variable created on heap was not deleted in C++ - c++

Question

I was looking into BFS search code provide in here:
// Program to print BFS traversal from a given
// source vertex. BFS(int s) traverses vertices
// reachable from s.
#include<iostream>
#include <list>
using namespace std;
// This class represents a directed graph using
// adjacency list representation
class Graph
{
int V; // No. of vertices
// Pointer to an array containing adjacency
// lists
list<int> *adj;
public:
Graph(int V); // Constructor
// function to add an edge to graph
void addEdge(int v, int w);
// prints BFS traversal from a given source s
void BFS(int s);
};
Graph::Graph(int V)
{
this->V = V;
adj = new list<int>[V];
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w); // Add w to v’s list.
}
void Graph::BFS(int s)
{
// Mark all the vertices as not visited
bool *visited = new bool[V];
for(int i = 0; i < V; i++)
visited[i] = false;
// Create a queue for BFS
list<int> queue;
// Mark the current node as visited and enqueue it
visited[s] = true;
queue.push_back(s);
// 'i' will be used to get all adjacent
// vertices of a vertex
list<int>::iterator i;
while(!queue.empty())
{
// Dequeue a vertex from queue and print it
s = queue.front();
cout << s << " ";
queue.pop_front();
// Get all adjacent vertices of the dequeued
// vertex s. If a adjacent has not been visited,
// then mark it visited and enqueue it
for (i = adj[s].begin(); i != adj[s].end(); ++i)
{
if (!visited[*i])
{
visited[*i] = true;
queue.push_back(*i);
}
}
}
}
// Driver program to test methods of graph class
int main()
{
// Create a graph given in the above diagram
Graph g(4);
g.addEdge(0, 1);
g.addEdge(0, 2);
g.addEdge(1, 2);
g.addEdge(2, 0);
g.addEdge(2, 3);
g.addEdge(3, 3);
cout << "Following is Breadth First Traversal "
<< "(starting from vertex 2) \n";
g.BFS(2);
return 0;
}
In the constructor of Graph class they created a adjacency list in the heap but they never use a delete to free the memory. My question is as following
a) Is there any chance of memory leak?
If there is any memory leak how could we solve the problem?

Answer 1

Yes there are leaks.
Leak 1 is new list<int>[V];
Leak 2 is new bool[V];
Looks like some guy with Java or C# background wrote this piece of code. To fix the leaks use delete[] in the function void Graph::BFS(int s) also use a destructor to delete the list.
Then, you might consider std::shared_ptr.

Answer 2

Is there any chance of memory leak?
Yes, there is a high chance for a memory leak.
If there is any memory leak how could we solve the problem?
Generally, this can be solved by implementing a destructor. Then again, according to the rule of three, we would need a copy-constructor as well, just in case an end user decides to copy one list to another.
But we could actually sidestep this by not dynamically-allocating in the first place! Let's reimplement with std::vector:
class Graph
{
int V;
vector<list<int>> adj;
public:
Graph(int V);
// ...
};
Graph::Graph(int V)
{
this->V = V;
adj.assign(V, list<int>()); // std::vector::assign
}
void Graph::BFS(int s)
{
// Mark all the vertices as not visited
vector<bool> visited(V); // see note below *
for(int i = 0; i < V; i++)
visited[i] = false;
// Create a queue for BFS
list<int> queue;
// Mark the current node as visited and enqueue it
visited[s] = true;
queue.push_back(s);
// 'i' will be used to get all adjacent
// vertices of a vertex
list<int>::iterator i;
while(!queue.empty())
{
// Dequeue a vertex from queue and print it
s = queue.front();
cout << s << " ";
queue.pop_front();
// Get all adjacent vertices of the dequeued
// vertex s. If a adjacent has not been visited,
// then mark it visited and enqueue it
for (i = adj[s].begin(); i != adj[s].end(); ++i)
{
if (!visited[*i])
{
visited[*i] = true;
queue.push_back(*i);
}
}
}
}
There are a ton of other minor issues with this code, but I shall leave that as an exercise for the reader.
*Note: vector<bool> isn't your normal vector.