CODEWITHSUNDEEP
c++c++11templates
Tony
Tony

Reputation: 1393

Compiler gives lambda function cast error

First, define a tree node, where the BFSTraversal is breath-first search, ReconstructBFS is to construct a tree based on a breadth-first search sequence.

template<typename T>
struct TreeNode
{
    T val;
    TreeNode *left;
    TreeNode *right;

    explicit TreeNode(T x) : val(x), left(nullptr), right(nullptr)
    {}

    static void BFSTraversal(TreeNode *node, void visit(T));

    static TreeNode<T> *ReconstructBFS(vector<T> &seq);
};

Then overload the << operator, which gives a compiler error that I do not understand. Is there something wrong with my lambda function?

template<typename T>

ostream &operator<<(ostream &os, TreeNode<T> *node)
{
    void (*visit)(T) = [&os](T v) -> void
    { os << v << ','; };

    // compiler error
    // cannot convert 'operator<<(std::ostream&, TreeNode<T>*) 
    // [with T = int; std::ostream = std::basic_ostream<char>]::<lambda(int)>' to 'void (*)(int)' in initialization
    TreeNode<T>::BFSTraversal(node, visit);

    return os;
}

enter image description here

I intend to achieve the following code.

auto bfs_seq = vector<int>{1,0,2,3}; // 0 represents `null`
TreeNode* root = ReconstructBFS(bfs_seq);
cout << root;

You can skip this. I am not sure but I think This is not directly relevant to my problem.

The implementation of BFSTraversal.

static TreeNode<T> *ReconstructBFS(vector<T> &seq)
{
    auto seq_size = seq.size();
    if (seq_size == 0) return {};
    auto root = new TreeNode<T>(seq[0]);

    queue<TreeNode<T> *> q;
    q.push(root);
    auto level_len = 1, i = 1;
    auto next_level_len = 0;
    auto null_val = T();

    while (!q.empty())
    {
        for (auto j = 0; j < level_len; ++j)
        {
            auto cur_parent = q.front();
            q.pop();

            auto v = seq[i++];
            if (v == null_val) cur_parent->left = nullptr;
            else
            {
                cur_parent->left = new TreeNode(v);
                ++next_level_len;
            }

            v = seq[i++];
            if (v == null_val) cur_parent->right = nullptr;
            else
            {
                cur_parent->right = new TreeNode(v);
                ++next_level_len;
            }
        }

        level_len = next_level_len;
        next_level_len = 0;
    }

    return root;
}

Upvotes: 1

Views: 69

Answers (1)

R Sahu
R Sahu

Reputation: 206577

A lambda function can be assigned to a function pointer only if does not capture. If it captures, it cannot be assigned to a function pointer.

See Passing capturing lambda as function pointer for more.

Use a std::function instead.

// Not this.
// void (*visit)(T) = [&os](T v) -> void { ... }

std::function<void(T)> visit = [&os](T v) -> void { ... }

Of course, that will require you to change the function BFSTraversal too. It should expect a std::function instead of a function pointer.

Upvotes: 5

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