C++ Lambda Local Variable Capturing Error

I am trying to use a lambda expression to make inorderTraversal function reusable, and I am having the following error while trying to capture the local parameters of printStandartDeviation function: No viable conversion from ‘(lambda at […]/AVLTree.cpp:[…])’ to ‘void (*)(AVLTree::TreeNode *)’.

void AVLTree::printStandartDeviation() {
    int totalWordCount = 0, totalWordFrequencies = 0;
    inorderTraversal(root, [&totalWordCount, &totalWordFrequencies](TreeNode *tNode) {
        totalWordCount++;
        totalWordFrequencies += tNode->frequency;
    });
}

void AVLTree::inorderTraversal(TreeNode *tNode, void (*visit)(TreeNode *)) {
    if (tNode == nullptr) {
        return;
    } else {
        inorderTraversal(tNode->left, visit);
        visit(tNode);
        inorderTraversal(tNode->right, visit);
    }
}

It fixes if I make my local variables static or make them global, but I do not want to do either actually.

I am also open to alternative solutions as well instead of using lambda.

I would be grateful if you can help.

>Solution :

A lambda is not a function pointer. A non-capturing lambda has a conversion to function pointer but one with captures not. It is somewhat similar to the difference between a pointer to free function and pointer to member function where the pointer to member function alone is of little use. You need an object to call a member function. Here you also need the data, the captured variables, to call the lambda, its not just a function.

You do not need a function pointer. Simply pass the lambda as it is, and because the lambdas type has no name, make the method a template to make this work:

template <typename F>
void AVLTree::inorderTraversal(TreeNode *tNode, F&& visit) {
    // all the same
}