返回指针时类型转换错误

Question

Id like to return a pointer(or reference to one) instead of a reference to my value. But I keep getting conversion errors.

I tried changing the return type to Node* but it doesnt recognize it as a known type since Node is a struct.

I'm having a hard time understanding these conversion erros clearly when using templates. I havent found a clear answer to these conversions yet.

error: cannot convert ‘TreeAVL<int>::Node*’ to ‘const int*’ in return

template<class T>
const T* TreeAVL<T>::find(Node* node, const T& element) const {
    if (node == nullptr)
        return nullptr;
    if (element == node->value)
        return &(node->value);
        //return &(node);            <<<< what I want to return if found.
    ...
}

Answer 1

To understand the conversion error,

error: cannot convert ‘TreeAVL<int>::Node*’ to ‘const int*’ in return

You have to look at the definition of the template that produces this error, which would be something like:

// A function returning const T*
template<class T>
const T* TreeAVL<T>::find(Node* node, const T& element) const {
    // OK, any T* can be a nullptr:
    if (node == nullptr)
        return nullptr;
    // But here the return value is a TreeAVL<T>::Node*, which is not
    // necessarily convertible to any T*
    if (element == node->value)
        return node;
    // ...
}

So you get an error as soon as you use this template with an incompatible T , such as int in your example. If you want to return a pointer to a node instead, the function must be declared to do so, and its return statements must all be compatible with that.

Also, if you want to avoid the cumbersome typename, you can also use const auto* , or make the definition inline and specify it as const Node* .