CODEWITHSUNDEEP
c++binary-search-tree
Jumper
Jumper

Reputation: 215

non-const lvalue reference to type 'int' cannot bind to a temporary of type 'int'

This is my node implementation for a binary tree.

template<typename T> class node{
    public:
        explicit node():data(0), left(NULL), right(NULL){}
        explicit node(const T& data):data(data),left(NULL), right(NULL){}

        template<typename E> friend class bst;

    private:
        T& data;
        node<T>* left;
        node<T>* right;
};

This is the binary search tree.

template<typename T> class bst{
    public:
        bst():root(NULL){}
        bst(node<T>* root):root(root){}
private:
        node<T>* root;
};

The calling class does something like this.

int main(){

    node<int>* root = new node<int>(17);
    bst<int>* t = new bst<int>(root);


    t->insert(root,21);
    t->insert(root,12);
    t->insert(root, 9);

    delete t;
}

I keep getting the error.

./node.hpp:3:19: error: non-const lvalue reference to type 'int' cannot bind to a temporary of type 'int'
                explicit node():data(0), left(NULL), right(NULL){}

Can someone please help me understand, what exactly is the issue here.

Upvotes: 0

Views: 7176

Answers (1)

Silvio Mayolo
Silvio Mayolo

Reputation: 70267

T& data;

There's your problem. That should be a T. There's no reason to make it a reference. In your default constructor, you try to assign a temporary value (the literal 0) to it, and since it's a reference type you can't give it a temporary value.

The literal 0 is still a poor choice for its default value, considering that 0 is an int, and your type is designed to work with all types. You should consider using a more polymorphic value, like T() or explicitly declaring that the type must be convertible from int, in a comment or documentation.

Upvotes: 1

Related Questions