check if the unique_ptrs are not empty before dereferencing them:

void print(const Node& root) {
    if (root.left_) print(*root.left_);   // check added
    std::cout << root.data_ << ' ';
    if (root.right_) print(*root.right_); // check added
}

Alternatively, use pointers:

void print(const Node* root) {
    if(root == nullptr) return;      // now this check makes sense
    print(root->left_.get());
    std::cout << root->data_ << ' ';
    print(root->right_.get());
}

...or pass a reference to the unique_ptr directly:

void print(const std::unique_ptr<Node>& root) {
    if (root == nullptr) return;
    print(root->left_);
    std::cout << root->data_ << ' ';
    print(root->right_);
}