C++ Lambda Function for BFS Traversal - c++

Question

Using the two functions below, how could I use a lambda function to BFS Traverse through a Graph of Thing's called x
i.e. Graph<Thing> x
which involves sending the display function to the bfsTraverse function so that the information in each Thing can be displayed.
void display(const Thing& c)
{
cout<<left<<setw(3)<<c.getKey()<<" "<<left<<setw(2)<<c.getLabel()<<endl;
}
template <typename E>
void Graph<E>::bfsTraverse(FuncType func)
{
/* some code not necessary to describe */
}
So basically, I just need to understand how to use a lambda function to tie these together here.

Answer 1

It pretty straightforward. For example, let;s print values of vector using lambda:
#include <algorithm>
#include <vector>
#include <iostream>
int main (int argc, char* argv[]) {
std::vector <int> data{1,2,3,4};
std::for_each (data.begin (), data.end (), [] (const int val) { std::cout << val << std::endl;});
return 0;
}
last argument [] (...) is the lambda.
A function, which accepts lambdas, might look like this:
template <typename E, typename FuncType>
void Graph<E>::bfsTraverse (FuncType func)
{
/* some code not necessary to describe */
}
UPDATE
In case of a graph you should do the following. Your graph's implementation should have a method to access vertices: it might be 'getRoot', 'getSource', 'getAllNodes', it's implementation defined. I'll stick with 'getRoot'. Each vertex/node should have a method like 'getAdjacentVertices', 'getChildren', whatever. Combining all together:
template <typename E, typename FuncType>
void Graph<E>::bfsTraverse (FuncType func)
{
std::queue<Node> q;
auto&& root = getRoot ();
q.push (root);
while (not q.empty ()) {
auto&& current = q.front ();
q.pop ();
func (current);
auto&& adjacent = current.getAdjacent ();
for (auto&& a: adjacent) {
q.push (a);
}
}
Be aware, that I deliberately omit keeping list of visited nodes and other things. But idea itself stays the same. Now you can call this function as follows:
Graph<E> g;
g.bfsTraverse ( [] (const Node& n) {
std::cout << n.to_str () << std::endl;
}
);